## Maxwell’s equations



My notes of the excellent lecture 33 by “Denis Auroux. 18.02 Multivariable Calculus. Fall 2007. Massachusetts Institute of Technology: MIT OpenCourseWare, https://ocw.mit.edu. License: Creative Commons BY-NC-SA.”

Explains how devices work, and how magnetic waves propagate. Maxwell’s equation tell you about divergence and curl of these fields.

This is not detailed enough to be really understandable, but hopeful makes you a bit curious about physics 8.02.

## Electric / Magnetic fields

The electric field $$\vec E$$ is a vector field that tells you wat force will be exerted on a charged particle. It is responsible for the flow of electrons when you have a voltage difference, because it is a gradient of a potential (=electric voltage). The force $$\vec F = q\,\vec E \nonumber$$

A magnetic field is a vector field that deflects the trajectory of a moving charged particle and make it rotate. The force $$\vec F = q\,\vec v \times \vec B \nonumber$$

### Electric field

#### Gauss-Coulomb law

$$\rho$$ here is electric charge density (=charge per unit volume), and $$\epsilon_0$$ is a physical constant $$\nabla\cdot\vec E = \frac{\rho}{\epsilon_0} \label{eq:max1a}$$

This says: the divergence of $$\vec E$$ is caused by the electric charge.

This is a partial differential equation satisfied by the electric charge. Not very intuitive. What is more intuitive is what we get when we apply the divergence theorem

To find the flux of the electric field out of the closed surface $$S$$, using equation $$\eqref{eq:max1a}$$ \newcommand{oiint}{\subset\!\!\supset\kern-1.65em\iint} \begin{align*} \oiint_S \vec E\cdot d\vec S &= \iiint_D \rm{div}(\vec E)\,dV \\ &= \frac{1}{\epsilon_0} \iiint_D \rho\,dV \end{align*} \nonumber

If we now integrate the charge density of the entire regions, I get the total electric charge $$Q$$ in the region $$D$$ \newcommand{oiint}{\subset\!\!\supset\kern-1.65em\iint} \begin{align*} \oiint_S \vec E\cdot d\vec S &= \frac{Q}{\epsilon_0} \end{align*} \nonumber

It tells us how electric charges influence the electric field around them.

One application is capacitors that store energy using a voltage between two plates. The voltage is obtained by integrating the electric field from one plate to the other plate. The charges in the plates are what causes the electric field between the plates. That gives you a relationship between voltage and charge in capacitor. (refer to Physics 802.2 Electricity and Magnetism)

The curl of the electric field (a gradient field) is $$0$$, but you can create voltage using magnetic fields. You have a failure of conservativity of the electric force if you have a magnetic field. $$\newcommand{pdv}[2]{\frac{\partial #1}{\partial #2}} \nabla\times\vec E = -\pdv{\vec B}{t} \label{eq:max2a}$$

Again, it is a strange partial differential equation relating these vector fields. To make sense of it, apply Stokes theorem to compute the work done y the electric field around a closed curve.

That means if we have a wire in there, and you want to find the voltage along the wire generated by the varying magnetic field.

An application is a transformer. It passes an input voltage through a loop of wire. Another loop of wire is intertwined. The magnetic field $$\vec C$$ generated by alternating current, varies over time. That causes curl of the electric field $$nabla\times\vec F$$, what will generate voltage over the intertwined loop.

The intertwined loop is a closed curve. The voltage generated in this circuit $$\oint_C \vec E\cdot\,d\vec r = \iint_S (\nabla\times\vec E)\cdot d\vec S \nonumber$$

Combined with equation $$\eqref{eq:max2a}$$ $$\newcommand{pdv}[2]{\frac{\partial #1}{\partial #2}} \oint_C \vec E\cdot\,d\vec r = \iint_S \left(-\pdv{\vec B}{t}\right) \cdot d\vec S \nonumber$$

If you don’t move the loop, and only the magnetic field changes then you can take the $$\newcommand{pdv}[2]{\frac{\partial #1}{\partial #2}} \pdv{\vec B}{t}$$ out of the integral.

It says, if the magnetic field changes over time, out of nowhere, it creates an electric field.

### Magnetic field

#### div

The divergence of the magnetic field is $$0$$ $$\nabla\cdot\vec B = 0$$

That is fortunate, otherwise you would run into trouble trying to understand surface independence when you apply Stokes’ theorem in here.

The curl of the magnetic field is caused by motion of charged particles $$\newcommand{pdv}[2]{\frac{\partial #1}{\partial #2}} \nabla\times\vec B = \mu_0\,\vec J + \epsilon_0\,\mu_0\pdv{\vec E}{t}$$

Here

• $$J$$ is the vector current density. It measures the flow of electrically charged particles.
• $$\mu_0$$ is a physics constant

This is how the transformer generates it magnetic field.

## Review



My notes of the excellent lectures 32 by “Denis Auroux. 18.02 Multivariable Calculus. Fall 2007. Massachusetts Institute of Technology: MIT OpenCourseWare, https://ocw.mit.edu. License: Creative Commons BY-NC-SA.”

## Completely different integrals

### Triple integrals

$$\iiint_R f\,dV \nonumber$$

Integrates a scalar quantity over a 3D region. The shape of the region is described in the integral’s bounds.

Volume element

• In Cartesian coordinates: $$dV = dx\,dy\,dz$$
• In cylindrical coordinates: $$dV = dz\,r\,dr\,d\theta$$
• In spherical coordinates: $$dV = \rho^2\sin\phi\,d\rho\,d\phi\,d\theta$$. Remember
• $$\rho$$ is the distance from the origin
• $$\phi$$ is the angle with the positive $$z$$-axis, where $$0\lt\phi\lt\pi$$
• $$\theta$$ is the angle around the $$z$$-axis, where $$0\lt\phi\lt 2\pi$$

Set the bounds of integration

• For Cartesian and cylindrical coordinates:
1. For $$\int dz$$, for a fixed point in the $$xy$$-plane, find the bounds for $$z$$. The bottom and top surface of your solid as a function of $$(x,y)$$.
2. For the outer integrals, look at the solid from above, to see its projection on the $$xy$$-plane. Set up a double integral in rectangular or polar coordinates for the bounds of $$x$$ and $$y$$. If you did $$z$$ first, the inner bounds are given by bottom and top, and the outer ones are given by looking at the shadow of the region.
• For spherical coordinates:
1. For $$\int d\rho$$, it is like you shoot a ray from the origin to space. You want to know what part of you beam is in the solid. You want to solve for the value of $$\rho$$ when you enter and when you leave the solid.
2. For $$\int d\phi$$, sketch the $$rz$$-plane to find the bounds for $$\rho$$.
3. For $$\int d\theta$$, ask yourself what values of $$\theta$$ will I be inside my region.

Applications

### Surface integrals

$$\iint_S \vec F\cdot\hat n\,dS \nonumber$$

Integrates a vector quantity over a 2D surface. The shape of the region is described in the integral’s bounds.

Area element

• In Carthesian coordinates: $$dA = dy\,dx$$:
• In polar coordinates: $$dA = r\,dr\,d\theta$$:

Formulas for $$\hat n\,dS$$ in various settings

That will turn it into a regular double integral.

You then set up the bounds in terms of the integration variables.

### Line integrals

$$\int_C \vec F\cdot\,d\vec r \nonumber$$

Integrates a vector quantity over a 1D curve. The shape of the curve is described in the integral’s bounds.

Evaluate as $$\int_C \vec F\cdot\,d\vec r = \int_C P\,dx + Q\,dy + R\,dz \nonumber$$

Approach

1. Parameterize the curve $$C$$ to express $$x,y,z$$ in terms of a single (parametric) variable.
2. Solve the single integral.

## Bridges between the integrals

Each of these theorems relates a quantity with a certain number of integral signs to a quantity with one more integral sign.

### Triple integral – Surface integral

The divergence theorem connects these.

For a closed surface $$S$$, you can replace the flux integral with a triple integral over the region inside. $$\iint_S\vec F\cdot\hat n\,dS = \iiint_D (\rm{div}\,\vec F)\,dV \nonumber$$

Here $$\vec F$$ is a vector field, and $$\rm{div}\,\vec F$$ is a function that relates to the vector field.

### Surface integral – Line integral

Stokes’ theorem connects these.

I can replace a line integral on closed curve $$C$$, with a double integral on the surface $$S$$ $$\oint_C \vec F\cdot d\vec r = \iint_S (\nabla\times\vec F)\cdot\hat n\,dS \nonumber$$

It relates a line integral for field $$\vec F$$ to a surface integral from another field, $$\nabla\times\vec F$$.

You compute it as any other surface integral. You find the formula for $$\hat n\cdot dS$$, do the dot-product, substitute and evaluate. The calculation doesn’t know it came from a curl, and is the same as any other flux integral.

### Line integral – Function

The line integral for a gradient of a function is equal to the change in value of the function. Given $$\vec F$$ with curl $$=0$$, find potential

The fundamental theorem of calculus says $$f(P_1) – f(P_0) = \int_C (\nabla f)\cdot d\vec r \nonumber$$

The line integral for the vector field given by the gradient of the function, is equal to the change in value of the function.

## Stokes’ for work (in space)


My notes of the excellent lectures 31 and 32 by “Denis Auroux. 18.02 Multivariable Calculus. Fall 2007. Massachusetts Institute of Technology: MIT OpenCourseWare, https://ocw.mit.edu. License: Creative Commons BY-NC-SA.”

## Stokes’ Theorem

The curl measures the value of the vector field to be conservative. In the plane, curl also came up as Green’s theorem to convert line integrals into double integrals. The same thing in space, is called Stokes’ theorem.

Stokes’ Theorem

The work done by a vector field along a closed curve, can be replaced with a double integral $$\oint_C \vec F\cdot d\vec r = \iint \underbrace{\left(\nabla\times\vec F\right)}_{\rm{curl}\left(\vec F\right)}\cdot\hat n\,dS \nonumber$$
Where
• $$C$$ is a closed curve in space
• $$S$$ is any surface bounded by $$C$$
The catch is that we have to figure out what orientation convention to use.

### Orientation

For the orientation of $$C$$ and $$S$$ to be compatible, follow either rule in the table below

Walking Right-hand rule
If I walk along $$C$$ in the positive direction, with $$S$$ to my left, then $$\hat n$$ is pointing up Thumb goes along $$C$$ positively; index finger tangent to $$S$$ (towards the interior of $$S$$); then the middle finger points parallel to $$\hat n$$.

#### Examples

It points up.

##### Two

Imagine you’re walking on the inside of the cone, following the boundary, then the surface is to your left. So the normal vector will be pointing up and slightly into the cone. Or using the rigt-hand rule, with the index finger pointing down along the cone, then the normal vector points up and in.

##### Three

The cylinder has two boundary curves. How should I choose the orientation of my curves?

• For $$C’$$: imagine you’re walking on the outside of the cylinder along $$C’$$. If you want to walk so, that the cylinder is to your left, then you have to go counterclockwise.
• For $$C$$: image walking along the outside of the cylinder. If you want to walk so, that the cylinder is to your left, then you have to go clockwise.

### Examples

#### One

Compute the line integral on the unit circle on the $$xy$$-plane.

You can choose any surface

Then calculate the flux of $$\rm{curl}(\vec F)$$

## Proof

### On a plane (comparing Stokes with Green)

Surface $$S$$ is a portion of $$xy$$-plane, bounded by a curve $$C$$ (counterclockwise). Let $$\vec F$$ be $$\left\langle P, Q, R\right\rangle$$.

The curve is only on the $$xy$$-plane $$\Rightarrow z=0 \Rightarrow dz=0$$ \begin{align*} \oint_C\vec F\cdot d\vec r &= \oint_C P\,dx + Q\,dy \\ &= \iint_S \left(\nabla\times\vec F\right)\cdot\hat n\,dS \tag{Stokes} \\ \end{align*} \nonumber

The normal vector $$\hat n$$ compatible with this $$C$$ and $$S$$ points up, $$\hat n=\hat k$$ \begin{align*} \oint_C\vec F\cdot d\vec r &= \iint_S \left(\nabla\times\vec F\right)\cdot\hat k\,dS \\ \end{align*} \nonumber

This means we will be integrating the $$z$$-component of $$\rm{curl}(\vec F)$$. Based on equation curl \begin{align*} \left(\nabla\times\vec F\right)\cdot\hat k &= (Qx – Py) \end{align*}

In the plane $$dS=dx\,dy$$, so the double integral becomes \begin{align*} \oint_C\vec F\cdot d\vec r &= \iint_S (Q_x-Py)\,dx\,dy \\ \end{align*} \nonumber

This shows Green’s theorem is the special case of Stokes in the $$xy$$-plane.

### Generalizing

• We know it for $$C$$, $$S$$ in the $$xy$$-plane. (Green’s theorem)
• Also for $$C,S$$ in any plane (using that work, flux, curl make sense independently of the coordinate system!)
• Given any $$S$$: decompose the surface into tiny, almost flat, pieces

Any surface we can cut into tiny pieces, and these pieces are basically flat.

Then you can use Stokes’ theorem on each small flat piece. It says that the line integral along, say, this curve is equal to the flux of a curl through this tiny piece of surface. If I add all of the small contributions to flux I get the total flux.

All the little line integrals on the inside will cancel out. The only ones that I go through only once are on the outermost pieces. So, summed together, I get the work done along the outer boundary.

In other words: sum of work around each little piece $$=$$ the work along $$C$$.

The sum of the flux through each piece $$=$$ flux through $$S$$.

## Examples

### One

Find the work of $$F=\left\langle z,x,y\right\rangle$$ around unit circle in $$xy$$-plane (counterclockwise).

#### Directly

Using the equation from \begin{align*} \oint_c \vec F\cdot d\vec r &= \oint_C\left( Pdx + Qdy + Rdz \right) \\ &= z\,dx + x\,dy + y\,dz \\ \end{align*}

On the circle $$z=0$$, and we can parametize $$x$$ and $$y$$ \left\{ \begin{align*} x = \cos t &\Rightarrow dx = -\sin t\,dt \\ y = \sin t &\Rightarrow dy = \cos t\,dt \\ z = 0 &\Rightarrow dz = 0 \end{align*} \right.

The range is counterclockwise around the circle \begin{align*} \oint_c \vec F\cdot d\vec r &= \int_0^{2\pi} 0\,dx + \overbrace{\cos t}^{x}\,\overbrace{\cos t\,dt}^{dy} + 0y \\ &= \int_0^{2\pi} \cos^2 t\,dt = \pi \end{align*}

#### Using Stokes’ theorem

The smart choice would be to use the flat unit disk. To convince you that we can take any surface, I am going to take a piece of paraboloid $$z = 1 – x^2 – y^2 \nonumber$$

With the normal vector pointing up

Compute \begin{align*} \iint_S\left(\nabla\times\vec F\right)\,\hat n\,dS \end{align*}

Let’s start with the curl \newcommand{pdv}[2]{\frac{\partial #1}{\partial #2}} \begin{align*} \nabla\times\vec F &= \left| \begin{array}{ccc} \hat\imath & \hat\jmath & \hat k \\ \pdv{}{x} & \pdv{}{y} & \pdv{}{z} \\ P & Q & R \end{array} \right| = \left| \begin{array}{ccc} \hat\imath & \hat\jmath & \hat k \\ \pdv{}{x} & \pdv{}{y} & \pdv{}{z} \\ z & x & y \end{array} \right| \\ &= \hat\imath \left| \begin{array}{cc} \pdv{}{y} & \pdv{}{z} \\ x & y \end{array} \right| – \hat\jmath \left| \begin{array}{cc} \pdv{}{x} & \pdv{}{z} \\ z & x \end{array} \right| + \hat k \left| \begin{array}{cc} \pdv{}{x} & \pdv{}{y} \\ z & x \end{array} \right| \\ &= (1-0)\hat\imath – (0-1)\hat\jmath + (1-0)\hat k = \left\langle 1, 1, 1\right\rangle \end{align*}

To find $$\hat n\,dS$$, let’s call $$z=1-x2-y^2=f(x,y)$$ and use the formula for the graph of a function \begin{align*} \hat n\,dS &= \left\langle -f_x,-f_y,1\right\rangle\,dx\,dy \\ &= \left\langle 2x, 2y, 1\right\rangle\,dx\,dy \\ \end{align*}

The flux over the shadow of the surface (the unit disk), and switch to polar coordinates \begin{align*} \iint_S\left(\nabla\times\vec F\right)\,\hat n\,dS &= \iint_S \left\langle 1, 1, 1\right\rangle \cdot \left\langle 2x, 2y, 1\right\rangle\,dx\,dy \\ &= \iint (2x + 2y + 1)\,dx\,dy \\ &= \int_0^1 \int_0^{2\pi} (2r\cos\theta + 2r\sin\theta + 1)\,r\,d\theta\,dr \\ &= \int_0^1 \int_0^{2\pi} (2r^2\cos\theta + 2r^2\sin\theta + r)\,d\theta\,dr \\ &= \int_0^1 \Big[ 2r^2\sin\theta – 2r^2\cos\theta + r\theta \Big]_0^{\theta=2\pi}\,dr \\ &= \int_0^1 2r\pi\,dr = \Big[ r^2\pi \Big]_0^{r=1} = \pi \end{align*}

## Path-independance

### Simply-connected

Define: a region is simply-connected if every closed loop inside it bounds a surface inside it.

