Overview

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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.

Closed surface \(S\) and region \(D\)

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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