Multivariable Exam 3 Review: Thomas 16.6

By Tyler Clarke in Calculus on 2025-4-11

Hello, everyone! Today we're going through Thomas 16.6: surface integrals. Fun! We've already done surface integrals in 2d and area integrals of a wide variety of shapes, as well as line integrals of a parameterized curve; now that we've seem 16.5 and have a good foundation in parameterized surfaces, we can start integrating over a parameterized surface!

The core concept here is unchanged from all of the other change of variable magic we've been doing: given an integral `int_S int G(x, y, z) dA`, we change variables with an appropriate item from the table:

An Increasingly Exhaustive Table of Area Elements

• `dV = dx dy dz` (or other orderings thereof)
• `dV = r dz dr d theta` (or other orderings thereof)
• `dV = rho^2 sin(phi) d rho d phi d theta` (or other orderings thereof)
• `dV = Jac(r(u, v)) dx dy` (or other orderings thereof; same pattern in 3d)
• `dV = |frac {dr}{dt}| dt`
• `dA = abs {frac {dr} {du} times frac {dr} {dv}} du dv` for a surface parameterization `r(u, v)`.

In this case, we want the last one. That means our integral becomes `int int G(x, y, z) abs {frac {dr} {du} times frac {dr} {dv}} du dv`. Quite simple. If you're still visualizing this as mass, then `G(x, y, z)` is the density function, and we've found the mass of a surface. Not much different from what we've already done. Let's do an example from the textbook. Integrate `G(x, y, z) = xyz` over the surface of a cone `z = sqrt(x^2 + y^2)` for `0 <= z <= 1`. Parameterizing this cone in radial coordinates is pretty easy: we get `r(u, v) = [vcos(u), vsin(u), v]` for `0 <= u <= 2pi`, `0 <= v <= 1` (if you aren't sure how I got that, you should study 16.5). Finding the area element is a bit less easy: we have `frac {dr} {du} = [-vsin(u), vcos(u), 0]` and `frac {dr} {dv} = [cos(u), sin(u), 1]`. The cross product of those is not terribly difficult to find to be `[vcos(u), vsin(u), -v]`. Finally, we need the magnitude of that: `sqrt(v^2cos^2(u) + v^2sin^2(u) + v^2) = vsqrt(2)`.

Sub'ing into an integral, we get `int_0^{2pi} int_0^1 xyz * vsqrt(2) dv du`. There's something wrong here - the integral is in terms of `u` and `v`, but we've a function in `x`, `y`, and `z`! Fortunately, we have a convenient function `r(u, v)` that we can use for substitutions. After substituting, we get `int_0^{2pi} int_0^1 v^3cos(u)sin(u) * vsqrt(2) dv du`. This looks complex, but is actually very easy to evaluate. The inner integral turns into `sqrt(2) |_0^1 frac {v^5} 5 cos(u)sin(u)`, which in turn becomes `frac {sqrt(2)} 5 cos(u)sin(u)`. Now we just have `frac {sqrt(2)} 5 int_0^{2pi} cos(u)sin(u) du`. It's pretty easy to use `u`-substitution to evaluate this, although you'll have to use another variable for the substitution to avoid confusion, and the result is... `0`!

Integrating a scalar function over a surface is all well and good, but what about a vector field? This is actually simpler than line integrals, if only because we don't have to consider the field integral tangent to the surface - that wouldn't be well-defined. All we have to consider is the flux. Given a unit normal vector `n` to a given surface, and a vector field `F`, then, the flux through the surface is `int_S int F cdot n dA`. But wait, there's more! Because the normal vector is `frac {dr} {du} times frac {dr} {dv}`, the unit normal is the normal divided by the magnitude of the normal `abs {frac {dr} {du} times frac {dr} {dv}}`, and `dA` is `abs {frac {dr} {du} times frac {dr} {dv}} du dv`, we have `int_S int F cdot frac {frac {dr} {du} times frac {dr} {dv}} {abs {frac {dr} {du} times frac {dr} {dv}}} abs {frac {dr} {du} times frac {dr} {dv}} du dv`. This simplifies nicely to just `F cdot (frac {dr} {du} times frac {dr} {dv}) du dv`. Convenient! Let's do a simple surface integral: the flux across a unit sphere `r(u, v) = [cos(u)sin(v), sin(u)sin(v), cos(v)]` of the vector field `F(x, y, z) = [x, y, z]`.

The normal of a radius 1 sphere is `n = [cos(u) sin^2 (v), sin(u) sin^2 (v), sin(v) cos(v)]` (see the section on 16.5 for a bit more detail). Note that there is another possible normal vector- `n * -1`. It's up to you to pick the correct one. In this case, I picked this one because it's radiating away from the surface. Because this is in spherical coordinates, our bounds are `0 <= u <= 2pi`, `0 <= v <= pi`. `n cdot F` is just `cos(u) sin^2 (v) * x + sin(u) sin^2 (v) * y + sin(v) cos(v) * z` - substituting gives us `cos(u) sin^2 (v) * cos(u)sin(v) + sin(u) sin^2 (v) * sin(u)sin(v) + sin(v) cos(v) * cos(v)`. This simplifies to `cos^2(u) sin^3 (v) + sin^2(u) sin^3 (v) + sin(v) cos^2(v)`, which simplifies to `sin^3 (v) + sin(v) cos^2(v)`, which simplifies to `sin(v) (sin^2(v) + cos^2(v)) = sin(v)`. Ah, trigonometry. `int_0^{2pi} int_0^{pi} sin(v) dv du` is really very easy to evaluate - it works out to `4pi`. If we'd chosen the other normal vector, the answer would be `-4pi`. It's quite important to pick the right normal!

The formulae for moments about axes and centers of mass are identical to the other ones, so I won't bother working through an example; to find the moment about an axis, multiply the integrand by that axis (so the first moment about the x-axis of `int_S int G(x, y, z) dA` is `int_S int x * G(x, y, z) dA`); for the second moments (moments of inertia), multiply by the squared distance from the axis (so `int_S int (z^2 + y^2)G(x, y, z) dA`). The center of mass about a given axis is the first moment about that axis divided by total mass.

That's all of 16.6. We're quickly nearing the end of the chapter! Stay tuned for 16.7 and 16.8. I might also do a quicker-than-usual (seriously, these only take about an hour to write for the hard ones) post on implicit surfaces, which I've ignored in this and the last post as they don't seem important to the class. They're pretty cool, though. Over in the Physics section, I'm going to be throwing up a bunch of worked-examples relevant to the test (some pulled from old tests, some from labs, some from homework, some made up on the spot), so if you're taking physics 2, keep an eye out over there. Sayonara for now, and good luck!