Examples with a unit circle in the $$xy$$-plane

• A region is the space with the origin removed: this is simply-connected because we can avoid the origin by making a little dome over it.
• A region is the space with the $$z$$-axis removed: not simply-connected, because if I try to find a surface that is bounded by this curve, it has to cross the $$z$$-axis somewhere. There is no surface whose only boundary is the curve, that doesn’t intersect the $$z$$-axis anywhere.

Recall: if $$\vec F = \nabla f$$ is a gradient field, then its curl $$=0$$.

### Theorem

Theorem: if $$\vec F$$ is defined in a simply-connected region, and $$\rm{curl}\vec F = 0$$, then $$\vec F$$ is a gradient field, and $$\int_C\vec F\cdot d\vec r$$ is path-independent.

#### Proof

Assume (\rm{curl}\vec F = 0\), and there are two curves that go from $$P_0$$ to $$P_1$$

We like to proof that the difference in line integrals in $$0$$ $$\int_{C_1} \vec F\cdot d\vec r – \int_{C_2} \vec F\cdot d\vec r \nonumber$$

Form a closed curve that is $$C_1$$ minus $$C_2$$.

$$C$$ is a closed curve, so we can use Stokes’ theorem To use Stoke’s theorem, we need to find a surface to apply it to. That is where the assumption of simply-connected is useful. We can find a surface, $$S$$, that bounds $$C$$ because the region is simply connected. \begin{align*} \int_{C_1} \vec F\cdot d\vec r – \int_{C_2} \vec F\cdot d\vec r &= \oint_C \vec F\cdot d\vec r \tag{Stokes’} \\ &= \iint_S \underbrace{\rm{curl}(\vec F)}_{=0} \cdot d\vec S \end{align*}

But now, the curl is zero. So, if I integrate zero, I will get zero. OK, so I proved that my two line integrals along C1 and C2 are equal. But for that, I needed to be able to find a surface which to apply Stokes theorem. And that required my region to be simply-connected.

If we had a vector field that was defined only outside of the $$z$$-axis and I took two paths that went on one side and the other side of the $$z$$-axis, I might have obtained, actually, different values of the line integral.

### Remark on simply-connected regions

Topology classifies surfaces in space, but trying to loop at loops.

#### Sphere

The surface of a sphere is simply-connected. Let’s take a closed curve $$C$$ on the surface of a sphere. You can always find a portion of the sphere’s surface that is bounded by it.

#### Torus

A torus (the surface of a doughnut), is not simply-connected. If you look at the $$C_1$$ loop, it bounds a surface in space, but that surface cannot be made to be just a piece of the donut. You have to go through the hole. You have to leave the surface of a torus.

The $$C_2$$ loop does not bound anything that’s completely contained in the torus. It bounds a disc, but that is inside of a torus, and not not a part of the surface. So, an topologists would say: there’s two “independent” loops that don’t bound surfaces, that don’t bound anything.

### Orientability

Say, I want to apply Stokes’ theorem to simplify a line integral along the curve below on the left. This curve goes twice around the $$z$$-axis.

One way to find the surface that is bounded by this curve, is to take a Möbius strip. It’s a one sided strip, where when you go around, you flip one side becomes the other. So, if you want to take a band of paper and glue the two sides with a twist, so, it’s a one sided surface. And, that gives us, actually, serious trouble if we try to orient it to apply Stokes theorem.

If we use the outside edge to orient the normal vector, it points up. But when I follow the curve, I end up on the inside where the convention for the normal vectora says it has to point down on that same surface. There’s no way to consistently choose a normal vector for the Mobius strip. So, that’s what we call a non-orientable surface. That just means it has only one side.

If it has only one side, that we cannot speak of flux, because we have no way of saying that we’ll be counting things positively one way, negatively the other way, because there’s only one. There’s no notion of sides. So, you can’t define a side towards which things will be going positively. So, that’s actually a situation where flux cannot be defined.

So, if we really wanted to apply Stokes theorem, because space is simply-connected, and I will always be able to apply Stokes theorem to any curve, what would I do? Well, this curve actually bounds another surface that is orientable. You can take a hemisphere, and you can take a small thing and twist it around. That spherical thing with a little slit going twisting into it, will border my loop. And, that one is orientable.

### Surface-independence

For a surface in space, you can take e.g. an upper half sphere $$S_1$$. Based on the orientation convention, you wil need to make the normal vector upwards. We could also take $$S_2$$, with according to the convention the normal vector will be going into the thing.

Stokes’ says $$\int_C \vec F\cdot d\vec r = \iint_{S_1} (\nabla\times\vec F)\cdot\hat n\,dS = \iint_{S_2} (\nabla\times\vec F)\cdot\hat n\,dS \nonumber$$

That means that curl $$\vec F$$, $$\nabla\times\vec F$$ has some soft of surface independent property for as long as the boundary is curve $$C$$. Why is that?

To compare the flux through $$S_1$$ and $$S_2$$ we subtract them and call it $$\iint_S$$ $$\iint_{S_1} (\nabla\times\vec F)\cdot\hat n\,dS – \iint_{S_2} (\nabla\times\vec F)\cdot\hat n\,dS = \iint_{S} (\nabla\times\vec F)\cdot\hat n\,dS \nonumber$$

Here $$S$$ is $$S_1$$ with its orientation minus $$S_2$$ with its reversed orientation. So $$S$$, is the whole closed surface, with the normal vector pointing out everywhere.

To find the flux through closed surface $$S$$, we can replace it with a tripple integral. That is the divergence theorem. Let’s call the region $$D$$ $$\iint_{S_1} (\nabla\times\vec F)\cdot\hat n\,dS – \iint_{S_2} (\nabla\times\vec F)\cdot\hat n\,dS = \iiint_D \rm{div}\left( \nabla\times\vec F \right)\,dV \nonumber$$

We can check: if you take the divergence of the curl of a vector field, you always get $$0$$ $$\rm{div\left(\nabla\times\vec F\right)} = 0 \nonumber$$

So $$\iint_{S_1} (\nabla\times\vec F)\cdot\hat n\,dS – \iint_{S_2} (\nabla\times\vec F)\cdot\hat n\,dS = \iiint_D 0\,dV = 0 \nonumber$$

That is why the flux for $$S_1$$ and $$S_2$$ were the same, and we could choice either surface for Stokes’.

#### Let’s check that $$\nabla\cdot\left(\nabla\times\vec F\right) = 0$$

Let vector field $$\vec F=\left\langle P,Q,R\right\rangle$$, where $$P(x,y,z), Q(x,y,z)$$ and $$Q(x,y,z)$$

Remember the curl

Let’s try to do the cross-product using the pseudo-determinant \begin{align*} \rm{curl}\left(\vec F\right) &= \nabla\times\vec F \\ &= \left| \begin{array}{ccc} \hat\imath & \hat\jmath & \hat k \\ \pdv{}{x} & \pdv{}{y} & \pdv{}{z} \\ P & Q & R \end{array} \right| \\ &= \left\langle R_y – Q_z, P_z-R_x, Q_x-P_y\right\rangle \end{align*}

The divergence of this, $$\nabla \cdot (\nabla\times\vec F)$$ \newcommand{pdv}[1]{\tfrac{\partial}{\partial #1}} \begin{align*} \nabla \cdot (\nabla\times\vec F) &= \left\langle \pdv{x}, \pdv{y}, \pdv{z} \right\rangle \cdot \left\langle R_y – Q_z, P_z-R_x, Q_x-P_y \right\rangle \\ &= \pdv{x}(R_y – Q_z) + \pdv{y}(P_z-R_x) + \pdv{z}(Q_x-P_y) \\ &= (R_y – Q_z)_x + (P_z-R_x)_y + (Q_x-P_y)_z \\ &= R_{yx} – Q_{zx} + P_{zy} – R_{xy} + Q_{xz} – P_{yz} \end{align*}

The mixed second derivatives are the same, no matter what order you take them \newcommand{pdv}[1]{\tfrac{\partial}{\partial #1}} \begin{align*} \nabla \cdot \nabla\times\vec F &= \xcancel{R_{yx}} – \cancel{Q_{zx}} + \bcancel{P_{zy}} – \xcancel{R_{xy}} + \cancel{Q_{xz}} – \bcancel{P_{yz}} \\ &= 0 \end{align*}

Note: if we had “real” vectors (not using the $$\nabla$$-operator) $$u \cdot \left( u \times v\right) \nonumber$$ $$u\times v$$ is $$\perp$$ to $$u$$ and $$v$$, so its dot-product will be $$0$$.

## Curl (in space)


My notes of the excellent lecture 30 by “Denis Auroux. 18.02 Multivariable Calculus. Fall 2007. Massachusetts Institute of Technology: MIT OpenCourseWare, https://ocw.mit.edu. License: Creative Commons BY-NC-SA.”

The curl measures the value of the vector field to be conservative. For a velocity field, curl measures the rotation component of the motion.

## Definition

Let $$\vec F$$ be a vector field with components $$P(x,y,z)$$, $$Q(x,y,z)$$ and $$R(x,y,z)$$, then \begin{align*} \vec F &= \left\langle P,Q,R\right\rangle \\ &= P\hat\imath+Q\hat\jmath+R\hat k \end{align*}

Then \newcommand{pdv}[2]{\frac{\partial #1}{\partial #2}} \begin{align} \rm{curl}\left(\vec F\right) &= \left\langle \pdv{R}{y} – \pdv{Q}{z}\,,\, \pdv{P}{z}-\pdv{R}{x}\,,\, \pdv{Q}{x}-\pdv{P}{y} \right\rangle \nonumber \\[.5em] &= \left\langle R_y – Q_z\,,\, P_z-R_x\,,\, Q_x-P_y \right\rangle \nonumber \\ &= (R_y – Q_z)\hat\imath + (P_z-R_x)\hat\jmath + (Q_x-P_y)\hat k \label{eq:curl1} \end{align}

If $$\vec F$$ is defined in a simply-connected region $$S$$, and differentiable everywhere on $$S$$, then $$\shaded{ \vec F \text{ is conservative} \Leftrightarrow \rm{curl}\left(\vec F\right)=0 } \nonumber$$

The difference with curl in the plane, is that in space the curl is again a vector field, not a scalar.

Each of the terms has to be $$0$$ for the field to be conservative.

### $$\nabla$$-notation

the symbolic $$\nabla$$ “del” notation for the operator used for the gradient and divergence \newcommand{pdv}[2]{\frac{\partial #1}{\partial #2}} \begin{align*} \nabla &= \left\langle\pdv{}{x}, \pdv{}{y}, \pdv{}{z}\right\rangle \\[0.5em] \nabla f &= \left\langle\pdv{f}{x}, \pdv{f}{y}, \pdv{f}{z}\right\rangle & \text{gradient} \\[0.5em] \nabla\cdot\vec F &= \left\langle\pdv{}{x}, \pdv{}{y}, \pdv{}{z}\right\rangle \cdot \left\langle P,Q,R\right\rangle \\ &= \pdv{P}{x} + \pdv{Q}{y} + \pdv{R}{z} & \text{divergence} \end{align*}

Let’s try to do the cross-product using the pseudo-determinant \newcommand{pdv}[2]{\frac{\partial #1}{\partial #2}} \begin{align*} \nabla\times\vec F &= \left| \begin{array}{ccc} \hat\imath & \hat\jmath & \hat k \\ \pdv{}{x} & \pdv{}{y} & \pdv{}{z} \\ P & Q & R \end{array} \right| \\ &= \hat\imath \left| \begin{array}{cc} \pdv{}{y} & \pdv{}{z} \\ Q & R \end{array} \right| – \hat\jmath \left| \begin{array}{cc} \pdv{}{x} & \pdv{}{z} \\ P & Q \end{array} \right| + \hat k \left| \begin{array}{cc} \pdv{}{x} & \pdv{}{y} \\ P & Q \end{array} \right| \\ &= \left( \pdv{R}{y} – \pdv{Q}{z} \right) \hat\imath – \left( \pdv{R}{x} – \pdv{P}{z} \right) \hat\jmath + \left( \pdv{Q}{x} – \pdv{P}{y} \right) \hat k \\ &= (R_y – Q_z)\hat\imath + (P_z-R_x)\hat\jmath + (Q_x-P_y)\hat k \end{align*}

This matches equation $$\eqref{eq:curl1}$$. So, the best way to remember the formula for curl is using this $$\nabla$$-notation \shaded{ \begin{align*} \rm{curl}\left(\vec F\right) &= \nabla\times\vec F \\ &= \left| \begin{array}{ccc} \hat\imath & \hat\jmath & \hat k \\ \pdv{}{x} & \pdv{}{y} & \pdv{}{z} \\ P & Q & R \end{array} \right| \\ &= \left\langle R_y – Q_z, P_z-R_x, Q_x-P_y\right\rangle \end{align*} } \label{eq:curl2}

## Geometrically

Curl measures the rotation component of a velocity field:

• the direction corresponds to the axis of rotation, and
• the magnitude corresponds to twice the angular velocity ($$\omega$$).

### Examples

#### One

Let $$\vec v$$ be a fluid that is rotating with angular velocity $$\omega$$ around the $$z$$-axis

So $$\vec v = \left\langle -\omega y, \omega x,0 \right\rangle \nonumber$$

The curl \newcommand{pdv}[2]{\frac{\partial #1}{\partial #2}} \begin{align*} \nabla\times\vec v &= \left| \begin{array}{ccc} \hat\imath & \hat\jmath & \hat k \\ \pdv{}{x} & \pdv{}{y} & \pdv{}{z} \\ -\omega y & \omega x & 0 \end{array} \right| \\ &= (0 – 0)\hat\imath – (0-0)\hat\jmath + (\omega-(-\omega)))\hat k \\ &= 2\omega\hat k = \left\langle 0,0,2\omega \right\rangle \end{align*}

The curl gives you the angular rotation ($$\times 2$$), and the axis of rotation (the vertical axis).

#### Two

Let $$\vec v$$ be a fluid moving in a straight direction

So $$\vec F = \left\langle a,b,c \right\rangle \nonumber$$

All the partial derivatives are $$0$$, so the curl is $$0$$.

#### Three

Let $$\vec v$$ be a vector field that stretches things along the $$x$$-axis (expanding)

So $$\vec F = \left\langle x,0,0 \right\rangle \nonumber$$

Again, the curl will be $$0$$. But the $$\rm{div}(\vec F)=1$$, because that measures stretching of a vector field.



My notes of the excellent lecture 30 by “Denis Auroux. 18.02 Multivariable Calculus. Fall 2007. Massachusetts Institute of Technology: MIT OpenCourseWare, https://ocw.mit.edu. License: Creative Commons BY-NC-SA.”

