# 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.

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