Multivariable Exam 3 Review: Thomas 16.8

By Tyler Clarke in Calculus on 2025-4-23

Hello, everybody! This is the very last section in the Thomas textbook, and it's (fortunately) not a very hard one. This section is on something called divergence theorem.

The critical idea behind divergence theorem can be summed up with this formula, where `S` is a closed smooth surface, `C` is the volume contained by it, and `F` is a differentiable vector field: `int int_S F cdot n dA = int int_C int grad cdot F dV`. The flux of a vector field through a closed smooth surface is equal to the volume integral of the divergence of the vector field in the volume contained by the surface.

This has obvious applications. In many cases, it's much easier to compute `grad cdot F` than `F cdot n`: finding a normal vector can be extremely difficult for complex surfaces. It also means you don't have to worry about orientation, and in some cases you can greatly simplify the bounds of integration. Let's do an example!

Straight from the textbook, we have nice spherical one: find the flux of the vector field `F = [x, y, z]` over the sphere `x^2 + y^2 + z^2 = a^2`. Finding this the conventional way is pretty easy, but I'll leave that as an exercise to the reader. This is very obviously a sphere with radius `a` centered at origin, so we set up our substitution with `0 <= theta <= 2 pi`, `0 <= phi <= pi`, `0 <= rho <= a`, and `dV = rho^2 sin(phi)`. It's easy enough to find that `grad cdot F = 3`. Substituting everything in gives us the integral `int_0^{2pi} int_0^{pi} int_0^a 3rho^2 sin(phi) d rho d phi d theta`. The inner integral is trivial - it works out to `int_0^a 3 rho^2 sin(phi) d rho = a^3 sin(phi)`. This gives us an integral in `phi` of `int_0^{pi} a^3 sin(phi) d phi = |_0^{pi} -a^3 cos(phi) = 2a^3`. Finally, because there is no `theta` term here, we just multiply by `2pi` to get `4pi a^3`.

This technique can also be used to make finding the flux in a nonsmooth surface much less terrible, if it can be reduced to smooth bounds. For instance: to find the flux of `F(x, y, z) = [cos^3(z), cos^3(z), z - e^{cos(x)sin(y)}]` through a cube described by the points `(0, 0, 0)` and `(1, 4, 9)`, we would normally have to calculate the flux through each individual face, which would be absolutely terrible; but with divergence theorem, we can turn this into a much more manageable triple integral over the volume `int_0^1 int_0^4 int_0^9 grad cdot F dz dy dx`. We're in the incredibly lucky position of having `grad cdot F = 1` - meaning the flux is just the volume, `1 * 4 * 9 = 36`. If you don't believe that that was easier, I invite you to calculate `int e^{cos(x)sin(y)} dx`. It's not pleasant.

That's the end of the whole textbook - my copy, at least. See you in the summer with differential equations! Also keep a look-out in the physics section - after finals, I'll be building railguns with the folks at Scrappy's Garage, and I'll definitely be posting the results of our experimentation here. For the very last time (well, not really), au revoir, and good luck!