The fundamental theorem of multivariable calculus tell us (using the symbolic $$\nabla$$-operator)

If you take the line integral of the gradient of a function, what you get back is the function.
$$\shaded{ \int_C\nabla f\cdot d\vec r = f(P_1) – f(P_0) } \label{eq:fundthm}$$

Similar to the plane, in space

When vector field $$\vec F$$ is a gradient of function $$f(x,y)$$, it is called a gradient field $$\newcommand{pdv}[1]{\tfrac{\partial}{\partial #1}} \shaded{ \vec F = \nabla f = \left\langle \pdv{x}f, \pdv{y}f, \pdv{z}f \right\rangle } \nonumber$$ where $$f(x,y,z)$$ is called the potential.

## When is a vector field a gradient field?

In the plane, to check if a vector field is a gradient field we only had to check the condition


In space, we want to know if $$\newcommand{pdv}[2]{\frac{\partial #1}{\partial #2}} \vec F = \left\langle P,Q,R\right\rangle \stackrel{?}{=} \nabla f = \left\langle \pdv{}{x}f, \pdv{}{y}f, \pdv{}{z}f \right\rangle \nonumber$$ where $$\vec F=\left\langle P,Q,R\right\rangle$$ be defined in a simply connected region.



Recall:


Based on this second partial derivative rule \begin{align*} \underline{P_y} = f_{xy} &= f_{yx} = \underline{Q_x} \\ \underline{P_z} = f_{xz} &= f_{zx} = \underline{R_x} \\ \underline{Q_z} = f_{yz} &= f_{zy} = \underline{R_y} \end{align*} \nonumber

Therefore, if $$\vec F=\left\langle P,Q,R\right\rangle$$, defined in a simply connected region is a gradient field, when \shaded{ \left\{ \begin{align*} P_y &= Q_x \\ P_z &= R_x \\ Q_z &= R_y \end{align*} \right. } \nonumber

### Exact differential

We can also think of it in terms of differentials.

If we have a differential of the form $$P\,dx + Q\,dy + R\,dz \nonumber$$ is an exact differential. That means it is going to equal to $$df$$ for some function $$F$$ exactly and of the same conditions. Just another way of saying it.

#### Example

For which $$a$$ and $$b$$ is the differential below exact? $$\underbrace{axy}_P\,dx + (\underbrace{x^2+z^3}_Q)dy + (\underbrace{byz^2+4z^3}_R)dz \nonumber$$

Or, for which $$a$$ and $$b$$ is $$\vec F$$ a gradient field? $$\vec F=\left\langle axy,x^2+z^3, byz^2-4z^3\right\rangle \nonumber$$

Compare $$\left. \begin{array}{lllll} P_y &= ax &= 2x &= Q_x &\Rightarrow a = 2 \\ P_z &= 0 &= 0 &= R_x \\ Q_z &= 3z^2 &= bz^2 &= R_y &\Rightarrow b=3 \end{array} \right\} \nonumber$$

For those values of $$a$$ and $$b$$ we can look for a potential. For any other values we would have to set up the line integral.

## Potential of a gradient field

Recall: for the plane

When the field is a gradient, and you know the function $$f$$, you can simplify the evaluation of the line integral for work. $$\shaded{ \int_C\nabla f\cdot d\vec r=f(P_1)-f(P_0) } \nonumber$$ where $$f(x,y)$$ is called the potential

Once again, you have two methods: computing line integrals, or using antiderivatives.

### Method 1: Computing line integrals

Similar to in the plane, apply the fundamental theorem, equation $$\eqref{eq:fundthm}$$, to find an expression for the potential at $$(x_1,y_1,z_1)$$ \begin{align} &\int_C\vec F\cdot d\vec r=f(x_1,y_1,z_1)-f(0,0,0) \nonumber \\ \Rightarrow & f(x_1,y_1,z_1) = \underbrace{\int_C\vec F\cdot d\vec r}_\text{work} + \underbrace{f(0,0,0)}_{\mathrm{constant}} \label{eq:method1} \end{align}

The work in a gradient field is path independent $$\Longrightarrow$$ find the easiest path

Apply the work differential, to find the work along $$C$$ in gradient field $$\vec F$$

Then add the line integrals together, to get the total work. $$\underline{\int_C\vec F\cdot d\vec r} = \int_{C_1}\ldots + \int_{C_2}\ldots + \int_{C_3}\ldots \nonumber$$

Substitute this back in equation $$\eqref{eq:method1}$$ and drop the subscripts $$f(x,y,z) = \underline{\int_C\vec F\cdot d\vec r} + \rm{c} \nonumber$$ If you would take the gradient of $$f(x,y,z)$$, you should get $$\vec F$$ back.

### Method 2: Using Antiderivatives

Similar to in the plane. No integrals, but you have to follow the procedure very carefully.

Continue with the earlier example, with $$a=2$$ and $$b=3$$. Solve \newcommand{pdv}[2]{\frac{\partial #1}{\partial #2}} \begin{align} \pdv{f}{x} &= f_x = 2xy \label{eq:potential1} \\ \pdv{f}{y} &= f_y = x^2 + z^3 \label{eq:potential2} \\ \pdv{f}{z} &= f_z = 3yz^2 – 4z^3 \label{eq:potential3} \end{align}

Integrate $$\eqref{eq:potential1}$$ in respect to $$x$$. The integration constant might depend on $$y$$ and $$z$$, so call it $$g(y,z)$$ $$\newcommand{pdv}[2]{\tfrac{\partial #1}{\partial #2}} f_x = 2xy \xrightarrow{\int dx} \underline{f = x^2y + g(y,z)} \nonumber$$

To get information on $$g(y,z)$$ we look at the other partials. Take the derivative of $$f$$ in respect to $$y$$ and compare to $$\eqref{eq:potential2}$$. The integration constant might depend on $$y$$, so call it $$h(z)$$ \newcommand{pdv}[2]{\frac{\partial #1}{\partial #2}} \begin{align*} x^2 + \pdv{g}{y} &= x^2 + z^3 \\ \Rightarrow \pdv{g}{y} &= z^3 \xrightarrow{\int dy} g = \underline{yz^3 + h(z)} \\ \end{align*}

Substituting $$g$$ in $$f$$ $$f = x^2y + \underline{yz^3 + h(z)} \nonumber$$

To get information on $$h(z)$$ we look at the other partials. Take the derivative of $$f$$ in respect to $$z$$ and compare to $$\eqref{eq:potential3}$$. The integration constant $$\rm{c}$$ is a true constant \newcommand{pdv}[2]{\frac{\partial #1}{\partial #2}} \begin{align*} 3yz^2 + \pdv{h}{z} &= 3yz^2 – 4z^3 \\ \Rightarrow \pdv{h}{z} &= – 4z^3 \xrightarrow{\int dz} h = \underline{-z^4 + \rm{c}} \\ \end{align*}

Substituting $$h$$ in $$f$$ $$f = x^2y + yz^3 \underline{-z^4\,(+ \rm{c})} \nonumber$$

If you want to find one potential, you can just forget about the constant. If you want to find all the potentials: they differ by this constant.

## Diffusion/heat (in space)



My notes of the excellent lecture 29 by “Denis Auroux. 18.02 Multivariable Calculus. Fall 2007. Massachusetts Institute of Technology: MIT OpenCourseWare, https://ocw.mit.edu. License: Creative Commons BY-NC-SA.”

The Diffusion equation governs motion of e.g. smoke in unmovable air, or dye in a solution.

## Definition

It is a partial differential equation. The unknown is a function, and the function relates the partial derivatives of that function to each other.


Let $$u$$ be the concentration of substance at a given point, $$u(x,y,z,t$$). Let $$\vec F$$ be the flow of substance.


For the heat equation $$u$$ would be the temperature.

## Understanding

Let $$\vec F$$ is the flow of smoke

1. Physics: smoke will flow from high concentration towards low concentration regions. So, in the direction where the concentration decreases the fastest. That is negative the gradient. So $$\vec F$$ is directed along $$-\nabla u$$. In fact, the flow is proportional to the gradient field $$\vec F = -k\,\nabla u \label{eq:understanding1}$$
2. Math: How does the flow affect the concentration? One place it will be decreasing, and another place it will be increasing. The divergence theorem relates $$\vec F$$ and $$\newcommand{pdv}[2]{\frac{\partial #1}{\partial #2}} \pdv{u}{t}$$.

Let’s try to understand that

Flux out of $$D$$ through $$S$$ measures the amount of smoke going through $$S$$ in unit time. $$\newcommand{oiint}{\subset\!\!\supset\kern-1.65em\iint} \oiint_S\vec F\cdot\hat n\,dS \nonumber$$


We can use the divergence theorem to compute this

$$\iiint_D\rm{div}(\vec F)\,dV =-\pdv{}{t}\left( \iiint_D u\,dV\right) \nonumber$$

Instead of taking the sum of all the amounts of smoke, and seeing how it changes over time, we can take all the little derivatives and then sum them. The derivative of the sum, is the sum of the derivatives

$$\newcommand{pdv}[2]{\frac{\partial #1}{\partial #2}} \frac{d}{dt}\left(\sum_i u(x_i,y_i,z_i,t)\,\Delta V_i\right) = \sum_i\pdv{u}{t}(x_i,y_i,z_i,t)\,\Delta V_i \nonumber$$

Thus $$\iiint_D\underline{\rm{div}(\vec F)}\,dV = -\iiint_D \underline{\pdv{u}{t}}\,dV \nonumber$$

For any region $$D$$. $$\newcommand{pdv}[2]{\frac{\partial #1}{\partial #2}} \tfrac{1}{\rm{Vol}(D)}\iiint_D \rm{div}(\vec F)\,dV = \tfrac{1}{\rm{Vol}(D)}\iiint_D-\pdv{u}{t}\,dV \\ \Rightarrow \text{avg. of }\rm{div}(\vec F)\text{ in }D = \text{avg. of }-\pdv{u}{t}\text{ in }D \nonumber$$

So, from the divergence theorem we derived $$\shaded{ \rm{div}(\vec F) = -\pdv{u}{t} } \nonumber$$

So, this is another way think about what divergence means: how much “stuff” is flowing out, how much you’re loosing.

If you combine that with equation $$\eqref{eq:understanding1}$$, you get the diffusion equation \newcommand{pdv}[2]{\frac{\partial #1}{\partial #2}} \left. \begin{align*} \vec F &= -k\,\nabla u \\ \pdv{u}{t} &= -\rm{div}(\vec F) \end{align*} \right\} \Rightarrow \shaded{ \pdv{u}{t} = +k\,\rm{div}(\nabla u)= k\,\nabla^2 u }

## Divergence (in space)



My notes of the excellent lectures 28 and 29 by “Denis Auroux. 18.02 Multivariable Calculus. Fall 2007. Massachusetts Institute of Technology: MIT OpenCourseWare, https://ocw.mit.edu. License: Creative Commons BY-NC-SA.”

Divergence measures how much the flow is expanding. It singles out the stretching component of motion.

AKA as the Gauss-Green theorem. The 3D analogue to Green’s theorem for flux.

## Definition

If $$S$$ is a closed surface, enclosing a region $$D$$, oriented with $$\hat n$$ outwards and $$\vec F$$ defined and differentiable everywhere in $$D$$, then $$\newcommand{oiint}{\subset\!\!\supset\kern-1.65em\iint} \shaded{ \oiint_S\vec F\cdot d\vec S = \iiint_D\rm{div}(\vec F)\,dV } \label{eq:divthm}$$ The circle in the double integral reminds us that it must be a closed surface. The normal vectors always point outwards.

Let $$\vec F$$ be a vector field, where $$P,Q,R$$ are functions of $$x,y,z$$. \newcommand{pdv}[2]{\frac{\partial #1}{\partial #2}} \shaded{ \begin{align*} \rm{div}\left(\vec F\right) &= \rm{div}\left(P\hat\imath+Q\hat\jmath+R\hat k\right) \\ &= \pdv{P}{x}+\pdv{Q}{y}+\pdv{R}{z} \\ &= P_x + Q_y + R_z \end{align*} } \nonumber

The divergence measures how much the flow is expanding. If you take a region of space, the total amount of water that flows out of it is the total amount of sources that you have in there minus the sinks.

### $$\nabla$$-notation

$$\nabla$$ “del” is a symbolic notation for the operator $$\newcommand{pdv}[2]{\frac{\partial #1}{\partial #2}} \nabla = \left\langle\pdv{}{x}, \pdv{}{y}, \pdv{}{z}\right\rangle \nonumber$$

That makes \newcommand{pdv}[2]{\frac{\partial #1}{\partial #2}} \begin{align*} \nabla f &= \left\langle\pdv{f}{x}, \pdv{f}{y}, \pdv{f}{z}\right\rangle & \text{gradient} \\[0.7em] \nabla\cdot\vec F &= \left\langle\pdv{}{x}, \pdv{}{y}, \pdv{}{z}\right\rangle \cdot \left\langle P,Q,R\right\rangle \\ &= \pdv{P}{x} + \pdv{Q}{y} + \pdv{R}{z} & \text{divergence} \end{align*}

This will come in very useful when we talk about curl in space.

## Physical interpretation


Claim that $$\rm{div}\left(\vec F\right) = \text{“source rate”} \nonumber$$

The amount of flux generated per unit volume. Think about an incompressible fluid, such as water. Incompressible fluid flow with velocity $$\vec F$$: (given mass occupies a fixed volume),

## Proof of divergence theorem

Simplifications

2. We will start with a vertically simple region. If region $$D$$ is vertically simple It lives above the shadow on the $$xy$$-plane, and is between two graphs $$z_1(x,y)$$ and $$z_2(x,y)$$.

$$\newcommand{oiint}{\subset\!\!\supset\kern-1.65em\iint} \oiint_S\left\langle 0,0,R\right\rangle \cdot \hat n\,dS = \iiint_D R_z\,dz \nonumber$$

We will compute the right- and left-hand sides of equation $$\eqref{eq:simple1}$$, and then compare them.

### Right-hand side

In right-hand side of equation $$\eqref{eq:simple1}$$, the range for $$z$$ is from the bottom face to the top face. \begin{align*} \iiint_D R_z\,dV &= \iint_U \underline{\int_{z_1(x,y)}^{z_2(x,y)} R_z\,dz}\,dx\,dy \end{align*}

When you integrate $$\newcommand{pdv}[2]{\frac{\partial #1}{\partial #2}} R_z=\pdv{R}{z}$$ in respect to $$z$$, you just get $$R$$ back \begin{align} \iiint_D R_z\,dV &= \iint_U \Big[R(x,y,z)\Big]_{z_1(x,y)}^{z=z_2(x,y)}\,dx\,dy \nonumber \\ &= \iint_U \Big[R(x,y,z_2(x,y))-R(x,y,z_1(x,y))\Big]\,dx\,dy \label{eq:divthmproofleft} \\ \end{align}

### Left-hand side

The surface $$S$$ of the vertically simple region, consists of the top, bottom and the sides.

We will calculate the flux on each side, and then sum them together \newcommand{oiint}{\subset\!\!\supset\kern-1.65em\iint} \begin{align} \oiint_S\left\langle 0,0,R\right\rangle \cdot \hat n\,dS &= \iint_\rm{top} + \iint_\rm{bottom} + \iint_\rm{sides} \label{eq:combine} \end{align}

#### Top

Recall equation $$\eqref{eq:fluxgraph}$$ for the flux of a graph

$$\hat n\,dS =\pm\left\langle -f_x,-f_y,1\right\rangle\,dx\,dy \nonumber$$

Apply this to the graph of $$z=z_2(x,y)$$ $$\newcommand{pdv}[2]{\frac{\partial #1}{\partial #2}} \hat n\,dS = \left\langle -\pdv{z_2}{x}, -\pdv{z_2}{y}, 1 \right\rangle\,dx\,dy \nonumber$$

Do the dot-product with the vector field that only has a $$z$$-component and enter it in the double integral $$\eqref{eq:simple1}$$ \begin{align*} \iint_\text{top}\left\langle 0,0,R\right\rangle\cdot\hat n\,dS &= \iint_\text{top}\left\langle 0,0,R\right\rangle\cdot \left\langle -\pdv{z_2}{x}, -\pdv{z_2}{y}, 1 \right\rangle\\ &= \iint_\text{top} R\,dx\,dy \end{align*}

$$R$$ is $$R(x,y,z)$$. That means we can substitute the equation using the graph equation $$z=z_2(x,y)$$. As far as the region is concerned, it is above shadow $$U$$ \newcommand{pdv}[2]{\frac{\partial #1}{\partial #2}} \begin{align} \iint_\rm{top} \left\langle 0,0,R\right\rangle\cdot\hat n\,dS &= \iint_U R(x,y,z_2(x,y))\,dx\,dy \label{eq:divthmproofright1} \\ \end{align}

#### Bottom

Similarly, apply equation $$\eqref{eq:fluxgraph}$$ to the graph of $$z=z_1(x,y)$$ $$\newcommand{pdv}[2]{\frac{\partial #1}{\partial #2}} \hat n\,dS = \left\langle -\pdv{z_1}{x}, -\pdv{z_1}{y}, 1 \right\rangle \nonumber$$

Be care with the orientation of the normal vector. At the top surface, we want $$\hat n$$ pointing up, but at the bottom surface, we want $$\hat n$$ pointing down.


Do the dot-product with the vector field that only has a $$z$$-component and enter it in the double integral $$\eqref{eq:simple1}$$ \begin{align*} \iint_\text{bottom}\left\langle 0,0,R\right\rangle\cdot\hat n\,dS &= \iint_\text{bottom}\left\langle 0,0,R\right\rangle\cdot \left\langle\pdv{z_2}{x},\pdv{z_2}{y},-1 \right\rangle\\ &= \iint_\text{bottom} -R\,dx\,dy \end{align*}

$$R$$ is $$R(x,y,z)$$. That means we can substitute the equation using the graph equation $$z=z_2(x,y)$$. As far as the region is concerned, it is above shadow $$U$$ \newcommand{pdv}[2]{\frac{\partial #1}{\partial #2}} \begin{align} \iint_\rm{bottom} \left\langle 0,0,R\right\rangle\cdot\hat n\,dS &= \iint_U -R(x,y,z_2(x,y))\,dx\,dy \label{eq:divthmproofright2} \\ \end{align}

#### Sides

$$\left\langle 0,0,R\right\rangle$$ is tangent to the sides $$\Longrightarrow$$ the flux through the sides $$=0$$. (that is why we took a vector field with only a $$z$$-component.)

In other words: the vector field $$\left\langle 0,0,R\right\rangle$$ is parallel to the $$z$$-axis, so at the sides there nothing going on.

#### Together

Combining equation $$\eqref{eq:divthmproofright1}$$ and $$\eqref{eq:divthmproofright2}$$ using $$\eqref{eq:combine}$$ \newcommand{oiint}{\subset\!\!\supset\kern-1.65em\iint} \begin{align} \oiint_S\left\langle 0,0,R\right\rangle \cdot \hat n\,dS = \iint_U &R(x,y,z_2(x,y))\,dx\,dy \nonumber \\ + \iint_U &-R(x,y,z_1(x,y))\,dx\,dy \label{eq:together} \end{align}

### Compare left- and right-hand side


Therefore, for a vertically simple region $$\newcommand{oiint}{\subset\!\!\supset\kern-1.65em\iint} \shaded{ \oiint_S\left\langle 0,0,R\right\rangle \cdot \hat n\,dS = \iiint_D R_z\,dV } \nonumber$$

### Generalization

1. If $$D$$ is not vertically simple, we cut it into vertically simple pieces. For instance, a solid doughnut is not vertically simple because above and below the hole makes two regions. However, if we slice it into four pieces, then each piece has a well defined top and bottom side (=vertically simple).
The slices will have vertical sides. Not only does this vector field have have no flux on vertical sides. But also for each slice the normal vectors are opposite, and the flux will cancel out.
2. We could also do the same proof for the $$x$$ and $$y$$-component. Then sum them together to get the general case.

## Example

Redo the earlier example: Find the flux of $$\vec F=\left\langle x,y,z\right\rangle$$ through sphere of radius $$a$$ centered at the origin. The normal vector of the sphere points out radially from the origin.

Apply the divergence theorem $$\eqref{eq:divthm}$$ \newcommand{oiint}{\subset\!\!\supset\kern-1.65em\iint} \begin{align*} \oiint_S\vec z\hat k\cdot d\vec S &= \iiint_D\rm{div}(z\hat k)\,dV \\ &= \iiint_D (0+0+1) \,dV \\ &= \underbrace{\iiint_D dV}_{\text{Volume}(D)} = \frac{4}{3}\pi a^3\\ \end{align*}

## Work (in space)



My notes of the excellent lecture 30 by “Denis Auroux. 18.02 Multivariable Calculus. Fall 2007. Massachusetts Institute of Technology: MIT OpenCourseWare, https://ocw.mit.edu. License: Creative Commons BY-NC-SA.”

Let $$\vec F$$ be a vector field representing a force $$\vec F = P\hat\imath + Q\hat\jmath + R\hat k = \left\langle P, Q, R \right\rangle \nonumber$$

Let $$C$$ be a curve in space, and space vector $$d\vec r$$ with components $$d\vec r=\left\langle dx,dy,dz\right\rangle$$. Then, the work done by the field $$\shaded{ \rm{Work} = \int_C\vec F\cdot d\vec r = \int_C P\,dx+Q\,dy+R\,dz } \nonumber$$

The method is the same as in the plane: evaluate by parameterizing $$C$$, express $$x,y,z,dx,dy,dz$$ in terms of the parameter.

## Examples

Let $$\vec F$$ be a force field. What work does the force exert on the particle in two different trajectories? $$\vec F=\left\langle yz,xz,xy\right\rangle$$

### Trajectory A

Let a particle move over curve $$C$$, in force field $$\vec F$$. What work does the force exert on the particle? $$C: x=t^2, y=t^2, z=t, 0 \le t \le 1$$

Express $$x,y,z$$ in terms of parametric variable $$t$$ \begin{align*} x=t^3 &\Rightarrow dx = 3t^2\;dt \\ y=t^2 & \Rightarrow dy = 2t\;dt \\ z=t & \Rightarrow dz = dt \end{align*}

Substituting $$x,y,z,dx,dy,dz$$ in the integral expression \begin{align} \int_C\vec F\cdot d\vec r &=\int_C yz\;dx+xz\;dy+xy\;dz \label{eq:example3dwork} \\ &=\int_0^1 t^3.3t^2dt+t^4.2t\;dt+t^5dt \nonumber \\ &= \int_0^1 6t^5dt = \left[t^6\right]_0^1=1 \nonumber \end{align}

### Trajectory B

Let a particle move over curve $$C$$ consisting of three line segments, in the same force field $$\vec F=\left\langle yz,xz,xy\right\rangle$$. What work does the force exert on the particle? \begin{align*} C_1&: \mathrm{from\ } (0,0,0) \mathrm{\ to\ }(1,0,0) \\ C_2&: \mathrm{from\ } (1,0,0) \mathrm{\ to\ }(1,1,0) \\ C_3&: \mathrm{from\ } (1,1,0) \mathrm{\ to\ }(1,1,1) \end{align*}

Curves $$C_1,C_2$$ are in the $$xy$$-plane, so $$z=0$$ and therefore $$dz=0$$.

Substituting $$z$$ and $$dz$$ in equation $$\eqref{eq:example3dwork}$$, only leaves \begin{align*} \int_{C_{1,2}}\vec F\cdot d\vec r &=\int_C y\bcancel{z}_0\;dx+x\bcancel{z}_0\;dy+xy\;\bcancel{dz}_0 \\ &=\int_C xz\;dy = 0 \end{align*}

For $$C_3$$: $$\begin{array}{l} x=1 \Rightarrow dx=0 \\ y=1 \Rightarrow dy=0 \\ z \mathrm{\ from\ } 0 \mathrm{\ to\ } 1 \end{array} \nonumber$$

Substituting $$x,y,dx,dy$$ in the line integral along $$C_3$$ \begin{align*} \int_{C_3}\vec F\;d\vec r &=\int_C yz\;dx+xz\;dy+xy\;dz \\ &=\int_C z.0+1.z.0+1.1\;dz \\ &=\int_0^1 dz = \left[z\right]_0^1=1 \end{align*}

Adding the 3 terms together $$\int_C\vec F\;d\vec r= \int_{C_1}\vec F\;d\vec r + \int_{C_2}\vec F\;d\vec r + \int_{C_3}\vec F\;d\vec r = 1 \nonumber$$

### Compare

Trajectory A and B have the same answer, because the vector field is a (conservative) gradient field. Both paths go from the origin to $$(1,1,1)$$.

Find a function, what’s gradient is this vector field. It might be $$xyz$$ \newcommand{pdv}[1]{\tfrac{\partial}{\partial #1}} \begin{align*} \vec F &= \nabla(xyz) \\ &= \left\langle \pdv{x}xyz, \pdv{y}xyz, \pdv{z}xyz \right\rangle \\ &= \left\langle yz, xz, xy \right\rangle \end{align*}

Knowing this, we could have used the fundamental theorem of calculus for line integrals (written using the symbolic $$\nabla$$-operator) $$\shaded{ \int_C\nabla f\cdot d\vec r = f(P_1)-f(P_o) } \nonumber$$

Substituting $$f(x,y,z)=xyz$$ \begin{align*} \int_C\nabla f\cdot d\vec r &= f(1,1,1) – f(0,0,0) \\ &= 1 – 0 = 1 \end{align*}

## Vector fields; surface integrals; flux (in space)



My notes of the excellent lectures 27 and 28 by “Denis Auroux. 18.02 Multivariable Calculus. Fall 2007. Massachusetts Institute of Technology: MIT OpenCourseWare, https://ocw.mit.edu. License: Creative Commons BY-NC-SA.”

We will learn about vector fields in space and determining the surface vector, using flux as an example.

## Vector fields in space

Let $$\vec F$$ be a vector field, where $$P,Q,R$$ are functions of $$x,y,z$$ $$\vec F=\left\langle P,Q,R\right\rangle \nonumber$$

Some (not mutually exclusive) examples of vector fields

• Force fields (e.g. gravitational attraction) of a solid mass at $$(0,0,0)$$ on a mass at $$(x,y,z)$$. $$\vec F$$ directed towards origin with magnitude ~ $$\frac{c}{\rho^2}$$, where $$\rho$$ is the distance to the origin. $$\vec F = -c\frac{\left\langle x,y,z\right\rangle}{\rho^3} \nonumber$$ Here $$\left\langle x,y,z\right\rangle$$ goes from the origin to the point. It magnitude is $$\rho$$. Also Electric fields, magnetic fields, etc.
• Gradient fields, such as $$u=u(x,y,z)$$, then $$\nabla u = \left\langle u_x,u_y,u_z\right\rangle \nonumber$$ E.g. Electric fields and gravitational field are given by the gradient of the electric or gravitational potential.
• Velocity fields (e.g. wind or fluid)

## Surface integrals in a vector field

Remember flux in a 2D plane

In a plane, flux is a measure of how much a vector field is going across the curve. $$\int_C\vec F\cdot\hat n\,ds \nonumber$$
In space, to have a flow through something you need a surface, e.g. a net. $$\Longrightarrow$$ flux will be measured through a surface $$\Longrightarrow$$ surface integral.

With the line integral in the plane, you had two variables, that you reduced to one using a parametric equation for the curve. In space, you have three variables that you will reduce to two variables by figuring out what the surface is.

### Definition of flux in space

Let $$\vec F$$ be a vector field, and $$S$$ a surface in space. Flux measures how much the vector field is going across the surface.

Let $$\hat n$$ be the unit normal vector to a piece of surface $$\Delta S$$.

There are two choices for $$\hat n$$. We need to choose a side of the surface (“orientation”). The usual choose is to take the normal vector pointing out of the region, because then you will be looking at flux that is coming out of that region of space. Or, if you have a surface that is not closed but you will want the flux going up through the region.

Once we have made that choice, we can define the flux integral $$\shaded{ \rm{Flux} = \iint_S\vec F\cdot \hat n\,dS = \iint_S\vec F\cdot d\vec S } \label{ex:flux}$$

Here

• The big $$dS$$ stands for the surface area element.
• Vector $$d\vec S=\hat n\,dS$$ points perpendicular to the surface element. Its length corresponds to the surface element $$dS$$.
• It is often easier to compute $$d\vec S$$, than computing $$\hat n$$ and $$dS$$ seperately.

#### Example

Use geometry, or set up $$\iint_S\vec F\cdot\hat n\,dS$$

##### One

Compute the flux of $$\vec F=\left\langle x,y,z\right\rangle$$ through a sphere of radius $$a$$ centered at the origin. The normal vector of the sphere points out radially from the origin.

Recall: the double integral for flux $$\eqref{ex:flux}$$

$$\iint_S\vec F\cdot d\vec S=\iint_S\vec F\cdot\hat n\,dS \nonumber$$

The normal vector $$\hat n$$ points radially out. Find the magnitude of $$\left\langle x,y,z\right\rangle$$, then scale that down by its magnitude. \begin{align*} |\left\langle x,y,z\right\rangle| &= \sqrt{x^2+y^2+z^2} =a \\ \Rightarrow \hat n &= \frac{\left\langle x,y,z\right\rangle}{a} \end{align*} \nonumber

The normal vector $$\hat n$$ and field $$\vec F$$ are parallel to each other \begin{align*} \vec F\cdot\hat n &= |\vec F|\,\bcancel{|\hat n|}_1 \\ &= \sqrt{x^2+y^2+z^2} = \underline{a} \end{align*} \nonumber

The alternate way is to compute this is \begin{align*} \vec F\cdot\hat n &= \left\langle x,y,z \right\rangle \cdot \frac{1}{a}\left\langle x,y,z \right\rangle \\ &= \frac{x^2+y^2+z^2}{a} \tag{a^2=x^2+y^2+z^2} \\ &= \frac{a^2}{a} = \underline{a} \end{align*} \nonumber

Substitute $$\vec F\cdot\hat n$$ in the double integral for flux $$\eqref{ex:flux}$$ \begin{align*} \iint_S\underline{\vec F\cdot \hat n}\,dS &= \iint_S \underline{a}\,dS \\ &= a \underbrace{\iint_S dS}_{\text{area of }S} = a (4\pi a^2) = 4\pi a^3 \end{align*} \nonumber

##### Two

Calculate the flux of $$\vec H=z\hat k=\left\langle 0,0,z \right\rangle$$ through the same sphere.

The normal vector $$\hat n$$ is the same $$\hat n = \frac{1}{a}\left\langle x,y,z\right\rangle \nonumber$$

Compute the inside of the flux integral $$\iint_S\underline{\vec H\cdot\hat n}\,dS$$ \begin{align*} \vec H\cdot\hat n &= \left\langle 0,0,z \right\rangle \cdot \frac{1}{a}\left\langle x,y,z \right\rangle \\ &= \underline{\frac{z^2}{a}} \end{align*} \nonumber

Substitute this dot-product in the double integral for flux $$\eqref{ex:flux}$$ $$\iint_S\underline{\vec H\cdot \hat n}\,dS = \iint_S \frac{z^2}{a}\,dS \label{eq:ex2int}$$

Express $$dS$$ in our favorite set of two variable used to integrate. In this case, we will use a part of the spherical coordinates, as $$\phi$$ and $$\theta$$ (we don’t use $$\rho$$) $$dS =\ldots\,\, d\phi\, d\theta \nonumber$$

The surface $$\Delta S$$ of this small piece of sphere \begin{align*} \Delta S &\approx (a\sin\phi\,\Delta\theta)(a\,\Delta\phi) \\ \Rightarrow dS &= \underline{a^2\sin\phi\,d\phi\,d\theta} \end{align*}

Express the $$z$$-coordinates on the surface, in terms of $$\phi$$ and $$\theta$$

The expression for $$z$$ on the surface $$z = \underline{a\,\cos\phi} \nonumber$$

Substitute $$dS$$ and $$z$$ in equation $$\eqref{eq:ex2int}$$. For the bounds $$\phi$$ goes from the north pole to the south pole, and $$\theta$$ goes all around. \begin{align*} \iint_S\vec H\cdot \hat n\,dS &= \int_0^{2\pi} \int_0^\pi \frac{a^2\cos^2\phi}{a}(a^2\sin\phi\,d\phi\,d\theta) \\ &= a^3 \int_0^{2\pi} \underline{\int_0^\pi \cos^2\phi\,\sin\phi\,d\phi}\,d\theta \end{align*} \nonumber

For the middle integral, substitute $$u=\cos\phi \Rightarrow du=-sin\phi\,d\phi$$ \begin{align*} \int_0^\pi \underbrace{\cos^2\phi}_{u^2}\,\underbrace{\sin\phi\,d\phi}_{-du} &= -\int_{u=\cos 0}^{\cos\pi}u^2\,du \\ &= -\Big[ \frac{u^3}{3}\Big]_{\cos 0}^{u=\cos\pi} = -\frac{2}{3} \end{align*}

The flux follows from the outer integral \begin{align*} \iint_S\vec H\cdot \hat n\,dS &= a^3\int_0^{2\pi}-\frac{2}{3}\,d\theta \\ &= -\frac{2}{3}a^3\Big[\theta\Big]_0^{\theta=2\pi} = \frac{4}{3}\pi a^3 \end{align*}

## $$\hat n\,dS$$ for various surfaces

### Elementary surfaces

#### Horizontal plane

Let surface $$S$$ be a horizontal plane, $$z=a$$

The normal vector $$\hat n$$ and surface element $$dS$$ on an horizontal plane \shaded{ \begin{align*} \hat n &= \pm\hat k = \left\langle 0,0,\pm1\right\rangle \\ dS &= dx\,dy \end{align*} } \nonumber

When we take $$\vec F\cdot\hat n$$, that will give the $$z$$-component which might involve $$x$$, $$y$$ and $$z$$. It is easy to get rid of the $$z$$ because $$z=a$$. The $$x$$ and $$y$$ will stay as variables. Then integrate that $$dx\,dy$$.

#### Vertical plane

Let surface $$S$$ be a vertical plane $$\newcommand{\parallelsum}{\mathbin{\!/\mkern-5mu/\!}} \parallelsum\,yz$$-plane, $$x=a$$

The variables for my position on the $$yz$$-plane would be $$y$$ and $$z$$. The normal vector $$\hat n$$ and surface element $$dS$$ on the vertical plane \shaded{ \begin{align*} \hat n &= \pm\hat\imath = \left\langle \pm 1,0,0\right\rangle \\ dS &= dy\,dz \end{align*} } \nonumber \shaded{ } \nonumber

#### Sphere

Let the surface $$S$$ be a sphere of radius $$a$$ at the origin

The normal vector $$\hat n$$ of this sphere is radial from the origin. As we saw earlier $$\shaded{ \hat n = \pm\frac{\left\langle x,y,z\right\rangle}{a} } \nonumber$$

The surface element $$dS$$ follows from the width $$\times$$ height of the surface element $$\Delta S$$ (see figure above) $$\shaded{ dS = a^2 \sin\phi\,d\phi\,d\theta } \nonumber$$

It is best to start with the dot product $$\vec F\cdot\hat n$$, because $$\vec F$$ will be in terms of $$x,y,z$$. If there is any kind of symmetry, you might end up with things that are easier to express in terms of $$\phi$$ and $$\theta$$. \shaded{ \left\{ \begin{align*} x &= a\sin\phi\,\cos\theta \\ y &= a\sin\phi\,\sin\theta \\ z &= a\cos\phi \end{align*} \right. } \nonumber

#### Cylinder

Let the surface $$S$$ be a cylinder of radius $$a$$ centered on the $$z$$-axis

The normal vector $$\hat n$$ of this cylinder sticks radially out in the horizontal directions $$\shaded{ \hat n = \pm\frac{\left\langle x,y,0\right\rangle}{a} } \nonumber$$

The surface element $$dS$$ $$\shaded{ dS = a\,dz\,d\theta } \nonumber$$

You would calculate $$\vec F\cdot\hat n$$, then expand $$dS$$ and express the remaining $$x,y$$ in terms of $$dz,d\theta$$ \shaded{ \left\{ \begin{align*} x &= a\cos\theta \\ y &= a\sin\theta \end{align*} \right. } \nonumber

### Graph of a function

Let surface $$S$$ be a graph of $$z=f(x,y)$$

To determine $$dS$$, we approach it similarly to the flux in the plane. Reduce surface $$S$$ to variables $$x$$ and $$y$$, by considering the shadow of surface $$S$$ on the $$xy$$-plane. Then, the point $$(x,y)$$ in the shadow $$R$$ corresponds to $$(x,y,f(x,y))$$ on surface $$S$$.

Take a small rectangle in the shadow $$R$$ on the $$xy$$-plane, that corresponds to $$\Delta x\Delta y$$. This corresponds to $$\Delta S$$ on the surface. If $$\Delta S$$ is small enough, it will roughly look like a parallelogram. Find the two vectors $$\vec u$$ and $$\vec v$$ from this point on $$S$$. The magnitude will be the surface $$\Delta S$$, and the direction will be the normal to the surface $$\hat n$$. $$\pm\,\vec u\times\vec v=\hat n\,\Delta S \nonumber$$

The vector $$\vec u$$ (see close up illustration above) $$\begin{array}{llll} &\vec u: & \text{from} & (x,y,f(x,y)) \\ & & \text{to} & (x+\Delta x, y, \underbrace{f(x+\Delta x,y)}_{\approx f(x,y)+\Delta x.f_x}) \\ \Rightarrow &\vec u &\approx &\left\langle \Delta x,\,0,\,f_x\Delta x \right\rangle \\ &&\approx & \underline{\left\langle 1,\,0,\,f_x\right\rangle\Delta x} \end{array} \nonumber$$

Similarly, the vector $$\vec v$$ $$\begin{array}{llll} &\vec v: & \text{from} & (x,y,f(x,y)) \\ & & \text{to} & (x, y+\Delta y, \underbrace{f(x,y+\Delta y)}_{\approx f(x,y)+\Delta y.f_y}) \\ \Rightarrow &\vec v &\approx &\left\langle 0,\,\Delta y,\,f_y\Delta y \right\rangle \\ &&\approx & \underline{\left\langle 0,\,1,\,f_x\right\rangle\Delta y} \end{array} \nonumber$$

The cross-product of $$\vec u$$ and $$\vec v$$ gives us two in one: the normal vector $$\hat n$$ multiplied with $$\Delta S$$. \begin{align*} \hat n\,\Delta S &\approx \pm\,\vec u\times\vec v \\ &\approx \pm\,\left| \begin{array}{ccc} \hat\imath & \hat\jmath & \hat k \\ 1 & 0 & f_x \\ 0 & 1 & f_y \end{array} \right| \Delta x\Delta y \\ &\approx \pm \left\langle -f_x, -f_y, 1 \right\rangle\Delta x\Delta y \end{align*}

For the limit $$\Delta x,\Delta y\to 0$$, $$d\vec S$$ follows $$\shaded{ d\vec S = \hat n\,dS = \pm\underbrace{\left\langle -f_x, -f_y, 1\right\rangle}_{\text{not }\hat n}\,\underbrace{dx\,dy}_{\text{not }dS} } \label{eq:fluxgraph}$$

If you take the $$+$$, the $$z$$-component is positive, so the normal vector $$\hat n$$ points up.

#### Example

Find the flux of vector field $$\vec F=z\hat k=\left\langle 0,0,z\right\rangle$$ through the portion of paraboloid $$z=x^2+y^2$$ above the unit disk with the normal pointing upwards.

Enter the vector field in the formula for the flux $$\eqref{ex:flux}$$ \begin{align*} \iint_S\vec F\cdot\hat n\,dS &= \iint_S\left\langle 0,0,z\right\rangle\cdot\hat n\,dS \end{align*}

Use equation $$\eqref{eq:fluxgraph}$$ to find $$\hat n\,dS$$ \newcommand{pdv}[2]{\frac{\partial #1}{\partial #2}} \begin{align*} \hat n\,dS &= \left\langle -\pdv{f}{x}, -\pdv{f}{y}, 1 \right\rangle\,dx,\,dy \\ &= \left\langle -2x, -2y, 1 \right\rangle\,dx,\,dy \end{align*}

The double integral fo flux becomes \begin{align*} \iint_S\vec F\cdot\hat n\,dS &= \iint_S\left\langle 0,0,z\right\rangle\cdot\left\langle -2x, -2y, 1 \right\rangle\,dx,\,dy \\ &= \iint_S \underline{z}\,dx\,dy \end{align*}

To get rid of $$z$$, use the expression for the surface $$z=x^2+y^2$$ \begin{align*} \iint_S\vec F\cdot\hat n\,dS &= \iint_S \underline{(x^2+y^2)}\,dx\,dy \end{align*}

The range for $$x$$ and $$y$$ is the shadow of the region (the unit disk). Switch to polar coordinates. For the unit disk: $$0\lt r\lt 1$$ and $$0\lt\theta\lt 2\pi$$ \begin{align*} \iint_S\vec F\cdot\hat n\,dS &= \int_0^{2\pi} \int_0^1 r^2 r\,dr\,d\theta \\ &= \int_0^{2\pi} \Big[\frac{r^4}{4}\Big]_0^{r=1}\,d\theta \\ &= \int_0^{2\pi} \left(\frac{1^4}{4}-0\right)\,d\theta = \frac{\pi}{2}\\ \end{align*}

### Parametric surface

Sometimes the function, for the surface, is so complicated that you can’t express $$z$$ in terms of $$x$$ and $$y$$. In that case, we express $$x,y,z$$ in terms of any two (parametric) variables, because on a surface you can only move in two independent directions. S: \left\{ \begin{align*} x &= x(u,v) \\ y &= y(u,v) \\ z &= z(u,v) \end{align*} \right.

In other words, position vector $$\vec r$$ is a function of $$u$$ and $$v$$ $$\vec r = \left\langle x,y,z\right\rangle = \vec r(u,v) \nonumber$$

We can express $$\hat n\,dS$$ in terms of $$du\,dv$$, similar to what we did before: change by $$\Delta u$$ and $$\Delta v$$ in shadow $$R$$.

Follow a similar approach as with $$\Delta \vec r$$ for line integrals along a curve. The sides follow \newcommand{pdv}[2]{\frac{\partial #1}{\partial #2}} \left\{ \begin{align*} \pdv{\vec r}{u}\,\Delta u &= \left\langle \pdv{x}{u}, \pdv{y}{u}, \pdv{z}{u}\right\rangle \Delta u \\ \pdv{\vec r}{v}\,\Delta v &= \left\langle \pdv{x}{v}, \pdv{y}{v}, \pdv{z}{v}\right\rangle \Delta v \\ \end{align*} \right. \nonumber

To find the surface vector $$d\vec S$$, take the cross-product \newcommand{pdv}[2]{\frac{\partial #1}{\partial #2}} \begin{align*} \hat n\Delta S &\approx \pm\,\left(\pdv{\vec r}{u}\Delta u\right) \times \left(\pdv{\vec r}{v}\Delta v \right) \\ &\approx \pm\,\left(\pdv{\vec r}{u} \times \pdv{\vec r}{v}\right)\Delta u\,\Delta v \\ \Rightarrow d\vec S = \hat n\,dS &= \pm\,\left(\pdv{\vec r}{u} \times \pdv{\vec r}{v}\right)\,du\,dv \\ \end{align*}

### Generic

This will be the generic geometric way to think about $$\hat n\,dS$$.

Say, we know a normal vector $$\vec N$$ (with a capital because not necessary a unit vector) to surface $$S$$. For example

1. You know how a plane is slanted, such as the plane \begin{align*} S &: ax+by+cz=d\\ \Rightarrow \vec N &= \left\langle a,b,c \right\rangle \end{align*}
2. you know an equation in the form $$g(x,y,z)=0$$ $$\Longrightarrow$$ you know the gradient of $$g$$ is perpendicular to the level surface \begin{align*} S &: g(x,y,z)=0\\ \Rightarrow \vec N &= \nabla g \end{align*}

To find the surface vector $$\vec S$$, express the surface in terms of two variables: $$x$$ and $$y$$. To do so, relate $$\Delta S$$ to a projection of $$S$$ on the $$xy$$-plane.

Let $$\alpha$$ be the angle between the plane and the $$xy$$-plane. If you project $$S$$ on the $$xy$$-plane, the bottom and top sides don’t change, but the sides gets shortened by a factor of $$\cos\alpha$$. $$\Delta A = \Delta S \cos\alpha \nonumber$$

To get rid of the $$\cos$$: the angle between two planes is the same as the angle between the normal vectors. This angle $$\alpha$$ is the same angle as between $$\hat k$$ and $$\vec N$$. $$\cos\alpha = \frac{\vec N\cdot\hat k}{|\vec n|.\bcancel{|\hat k|}_1} \nonumber$$

Combining the last two equations $$\Delta S = \frac{1}{\cos\alpha}\Delta A = \frac{|\vec N|}{\vec N\cdot\hat k} \Delta A \nonumber$$

Multiply this by the unit normal vector $$\hat n$$. The term $$\left|\vec N\right|\,\hat n$$ equals $$\vec N$$ again. The only thing we don’t know is the $$\pm$$-direction. \begin{align*} \hat n\,\Delta S &= \frac{\cancel{|\vec N|\,\hat n}^{=\pm\vec N}}{\vec N\cdot\hat k}\Delta A \\ &= \pm\frac{\vec N}{\vec N\cdot\hat k}\Delta A \end{align*}

This brings us to $$\shaded{ \hat n\,dS = \pm\frac{\vec N}{\vec N\cdot\hat k}\,dx\,dy } \nonumber$$

FYI You could also have projected to the $$yz$$-plane, and ended up with $$\hat\imath$$ and $$dy\,dz$$

#### Example

Surface given by the equation $$\underbrace{z-f(x,y)}_{=g(x,y,z)} = 0 \nonumber$$

The normal vector $$\vec N$$ would be \begin{align*} \vec N &= \nabla g = \left\langle -f_x,-f_y,1 \right\rangle \\[1ex] \Rightarrow \frac{\vec N}{\vec N\cdot\hat k}\,dx\,dy &= \frac{\left\langle -f_x,-f_y,1 \right\rangle}{\left\langle -f_x,-f_y,,1 \right\rangle\cdot\left\langle 0,0,1\right\rangle} \\[1ex] &= \left\langle -f_x,-f_y,1 \right\rangle\,dx\,dy \end{align*}

That is matches formula $$\eqref{eq:fluxgraph}$$. This one is actually more general because you don’t need to solve for $$z$$.

## Triple integrals



My notes of the excellent lectures 25 and 26 by “Denis Auroux. 18.02 Multivariable Calculus. Fall 2007. Massachusetts Institute of Technology: MIT OpenCourseWare, https://ocw.mit.edu. License: Creative Commons BY-NC-SA.”

Recall: using a double integral of a you can find the volume between a certain region $$R$$ in the $$xy$$-plane, and a function $$f(x,y)$$.

$$\iint_R f(x,y)\,dA \nonumber$$

Using triple integrals we can find volume between two surfaces. We can even find the mass of a 3D object, when the volume of the region we’re interested in has a variable density.

## Definition

Triple integrals are just like double integrals, but in three dimensions. This extra dimension let’s us express the region $$R$$ in $$xyz$$-space.

$$\shaded{ \iiint_R f(x,y,z)\,dV } \nonumber$$

Here

• The shape of the region $$R$$ is described in the integrals’ bounds.
• When calculating mass, a non-uniform density will be described by function $$f$$ as $$\delta(x,y,z)$$.

Triple integrals are evaluated as three embedded integrals. The bounds of the inner integrals might be functions of the outer variables. $$\int_{z_{min}}^{z_{max}} \underbrace{ \int_{y_{min}(z)}^{y_{max}(z)} \overbrace{ \int_{x_{min}(y,z)}^{x_{max}(y,z)} f(x,y,z)\,dx }^{\text{function of only }y,z} \,dy }_{\text{function of only }z} \,dz \nonumber$$

## In Cartesian coordinates

### Volume element

The volume element for Cartesian coordinates is $$\shaded{ dV=dx\,dy\,dz } \nonumber$$

### Examples

#### One

Find the region between the paraboloids \left\{ \begin{align*} z &= x^2+y^2 \\ z &= 4-x^2-y^2 \end{align*} \right.

To calculate volume, we use $$f=1$$ $$\shaded{ \rm{Volume}=\iiint_R 1\,dV } \nonumber$$

Draw the region

The equations for the paraboloids are given as $$z(x,y)$$, so make $$\int dz$$ the inner integral.

Find the integrals’ bounds:

1. $$\int dz$$: for a fixed point in the $$xy$$-plane, find the bottom and top surface of your solid as $$z(x,y)$$. The $$z$$-bounds follow as $$z_{min}=x^2+y^2$$ and $$z_{max} = 4-x^2-y^2$$.
2. $$\int dy$$: reduce the equation to two variables by looking at the shadow that the solid casts on the $$xy$$-plane.

Find the radius of the shadow disk in the $$xy$$-plane \begin{align*} z_\text{bottom} &\lt z_\text{top} \\ \Rightarrow x^2+y^2 &\lt 4-x^2-y^2 \\ \Rightarrow 2x^2+2y^2 &\lt 4\\ \Rightarrow x^2+y^2 &\lt 2 \\ \Rightarrow \text{disk }&\text{of radius }\sqrt{2} \end{align*} Consider all $$(x,y)$$ in the shadow. For a given $$x$$, expressing the bounds of $$y$$ gives $$-\sqrt{2-x^2}\lt y\lt \sqrt{2-x^2}$$
3. $$\int dx$$: Consider all $$x$$ in the shadow. The $$x$$-bounds follow as $$-\sqrt{2}\lt x\lt\sqrt{2}$$

Fill in the bounds of the integral $$\int_{-\sqrt{2}}^{\sqrt{2}} \int_{-\sqrt{2-x^2}}^{\sqrt{2-x^2}} \int_{x^2+y^2}^{4-x^2-y^2}\,dz \,dy\,dx \label{eq:ex1a}$$

Evaluate the inner integral \begin{align*} \int_{x^2+y^2}^{4-x^2-y^2}\,dz &= \Big[z\Big]_{z=x^2+y^2}^{4-x^2-y^2} \\ &= 4-x^2-y^2-(x^2+y^2) \\ &= \underline{4-2x^2-2y^2} \end{align*} \nonumber

Evaluate the remaining integral \begin{align*} &\int_{-\sqrt{2}}^{\sqrt{2}}\int_{-\sqrt{2-x^2}}^{\sqrt{2-x^2}} \underline{4-2x^2-2y^2}\,dy\,dx \end{align*} \nonumber

To evaluate this, switch to cylindrical coordinates as described in the next section.

## In cylindrical coordinates

In cylindrical coordinates

• $$r$$ is the distance from the $$z$$-axis
• $$\theta$$ is the angle with the positive $$x$$-axis counterclockwise.
• $$z$$ is the height above the $$xy$$-plane

Some expressions

• $$r=a$$ is a surface in space: a cylinder.
• setting a value for $$\theta$$ will be a vertical half plane starting at the $$z$$-axis (because $$r$$ by convention is always positive)

To convert to $$xyz$$-coordinates \shaded{ (r,\theta,z) \Rightarrow \left\{ \begin{align*} x &= r\,\cos\theta \\ y &= r\,\sin\theta \\ z &= z \end{align*} \right. } \nonumber

### Volume element

A small piece of surface area $$\Delta S$$ at radius $$r$$

The horizontal side of $$\Delta S$$ is $$\Delta r$$. The vertical side is a piece of circle with angle $$\Delta\theta$$ and radius $$r$$ $$\Longrightarrow$$ the length is $$r.\Delta\theta$$. $$\Delta S \approx \Delta r.(r.\Delta\theta) = r\,\Delta r\,\Delta\theta \nonumber$$

With thickness $$\Delta z$$, this small piece has volume $$\Delta V \approx r\,\Delta r\,\Delta\theta\,\Delta z \nonumber$$

The volume element for cylindrical coordinates follows as $$\shaded{ dV = r\,dr\,d\theta\,dz } \nonumber$$

### Examples

#### One

Continue equation $$\eqref{eq:ex1a}$$ from the previous example with the shadow disk $$x^2+y^2\lt 2$$. Keep the inner integral, but switch to the outside integrals to polar coordinates.

The bounds change as

• The $$\int dz$$-bounds change based on $$x^2+y^2=r^2 \Longrightarrow$$ the upper bound $$4-x^2-y^2=-4-r^2$$.
• The $$\int dr$$ bounds cover the radius of the circle $$0\lt r\lt\sqrt{2}$$.
• The $$\int d\theta$$ bounds cover the angles of the circle $$0\lt \theta\lt 2\pi$$.

The tripple integral in cylindrical coordinates becomes $$\int_0^{2\pi}\int_0^{\sqrt{2}}\int_{r^2}^{4-r^2} \,dz\,r\,dr\,d\theta \nonumber$$ Now, the evaluation is much easier.

## Applications

### Find the mass of a solid

A mass with density $$\delta=\frac{\Delta m}{\Delta v}$$. The mass element $$dm = \delta\,dV \nonumber$$ The mass of a solid $$\shaded{ \rm{mass}=\iiint_R\delta\,dV } \nonumber$$

### Find the average value

The average value of $$f(x,y,z)$$ in $$R$$ $$\shaded{ \bar f=\frac{1}{\rm{Volume}(R)}\iiint_R f\,dV } \nonumber$$

or, the weighted average $$\shaded{ \bar f=\frac{1}{\rm{mass}}\iiint_R f\,\delta\,dV } \nonumber$$

or, center of mass $$(\bar x, \bar y, \bar z)$$, where \shaded{ \begin{align*} \bar x &= \frac{1}{\rm{mass}}\iiint_R x\,\delta\,dV \\ \bar y &= \frac{1}{\rm{mass}}\iiint_R y\,\delta\,dV \\ \bar z &= \frac{1}{\rm{mass}}\iiint_R z\,\delta\,dV \end{align*} } \nonumber

### Find the moment of inertia

The moment of inertia of a solid in respect to an axis $$\iiint_R (\text{distance to axis})^2\,\delta\,dV \nonumber$$

The moment of inertia around the $$x,y,z$$-axis is \shaded{ \begin{align*} I_x &= \iiint_R (y^2+z^2)\,\delta\,dV \\ I_y &= \iiint_R (x^2+z^2)\,\delta\,dV \\ I_z &= \iiint_R (x^2+y^2)\,\delta\,dV \end{align*} } \nonumber Note: rotating around the $$z$$-axis corresponds to rotation around the origin in the plane that we saw in double integrals. The distance to the $$z$$-axis is $$r$$.

### Examples

#### One

Find $$I_z$$ of the solid cone between \left\{ \begin{align*} z &= ar \\ z &= b \end{align*} \right.

The equation $$z=ar$$ implies that $$z$$ is proportional with $$r$$, so a line with slope $$a$$ around the $$z$$-axis. The equation $$z=b$$ represents a plane parallel to the $$xy$$-plane.

Use the equation for the “moment of initia” $$I_z$$ with $$\delta=1$$.

$$I_z = \iiint_R (x^2+y^2)\,dV \nonumber$$

Use cylindrical coordinates, so $$dV=\underline{r\,dr\,d\theta\,dz}$$. \begin{align*} I_z &= \iiint_R (x^2+y^2)\,\underline{r\,dr\,d\theta\,dz} \end{align*}

For the bounds:

• For $$\int dr$$: with $$z$$ constant, we’re slicing the cone in disks of radius $$\frac{z}{a}$$.
• For $$\int d\theta$$, the bounds cover the full circle $$0\lt\theta\lt 2\pi$$.
• For $$\int dz$$ the bounds are limited by the bottom of the cone and the plane at $$z=b$$

Place the bounds on the triple integral and compute \begin{align*} I_z &= \int_0^b \int_0^{2\pi} \int_0^{z/a} r^2\,r\,dr\,d\theta\,dz \\ &= \int_0^b \int_0^{2\pi} \Big[\frac{r^4}{4}\Big]_0^{z/a} \,d\theta\,dz = \frac{1}{4a^4}\int_0^b \int_0^{2\pi}z^4\,d\theta\,dz \\ &= \frac{1}{4a^4}\int_0^b \Big[z^4\theta\Big]_0^{2\pi}\,dz = \frac{2\pi}{4a^4}\int_0^b z^4\,dz \\ &= \frac{2\pi}{4a^4}\Big[\frac{z^5}{5}\Big]_0^b = \frac{\pi}{2a^4}\frac{b^5}{5} = \frac{\pi b^5}{10a^4} \end{align*}

#### Two

Setup a triple integral for the volume inside a unit sphere centered at the origin, and above the plane $$z\gt 1-y$$. $$\left\{ \begin{array}{c} x^2 + y^2 + z^2 \lt 1 \\ z \gt 1-y \end{array} \right. \nonumber$$

The plane $$z\gt 1-y$$, is independent of $$x$$, therefore parallel to the $$x$$-axis. At $$y=0$$, the $$z=1$$. The slope is $$1$$.

The equations for the surfaces are given as z(x,y), so make $$\int dz$$ the inner integral. $$\iint\int_{1+y}^{\sqrt{1-x^2+y^2}} dz\,dx\,dy \nonumber$$

Projected on the on the $$yz$$- and $$xy$$-plane

For the bounds:

• For $$\int dz$$: for a given $$(x,y)$$, $$z_{min}$$ is set by the plane $$z=1+y$$, and $$z_{max}$$ is determined by the sphere $$z=\sqrt{1-x^2+y^2}$$.
• For $$\int dx$$: looked from above, in the $$xy$$-plane, the shadow of the solid is an ellipse. For a given $$y$$, the bounds $$x_{min}$$ and $$x_{max}$$ are when the plane intersects the sphere. \require{cancel} \begin{align*} 1-y &= \sqrt{1-x^2-y^2} \\ \Rightarrow (1-y)^2 &= 1-x^2-y^2 \\ \Rightarrow \bcancel{1}-2y+y^2 &= \bcancel{1}-x^2-y^2 \\ \Rightarrow x^2 &= 2y+2y^2 \\ \Rightarrow x &= \pm\sqrt{2y+2y^2} \end{align*} \nonumber
• For $$\int dy$$, the bounds are $$0$$ and $$1$$.

Place the bounds on the triple integral $$\int_0^1 \int_{-\sqrt{2y+2y^2}}^{\sqrt{2y+2y^2}} \int_{1-y}^{\sqrt{1-x^2-y^2}} dz\,dx\,dy \nonumber$$

## In spherical Coordinates

We use spherical coordinates where there is a lot of symmetry. Best when the solid is centered on the $$z$$-axis.

For spherical coordinates: imagine a vertical half plane through the origin and a point, the $$rz$$-plane.

Spherical coordinates are like “polar coordinates” in this $$rz$$-plane

• $$\rho$$ (rho) is the distance from the origin.
• $$\phi$$ (phi) is the angle form the positive $$z$$-axis, $$0\leq\phi\leq\pi$$.
• $$\theta$$ (theta) is the angle counterclockwise from the positive $$x$$-axis.

Converting spherical $$(\rho,\phi,\theta)$$ $$\to$$ cylindrical coordinates $$(r,\theta,z)$$ \shaded{ (\rho,\phi,\theta) \Rightarrow \left\{ \begin{align*} r &= \rho\sin\phi \\ \theta &= \theta \\ z &= \rho\cos\phi \end{align*} \right. } \nonumber Converting spherical $$(\rho,\phi,\theta)$$ $$\to$$ Cartesian coordinates $$(x,y,z)$$ \shaded{ (\rho,\phi,\theta) \Rightarrow \left\{ \begin{align*} x &= r\cos\theta = \rho\sin\phi\cos\theta \\ y &= r\sin\theta = \rho\sin\phi\sin\theta \\ z &= \rho\cos\phi \end{align*} \right. } \nonumber Converting Cartesian $$(x,y,z)$$ $$\to$$ spherical coordinates $$(\rho,\phi,\theta)$$ \shaded{ (x,y,z) \Rightarrow \left\{ \begin{align*} \rho &= \sqrt{r^2+z^2} = \sqrt{x^2+y^2+z^2} \\ \phi &= \rm{atan}\,\frac{\sqrt{x^2+y^2}}{z} \\ \theta &= \rm{atan}\,\frac{y}{x}\\ \end{align*} \right. } \nonumber

Some expressions

• $$\rho=a$$: a sphere of radius $$a$$, centered at the origin.
• $$\phi=\frac{\pi}{4}$$: a cone. In cylindrical that would $$z=r$$.
• $$\phi=\frac{\pi}{2}$$: the $$xy$$-pane.

### Volume element

To find the volume element $$\Delta V\approx\Delta\rho\,\Delta\phi\,\Delta\theta$$, start with the surface area on a sphere with radius $$a$$. If it is small enough it looks like a rectangle. So, what are the sides?

Start with a surface area element $$\Delta S$$

• To move north-south: you always have to traverse half a circle $$\Longrightarrow$$ the vertical side of $$\Delta S$$ is a piece of circle with radius $$a$$ $$\Rightarrow$$ the length is $$a.\Delta\phi$$.
• To move east-west: your distance depends on your distance to the $$z$$-axis, called $$r$$ $$\Longrightarrow$$ the horizontal side of $$\Delta S$$ is a piece of circle of radius $$r=a\sin\phi$$. That makes its length $$a.\sin\phi.\Delta\theta$$.

The surface $$\Delta S$$ of this small piece of sphere follows \begin{align*} \Delta S &\approx (a.\sin\phi.\Delta\theta) . (a.\Delta\phi) \\ &\approx a^2\,\sin\phi\,\Delta\phi\,\Delta\theta \end{align*} \nonumber

Moving this area from the surface at radius $$a$$ to a distance $$\rho$$ from the origin $$\Delta S’ \approx \rho^2\,\sin\phi\,\Delta\phi\,\Delta\theta \nonumber$$

Give area $$\Delta S’$$ a thickness $$\Delta\rho$$

This small piece has volume $$\Delta V$$ \begin{align*} \Delta V &\approx \Delta\rho.\Delta S’ \\ &\approx \Delta\rho\,(\rho^2\,\sin\phi.\Delta\phi.\Delta\theta) \end{align*} \nonumber

The volume element for spherical coordinates follows as $$\shaded{ dV = \rho^2\,\sin\phi\,\,d\rho\,\,d\phi\,\,d\theta } \nonumber$$

### Examples

#### One

Redo a previous example using spherical coordinates: Setup a triple integral for the volume inside $$\left\{ \begin{array}{c} z \gt 1-y \\ x^2 + y^2 + z^2 \lt 1 \end{array} \right. \nonumber$$

To make it easier, we rotate it so that the plane is horizontal, and there is symmetry around the $$z$$-axis $$\left\{ \begin{array}{c} z \gt \frac{1}{\sqrt{2}} \\ x^2 + y^2 + z^2 &\lt 1 \end{array} \right. \nonumber$$

Picture

Setup triple integral for volume in spherical coordinates $$\iiint_R 1\overbrace{\,\rho^2\,\sin\phi\,d\rho\,d\phi\,d\theta}^{dV} \nonumber$$

For the bounds:

• For $$\int d\rho$$: angles $$\phi$$ and $$\theta$$ are fixed, and $$\rho$$ is the integration variable. It is like shooting a ray from the origin, and seeing where it enters and leaves the solid. We exit the solid cap, when $$\rho=1$$. We enter it when it hits the plane at $$z =\frac{1}{\sqrt{2}}$$. In spherical coordinates that is \begin{align*} z &=\frac{1}{\sqrt{2}} & (z =\rho\cos\phi)\\ \Rightarrow \rho\cos\phi &= \frac{1}{\sqrt{2}} \\ \Rightarrow \rho &= \frac{1}{\sqrt{2}\cos\phi} \end{align*}
• For $$\int d\phi$$: sketch the $$rz$$-plane
The minimum is just the north pole at $$\phi=0$$. The maximum is where it hits the boundary between the sphere and the plane. \begin{align*} \left. \begin{array}{lc} \rm{sphere:} & \rho = 1 \\ \rm{plane:} & \rho\cos\phi = \frac{1}{\sqrt{2}} \end{array} \right\} \Rightarrow \cos\phi &= \frac{1}{\sqrt{2}} \\ \Rightarrow \phi &= \frac{\pi}{4} \end{align*}
• For $$\int d\theta$$: angle $$\theta$$ goes all around $$\Longrightarrow$$ $$0\lt\theta\lt 2\pi$$

Place the bounds on the triple integral $$\int_0^{2\pi} \int_0^{\frac{\pi}{4}} \int_{\sec\phi/\sqrt{2}}^1 \,\rho^2\,\sin\phi\,d\rho\,d\phi\,d\theta \nonumber$$

This is quite evil, but eventually evaluates to $$\rm{Volume} = \frac{2\pi}{3}-\frac{5\pi}{6\sqrt{2}} \nonumber$$

### Gravitational attraction

#### In Cartesian coordinates

Express the gravitational force exerted by solid $$M$$ at $$(x,y,z)$$ on a mass $$m$$ at the origin.

With $$G$$ being the gravitational constant, physics tells us

$$|\vec F| = \frac{G.m_1.m_2}{\rm{distance}^2} \nonumber$$

For a tiny mass $$\Delta M$$, substitute \left\{ \begin{align*} m_1 &= m \\ m_2 &\approx \Delta M \\ \rm{distance} &= \rho \end{align*} \right.

Combine this with the direction of this force (unit vector) \left. \begin{align*} |\vec {F_\Delta}| &\approx \frac{G.m.\Delta M}{\rho^2} \\ \text{dir}\left(\vec F_\Delta\right) &= \frac{\left\langle x,y,z \right\rangle}{\rho} \nonumber \end{align*} \right\} \Rightarrow \vec F_\Delta \approx \frac{G.m}{\rho^3}\left\langle x,y,z \right\rangle\,\Delta M

Let the density be $$\delta$$, and substitute $$\Delta M=\delta\,\Delta V$$ $$\vec F_\Delta \approx Gm\frac{\left\langle x,y,z \right\rangle}{\rho^3}\,\delta\,\Delta V \nonumber$$

Sum all $$\Delta V$$, and take the limit for $$\Delta V\to 0$$ gives the gravitational force $$\shaded{ \vec F = Gm\iiint \frac{\left\langle x,y,z \right\rangle}{\rho^3}\,\delta\,dV } \nonumber$$ Here integrating a vector quantity, just means: integrate component by component to get each component of the force. E.g. for the force in direction of the $$x$$-axis $$F_{\color{red}x}=Gm\iiint \frac{\color{red}x}{\rho^3}\,\delta\,dV \nonumber$$

The $$\rho^3$$ wrecks havoc with rectangular coordinates, and it will cause a factor like $$(x^2+y^2+z^2)^{3/2}$$. On the other hand, with spherical coordinates it might just disappear as we will see in the following section.

#### In spherical coordinates

Redo the previous example in spherical coordinates: Place the solid so that $$z$$-axis is an axis of symmetry.

Because of the symmetric in the $$z$$-axis, the force $$\vec F$$ will only have a $$z$$-component \left\{ \begin{align*} \vec F &= \left\langle 0,0,F_z\right\rangle \\ F_z &= Gm\iiint \frac{z}{\rho^3}\delta\,dV \end{align*} \right.

Using spherical coordinates, $$z=\rho\cos\phi$$ and $$dV=\rho^2\,\sin\phi\,\,d\rho\,\,d\phi\,\,d\theta$$ \begin{align*} F_z &= Gm\iiint \frac{\bcancel{\rho}\cos\phi}{\bcancel{\rho^3}} \delta\,\underbrace{\bcancel{\rho^2}\,\sin\phi\,\,d\rho\,\,d\phi\,\,d\theta}_{dV} \\ &= Gm\iiint \delta\,\cos\phi\,\sin\phi\,\,d\rho\,\,d\phi\,\,d\theta \end{align*} \nonumber

### Proof Newton’s theorem

Continuing from the gravitational attration example: proof Newton’s theorem.

The gravitational attraction of a spherical planet with uniform density is equal to that of a point mass (with the same total mass) at its center.

Setup: the sphere has radius $$a$$ and is centered on the positive $$z$$-axis, tangent to the $$xy$$-plane at the origin

Let $$M$$ be the mass of the planet, and $$m$$ be the test mass at the origin $$F_z = Gm\iiint \delta\,\cos\phi\,\sin\phi\,\,d\rho\,\,d\phi\,\,d\theta \nonumber$$

For the bounds

• For the inner integral $$\int d\rho$$, sketch a slice in the $$rz$$-plane to find the upper bound for $$\rho$$ using the right triangle, $$\rho_{max}=2a\cos\phi$$
• For the middle integral $$\int d\phi$$, the bounds are $$0$$ to $$\pi/2$$.
• For the outer integral $$\int d\theta$$, the bounds are $$0$$ to $$2\pi$$.

Enter the bounds and evaluate $$F_z = Gm\int_0^{2\pi} \int_0^{\pi/2} \underline{\int_0^{2a\cos\phi} \delta\,\cos\phi\,\sin\phi\,\,d\rho}\,\,d\phi\,\,d\theta \nonumber$$

The inner integral \begin{align*} \int_0^{2a\cos\phi} \delta\,\cos\phi\,\sin\phi\,\,d\rho &= \delta\cos\phi\,\sin\phi\,\Big[\rho\Big]_{\rho=0}^{2a\cos\phi} \\ &= \delta\cos\phi\sin\phi\,2a\cos\phi \\ &= \underline{2a\delta\cos^2\phi\sin\phi} \end{align*}

For the middle integral, substitute $$u=\cos\phi \Rightarrow du=-\sin\phi\,d\phi$$ \begin{align*} \int_0^{\pi/2}\underline{2a\delta\,\cos^2\phi\,\sin\phi}\,\,d\phi &= \int_0^{\pi/2}-2a\,\delta\,\overbrace{\cos^2\phi}^{u^2}\,(\overbrace{-\sin\phi\,\,d\phi}^{du}) \\ &= -2a\delta \int u^2\,du = -2a\delta \Big[\frac{u^3}{3}\Big]_{\cos0}^{u=\cos\pi/2} \\ &= -2a\delta\left(0-\frac{1}{3}\right) = \underline{\frac{2}{3}a\delta} \end{align*}

The outer integral \begin{align*} F_z &= Gm\int_0^{2\pi}\underline{\frac{2}{3}a\delta}\,d\theta = Gm\,\frac{2}{3}a\delta\Big[\phi\Big]_{0}^{\phi=2\pi}\\ &=\frac{2}{3}a\delta\,Gm\,2\pi = \underline{\frac{4}{3}\pi a\delta}\,Gm \end{align*}

The volume of a spherical planet with radius $$a$$ is $$\frac{4}{3}\pi a^3$$ $$\Longrightarrow$$ the mass of this planet with uniform density $$\delta$$ $$=$$ volume $$\times$$ density. \begin{align*} M &= \frac{4}{3}\pi a^3 \delta \Longleftrightarrow \underline{\frac{4}{3}\pi a\delta=\frac{M}{a^2}} \end{align*} \nonumber

Substituting $$\frac{M}{a^2}$$ in the outer integral proves Newton’s theorem $$\shaded{ F_z = \frac{GmM}{a^2} } \nonumber$$

## Flux; Green’s (in plane)



My notes of the excellent lecture 23 of “Denis Auroux. 18.02 Multivariable Calculus. Fall 2007. Massachusetts Institute of Technology: MIT OpenCourseWare, https://ocw.mit.edu. License: Creative Commons BY-NC-SA.”

Flux is the amount of something (water, wind, electric field, magnetic field) passing through a surface.

## Flux in a plane

Let $$C$$ be a plane curve, and $$\vec F$$ be a vector field in that plane.

The flux of $$\vec F$$ across $$C$$ is defined as $$\shaded{ \rm{flux}=\int_C\vec F\cdot\hat n\;ds } \nonumber$$

Where $$\hat n$$ is the unit normal vector to curve $$C$$ pointing 90° clockwise from $$\hat T$$. So, it is to the right of the curve, as you travel along the curve.

### Compared to work

Flux compared to the line integral for work

Flux versus line integral for work
Flux Work
$$\lim_{\Delta s\to 0}\sum\vec F\cdot\hat n\;\Delta s \nonumber$$ $$\lim_{\Delta s\to 0}\sum\vec F\cdot\hat T\;\Delta s \nonumber$$
Sums the tangent component of $$\vec F$$. Sums the normal component of $$\vec F$$.
Measures how much the curve goes with $$\vec F$$. Measures how much $$\vec F$$ goes across the curve.
$$\vec F$$ is a velocity field. $$\vec F$$ is a force field.

### Physics Interpretation

Let $$\vec F$$ be a velocity field of water, and $$C$$ being a curve, then flux measures how much fluid passes through $$C$$ per unit of time.

If you imagine that you have a river and you are building a damn with holes, then flux measures how much water passes through your membrane per unit time.

Zoomed in enough, the flow will be the same everywhere. In respect to the flow, the curve $$C$$ moves left. The parallelogram represents the fluid that passes through $$C$$.

The size of the parallelogram is base $$\times$$ height. The height is the normal component of $$\vec F$$ \begin{align*} \rm{area} &= \rm{base}\;.\rm{height} \\ &= \Delta s\,(\vec F\cdot\hat n) \nonumber \end{align*}

The flux is the summation \begin{align*} \rm{flux} &= \lim_{\Delta s\to 0}\sum\vec F\cdot\hat n\;\Delta s \\ \Rightarrow \rm{flux} &= \int_S \vec F\cdot\hat n\,ds \end{align*}

Note

• what flows across a small segment of $$C$$ from left-to-right is counted positively, while
• what flows left-to-right is counted negatively.

### Examples using geometry

#### One

\begin{align*} C &: \rm{circle\ of\ radius\ } a {\ at\ }(0,0),\rm{\ counterclockwise} \\ \vec F &= x\hat\imath+y\hat\jmath \end{align*}

Along $$C$$ \require{cancel} \newcommand{\parallelsum}{\mathbin{\!/\mkern-5mu/\!}} \begin{align*} \vec F\;&\parallelsum\;\hat n \\ \Rightarrow\ \vec F\cdot\hat n &=\left|\vec F\right|\;\bcancel{\left|\hat n\right|}_1=\left|\vec F\right| \\ &=\sqrt{x^2+y^2}=a \\ \Rightarrow \int_C\vec F\cdot\hat n\;ds &=\int_C a\;ds =a\;.\rm{length}(C)=2\pi a^2 \end{align*} There would be water added uniformly. Because as you go from the origin, it is going faster and faster.

#### Two

\begin{align*} C &: \rm{circle\ of\ radius\ } a {\ at\ }(0,0),\rm{\ counterclockwise} \\ \vec F &= \left\langle -y,x\right\rangle \end{align*}

When $$\vec F$$ is tangent to $$C$$ \require{cancel} \begin{align*} \vec F&\perp\hat n \\ \Rightarrow\ \vec F\cdot\hat n &=0 \\ \Rightarrow \int_C\vec F\cdot\hat n\;ds &=\int_C 0\;ds = 0 \end{align*}

### Compute using components

Let $$\vec F$$ be a vector field with components $$M(x,y)$$, and $$N(x,y)$$, and let $$\Delta s$$ be a tiny length of curve $$C$$.

Remember: for work we sum how much the vector field goes along the curve $$\Longrightarrow$$ it uses the dot-product between $$\vec F$$ and the tangent vector of the curve $$\hat T$$

$$\left. \begin{array}{l} d\vec r=\underline{\hat T\;ds} = \left\langle dx ,dy\right\rangle \\ \vec F = \left\langle M, N\right\rangle \end{array} \right\} \Rightarrow \vec F\cdot \hat T\,ds = Mdx+Ndy \nonumber$$

For flux, we sum how much the vector field goes across the curve (=along the normal vector) $$\Longrightarrow$$ the dot-product between $$\vec F$$ and the normal vector of the curve $$\hat n$$. Where $$\hat n$$ equals the tangent vector $$\hat T$$ rotated by 90° clockwise.

Rotating $$\hat T\,ds=\left\langle dx ,dy\right\rangle$$ by 90&deg clockwise, makes the $$x$$-component $$y$$, and the $$y$$-component $$-x$$ $$\left.\begin{array}{l} \underline{\hat n\;ds}=\left\langle dy,-dx \right\rangle \\ \vec F=\left\langle M,N\right\rangle \end{array}\right\} \Rightarrow \vec F\cdot\hat n\;ds = -N\,dx + M\,dy \nonumber$$

The flux integral can then be written as the differential $$\shaded{ \int_C \vec F\cdot\hat n\;ds = \int_C -N\,dx + M\,dy } \nonumber$$

### Green’s theorem in normal form

For work, Green’s theorem, replaces a line integral along a closed curve with a double integral.

If $$C$$ is a closed curve, enclosing region $$R$$, counterclockwise, and vector field $$\vec F$$ is defined and differentiable everywhere in $$R$$, then $$\shaded{ \oint_C\vec F\cdot d\vec r = \iint_R\mathrm{curl}(\vec F)\;dA } \nonumber$$

For flux, it is similar: Green’s theorem for flux

If $$C$$ is a closed curve, enclosing region $$R$$, counterclockwise, and vector field $$\vec F$$ is defined and differentiable everywhere in $$R$$, then $$\oint_C\vec F\cdot\hat n\,ds = \iint_R\mathrm{div}(\vec F)\,dA \nonumber$$

Recall: divergence for vector field $$\vec F=\left\langle P,Q\right\rangle$$, where $$P(x,y)$$ and $$Q(x,y)$$

$$\newcommand{pdv}[2]{\frac{\partial #1}{\partial #2}} \rm{div}\left\langle P,Q\right\rangle = \pdv{}{x}P + \pdv{}{y}Q = P_x + Q_y \nonumber$$

Green’s theorem in coordinates $$\shaded{ \oint_C -Q\,dx + P\,dy = \iint_R(P_x +Q _y)\;dA } \label{eq:greens}$$

Interpretation of divergence:

1. Measures how much the flow is “expanding”. E.g. how much a gas expands.
2. “Source rate”. Amount of fluid added to the system per unit time and area.

#### Proof

In equation $$\eqref{eq:greens}$$, substitute $$M=-Q$$ and $$N=P$$ $$\oint_C \underbrace{-Q}_{M}\,dx + \underbrace{P}_{N}\,dy = \iint_R(\underbrace{P_x}_{N} +\underbrace{Q _y}_{-M_y})\;dA \nonumber$$

$$\oint_C M\,dx+N\,dy = \iint_R (N_x-M_y)\,dA \nonumber$$

By renaming the components, we go from tangential form to the normal form. Therefore it is really the same theorem.

#### Examples

##### One

Look at the earlier example \begin{align*} C&:\rm{circle\ of\ radius\ } a {\ at\ }(0,0),\rm{\ counterclockwise} \\ \vec F&= x\hat\imath+y\hat\jmath = \left\langle x,y\right\rangle \end{align*}

Picture

Start with $$\rm{div}(\vec F)$$ $$\newcommand{pdv}[2]{\frac{\partial #1}{\partial #2}} \rm{div}(\vec F) = \pdv{}{x}x+\pdv{}{y}y=1+1=\underline{2} \nonumber$$

Apply Green’s theorem \begin{align*} \oint_C\vec F\cdot\hat n\,dx &= \iint_R\underline{2}\,dA = 2\overbrace{\iint_R\,dA}^{\text{Area}(R)} \\ &= 2\pi a^2 \end{align*} \nonumber

#### Two

Determine the flux for \begin{align*} C&:\rm{circle\ of\ radius\ } a {\ NOT\ at\ }(0,0),\rm{\ counterclockwise} \\ \vec F&= x\hat\imath+y\hat\jmath \end{align*} This will be the same, because we never used the location of the circle. We only used its area. The flux is the same everywhere.

## Curl; Green’s (in plane)



My notes of the excellent lectures 22, 27 and 33 by “Denis Auroux. 18.02 Multivariable Calculus. Fall 2007. Massachusetts Institute of Technology: MIT OpenCourseWare, https://ocw.mit.edu. License: Creative Commons BY-NC-SA.”

For a velocity field, curl measures the rotation component of the motion.

Curl also measures how far the vector field is from being conservative.

## Definition

Let $$\vec F=\left\langle M,N\right\rangle$$ be a vector field with components $$M(x,y)$$, and $$N(x,y)$$.

Curl is defined as (written using the symbolic $$\nabla$$-operator) $$\newcommand{pdv}[2]{\frac{\partial #1}{\partial #2}} \shaded{ \mathrm{curl}\left(\vec F\right) = \nabla\times\vec F = \pdv{N}{x}-\pdv{M}{y} = N_x-M_y } \label{eq:curldef}$$

It measures how far the vector field is from being conservative.

A conservative vector field is a vector field that is the gradient of some function $$f(x,y)$$. The line integral of a conservative vector field is path independent; the choice of any path between two points does not change the value of the line integral.

Recall: a conservative vector field

If $$\vec F$$ is defined and differentiable everywhere, and $$M_y=M_x$$, then $$\vec F$$ is a gradient field, a conservative vector field.

### Some expressions

1. If the curl is $$0$$ and defined everywhere, then the field is conservative $$\Longrightarrow$$ a line integral on a closed curve will be $$0$$. $$\newcommand{pdv}[2]{\frac{\partial #1}{\partial #2}} \oint_C\vec F\cdot d\vec r = \iint_R\underbrace{\mathrm{curl}(\vec F)}_{=0}\;dA = 0 \nonumber$$
2. For a gradient field, $$N_x=M_y \Longrightarrow\mathrm{curl}(\vec F)=0$$. $$\newcommand{pdv}[2]{\frac{\partial #1}{\partial #2}} \rm{curl}\left(\vec F\right) = \pdv{N}{x}-\pdv{M}{y} = 0 \nonumber$$
3. For $$\vec F=\left\langle a,b\right\rangle$$, where $$a$$ and $$b$$ are constants $$\newcommand{pdv}[2]{\frac{\partial #1}{\partial #2}} \rm{curl}\left(\vec F\right) = \pdv{b}{x}-\pdv{a}{y} = 0 \nonumber$$
4. In a radial vector field, $$\vec F=\left\langle x,y\right\rangle$$, there is no rotation $$\newcommand{pdv}[2]{\frac{\partial #1}{\partial #2}} \mathrm{curl}\left(\vec F\right)=\pdv{}{x}y-\pdv{}{y}x=0 \nonumber$$
5. In a rotation vector field $$\vec F=\left\langle -y,x\right\rangle$$, $$\newcommand{pdv}[2]{\frac{\partial #1}{\partial #2}} \mathrm{curl}\left(\vec F\right)=\pdv{}{x}x-\pdv{}{y}(-y)=1+1=2 \nonumber$$

## Topological considerations (Lec. 33)

### Curl for a velocity field

In a velocity field, curl measures (twice) the angular velocity ($$\omega$$) of the rotation component of the field. If you would put something light in a fluid, the curl tells you how fast it will spin at a given time ($$\times 2$$).

The curl measures how strong the rotation effect of a velocity field is at a given point. For instance, if you’re looking at wind vector fields, the regions with high curl trend to be tornedos or hurricanes.

Curl singles out the rotation component of motion (while divergence singles out the stretching component).

### Curl for a force field

We will continue to use $$\vec F$$ for the force field, and will use $$\vec f$$ for force.

Recall: Newton’s Second Law and its angular analogue (see moment of inertia)

Linear motion Circular motion
Mass $$=m$$ Moment of inertia $$=I=m\,r^2$$
Force $$=\vec F$$ Torque $$=\vec\tau = \vec r\times\vec f$$
Velocity $$=\vec v$$ Angular velocity $$=\vec\omega=\displaystyle\frac{d}{dt}\theta$$
Acceleration $$=\vec a = \displaystyle\frac{\vec F}{m} = \frac{d}{dt}\vec v$$ Angular acceleration $$=\vec\alpha = \displaystyle\frac{\vec\tau}{I} = \displaystyle\frac{d}{dt}\vec\omega$$

Let point mass $$m$$ be in a force field (or rather acceleration field) $$\vec F$$. The field is the force per unit mass that would be exerted on a small mass at that point. So, the force $$\vec f$$ exerted on $$m$$ is $$\vec f = \vec F\,m \nonumber$$

Let $$m$$ be connected with a light rod to the origin. The $$\vec f$$ exerts a torque $$\vec\tau$$ at the origin $$\vec\tau = \vec r \times \vec f = \vec r \times \vec F\,m \nonumber$$

#### Translation vs. rotation movement

Applying Newton’s 2nd law, to both translation movement and rotation moment (ignoring the factor $$2$$) $$\begin{matrix} \frac{\text{Force}}{\rm{Mass}} &=& \text{Acceleration} &=& \frac{d}{dt}(\text{Velocity}) \\ \downarrow\rlap{\color{blue}{\text{curl}}} && \downarrow\rlap{\color{blue}{\text{curl}}} && \downarrow\rlap{\color{blue}{\text{curl}}} \\ \frac{\text{Torque}}{\text{Moment of inertia}} &=& \text{Angular acceleration} &=& \frac{d}{dt} (\text{Angular velocity}) \end{matrix} \nonumber$$

Curl connects the translation and rotation moments. Thinking of curl as an operation

• We know that the curl of a velocity field, gives the angular velocity.
• Then, the curl of an acceleration field, gives the angular acceleration (in the rotation part of the acceleration effect).
• Therefore, the curl of a force field measures the torque per unit moment of inertia.

So, it measures how much torque the force field exerts on a small solid. It will measure how much it starts spinning if you leave it in this force field. In particular, a force field with no curl does not generate rotation movement. It may accelerate in some direction, but it will not start spinning.

Just as the curl of a velocity field measures the angular velocity of its rotation, the curl of a force field measures the torque it exerts on a mass per unit moment of inertia.

The curl of a force field tells you how quickly the angular velocity ($$\omega$$) is going to increase or decrease.

#### Consequence

If force field $$\vec F$$ derives from a potential (=conservative), then its curl $$\nabla\times\vec F = 0$$ $$\nabla\times\vec F = \nabla\times(\nabla f) = 0 \nonumber$$

$$\Longrightarrow$$ $$\vec F$$ does not induce any rotational motion.

That means that the earth by itself doesn’t spin because of gravitational attraction. (It spins because it was formed spinning.). It didn’t start spinning because of gravitational effects.

(This is however not strictly true, because actually the Earth is deformable. Friction and tidal effects due to Earth’s gravitational attraction explain why the Moon’s rotation and revolution around Earth are synchronous. This is why we always see the same side of the moon).

## Green’s Theorem in tangential form

Green’s theorem:

If $$C$$ is a closed curve, enclosing region $$R$$, counterclockwise, and vector field $$\vec F$$ is defined and differentiable everywhere in $$R$$, then the work $$\shaded{ \oint_C\vec F\cdot d\vec r = \iint_R\mathrm{curl}(\vec F)\;dA } \nonumber$$

Substitute the work integral in differential form, and equation $$\eqref{eq:curldef}$$ for curl $$\shaded{ \oint_C Mdx+Ndy = \iint_R(N_x-M_y)\;dA } \nonumber$$

Comparing the left- and right-hand side

$$\displaystyle\oint_C Mdx+Ndy$$ $$\displaystyle\iint_R(N_x-M_y)\;dA$$
Lives only on the curve $$C$$ Lives everywhere inside region $$R$$
$$(x,y)$$ are related through the curve $$C$$. $$(x,y)$$ are independent within boundary $$R$$.

FYI: A interesting practical application is the Planimeter.

### Example

Let curve $$C$$ be a circle of radius $$1$$ centered at $$(2,0)$$, counterclockwise. Let $$\vec F$$ be the vector field $$\vec F= \left\langle \underbrace{\rm ye^{-x}}_M, \underbrace{\tfrac{1}{2}x^2-\rm e^{-x}}_N \right\rangle \nonumber$$

Compute the work

#### Using the contour integral

$$\oint_C\vec F\cdot d\vec r =\oint_C M\,dx+N\,dy = \oint_C y\rm e^{-x}\,dx+\left(\tfrac{1}{2}x^2-\rm e^{-x}\right)dy \nonumber$$

Picture

To do it directly, you would parametrize the curve. \left\{ \begin{align*} x=2+\cos\theta &\Rightarrow dx=-\sin\theta\;d\theta \\ y=\sin\theta &\Rightarrow dy=\cos\theta\;d\theta \\ 0 \leq \theta\leq 2\pi \end{align*} \right. \nonumber Instead, we will use Green’s theorem

#### Using Green’s theorem

\newcommand{pdv}[2]{\frac{\partial #1}{\partial #2}} \begin{align*} \oint_C\vec F\cdot d\vec r &= \iint_R \left( N_x-M_y \right)\;dA \\ &= \iint_R\left( \pdv{}{x} \left( \tfrac{1}{2}x^2-\rm e^{-x} \right) – \pdv{}{y}ye^{-x} \right)\;dA \\ &= \iint_R\left( \left( x+\rm e^{-x} \right) – e^{-x} \right)\;dA \\ &= \iint_R x\;dA \end{align*} \nonumber

### Proof

Let $$\vec F=\left\langle M,N\right\rangle$$ be a vector field with components $$M(x,y)$$, and $$N(x,y)$$; and $$C$$ be closed curve around region $$R$$.

We want to proof

$$\oint_C Mdx+Ndy = \iint_R(N_x-M_y)\;dA \nonumber$$

#### Observations

1. Proof the special case, where $$N=0$$. $$\oint_C Mdx = \iint_R-M_y\;dA \nonumber$$ A similar argument will show $$\oint_C Ndy = \iint_R N_x\;dA \nonumber$$ Then summing, we get Green’s theorem
2. We can decompose $$R$$ into simpler regions,
if we can proof \left\{ \begin{align*} \oint_{C_1} Mdx &= \iint_{R_1}-M_y\;dA \\ \oint_{C_2} Mdx &= \iint_{R_2}-M_y\;dA \end{align*} \right. \nonumber then because the boundary between $$R_1$$ and $$R_2$$ is traversed twice, but in opposite directions. When you sum, them these pieces will cancel. $$\oint_C Mdx = \oint_{C_1} + \oint_{C_2} = \iint_{R_1} + \iint_{R_2} = \iint_R -M_ydA \nonumber$$

#### Proof the special case

Let’s proof $$\shaded{ \oint_C Mdx = \iint_R-M_y\;dA } \nonumber$$

Start with the left hand side. Cut $$R$$ into “vertically simple” regions.

$$a\lt x\lt b, f_1(x)\lt y\lt f_2(x) \nonumber$$

If $$R$$ is vertically simple, and $$C$$ is the boundary of $$R$$ counterclockwise,

then $$\oint_{C_i}M(x,y)\,dx$$ for the curve segments $$C_i$$ \left. \begin{array}{ll} C_1:& y=f_1(x), x\ \rm{from}\ a\ \rm{to}\ b \\ C_2:& x=b \Rightarrow dx=0 \\ C_3:& y=f_2(x), x\ \rm{from}\ b\ \rm{to}\ a \\ C_4:& x=a \Rightarrow dx=0 \end{array} \right\} \Rightarrow \ \left\{ \begin{align*} \oint_{C_1} M(x,y)\;dx &= \int_a^b M\left(x,f_1(x)\right) dx \\ \oint_{C_2} M(x,y)\;dx &= 0 \\ \oint_{C_3} M(x,y)\;dx &= \int_b^a M\left(x,f_2(x)\right) dx \\ &= -\int_a^b M\left(x,f_2(x)\right) dx \\ \oint_{C_4} M(x,y)\;dx &= 0 \end{align*} \right.

Combining the curve segments $$C_{1,2,3,4}$$ \begin{align} \oint_C M\;dx &= \int_a^b M\left(x,f_1(x)\right) dx – \int_a^b M\left(x,f_2(x)\right) dx \nonumber \\ &= \int_a^b \Big(M\left(x,f_1(x)\right) – M\left(x,f_2(x)\right)\Big)\ dx \label{eq:greenspr1} \end{align}

Now, evaluate the right hand side \newcommand{pdv}[2]{\frac{\partial #1}{\partial #2}} \begin{align*} \iint_R -M_y\;dA &= -\int_a^b\underbrace{\int_{f_1(x)}^{f_2(x)} \frac{\partial M}{\bcancel{\partial y}}\;\bcancel{dy}}_{\rm{}inner\ integral}\;dx \\ &= -\int_a^b\underbrace{\Big[ M\Big]_{f_1(x)}^{f_2(x)}}\ dx \\ &= -\int_a^b\Big(M(x,f_2(x)) – M(x,f_1(x))\Big)\ dx \\ &= \int_a^b \Big(M(x,f_1(x)) – M(x,f_2(x))\Big)\ dx \end{align*} \nonumber

This matches equation $$\eqref{eq:greenspr1}$$. For this special case, when we have only an $$x$$-component and a vertically simple region, things work.

#### Generalize

Now, we can remove the assumption that things are vertically simple using this second observation. We can just glue the various pieces together, and prove it for any region. Then, we do same thing with the $$y$$-component. That’s the first observation. When we add things together, we get Green’s theorem in its full generality.

### More about validity (Lec. 27)

Let $$\vec F$$ be vector field $$\vec F(x,y) = \frac{-y\hat\imath+x\hat\jmath}{x^2+y^2} = \left\langle \frac{-y}{x^2+y^2}, \frac{x}{x^2+y^2} \right\rangle \nonumber$$

The vector field is not defined at origin, so we can’t use Green’s theorem directly.

$$\int_C\vec F\cdot d\vec r=\iint_R\rm{curl}\left(\vec F\right)\,dA \nonumber$$

Everywhere, except at the origin, its curl is $$0$$ \newcommand{pdv}[2]{\frac{\partial #1}{\partial #2}} \begin{align*} \rm{Curl}\left(\vec F\right) &= \pdv{}{x}\left(\frac{x}{x^2+y^2}\right) – \pdv{}{y}\left(\frac{-y}{x^2+y^2}\right) \\ &= \frac{ (x^2+y^2)\pdv{x}{x} – x\pdv{x^2+y^2}{x}}{\left(x^2+y^2\right)^2} – \frac{ (x^2+y^2)\pdv{(-y)}{y} – (-y)\pdv{x^2+y^2}{y}} {\left(x^2+y^2\right)^2} \\ &= \frac{(x^2+y^2)-2x^2}{\left(x^2+y^2\right)^2} – \frac{-(x^2+y^2)+2y^2}{\left(x^2+y^2\right)^2} = \frac{0}{\left(x^2+y^2\right)^2} \end{align*}

If the curve wouldn’t include the origin, we could use Green’s theorem. So let’s cut out a smaller region $$C”$$ clockwise around the origin.

In this region with a hole at the origin, the curl is $$0$$. This tells us that the line integrals are equal to each other, but we don’t know the values. $$\int_{C’}\vec F\,dr-\int_{C”}\vec F\,dr = \iint_R\rm{curl}\left(\vec F\right)\,dA = 0 \nonumber$$

To make a single closed curve, introduce a slit to connect the outer and inner curves. Now, Green’s theorem is well defined.

Total line integral, with $$C”$$ now going clockwise, and the two horizontal segments cancelling out $$\int_{C’}\vec F\,dr – \int_{C”}\vec F\,dr = \iint_R\rm{curl}\left(\vec F\right)\,dA \nonumber$$

This same argument applies to Green’s theorem for flux.

#### Simply connected region

Definition

A connected region $$R$$ in the plane is simply connected if the interior of any closed curve in $$R$$ is also contained in $$R$$.

In other words: the region $$R$$ doesn’t have any holes inside it.

In the example above, the domain of definition is the plane minus the origin. The region is not simply connected, because there is a hole around the origin.

Condition for Green’s theorem:

If domain where $$\vec F$$ is defined (and differentiable) is simply connected, then we can always apply Green’s theorem.

If a vector field is defined in a simply connected region, and its curl is $$0$$, then the vector field is conservative and is a gradient field. Thus we need to update:

If $$\rm{curl}\,\vec F=0$$ and domain where $$\vec F$$ defined is simply connected, then $$\vec F$$ is conservative (=is a gradient field).