Multivariable Exam 3 Review: Thomas 15.6

By Tyler Clarke in Calculus on 2025-3-30

Welcome once more to another thrilling textbook section review. This one covers Thomas 15.6, which is a truly strange section- called simply "Applications", it doesn't cover any new material as much as demonstrate some interesting possibilities with triple integrals.

The first thing to consider is a geometric interpretation of the integral of `F` over a volume - if `F` is a function for density, the result of the integration is mass! This has some surprising consequences. For instance: with triple integrals, you can find the first moment about any given axis. If your mass integration is `int int_D int F(x, y, z) dV`, then the first moment about the x-axis is `int int_D int xF(x, y, z) dV`, the moment about the y axis `int int_D int yF(x, y, z) dV`, etc. These are handy for physical reasons, but the immediate useful point is finding the center of mass: the center of mass about an axis is the first moment about that axis divided by total mass, so the center of mass x-component is `frac {int int_D int xF(x, y, z) dV} {int int_D int F(x, y, z) dV}`, etc.

Let's do an example. Given a constant density function `F(x, y, z) = 2`, what is the center of mass of a shape bounded below by the xy plane and above by the function `z = 4 - x^2 - y^2`?

To find this, we first need to find the bounds of integration. The z bounds are obvious, but what about x and y? Fortunately, fate smiles upon us, and we can just take the intersection of the z bounds `z=0` and `z=4-x^2-y^2`. A quick substitution and some algebra gives us `x^2 + y^2 = 4`, which is just a circle with radius 2! Assuming we want to integrate in `dz dy dx` (`dz dx dy` would also be valid), this means the x-bounds are from -2 to 2, and the y bounds are from `-sqrt(4 - x^2)` to `sqrt(4 - x^2)`. Our mass is thus given by `int_{-2}^{2} int_{-sqrt(4 - x^2)}^{sqrt(4 - x^2)} int_{0}^{4-x^2-y^2} 2 dz dy dx`. The inner integral in `dz` evaluates nicely to `8 - 2x^2 - 2y^2`. This leaves us with `int_{-2}^{2} int_{-sqrt(4 - x^2)}^{sqrt(4 - x^2)} 8 - 2x^2 - 2y^2 dy dx`. Ew. It's useful to learn how to do these thorny ones properly, but without the much-desired help of cylindrical coordinates it's also very annoying, and out of the scope of this post. Wolfram|Alpha to the rescue! Our mass is 50.2655 (note that this is unitless).

Because this shape is symmetrical in the x and y directions and the density is constant, the first moment and thus center of mass in both directions should be zero. This leaves the z direction. To find that, we first need to find the first moment in z: `int_{-2}^{2} int_{-sqrt(4 - x^2)}^{sqrt(4 - x^2)} int_{0}^{4-x^2-y^2} 2z dz dy dx`. I categorically refuse to do this in Cartesian. For those who are interested, the integral in cylindrical coordinates is just `int_{0}^{2pi} int_{0}^{2} int_{0}^{4-r^2} 2zr dz dr d theta`, which is a lot easier to evaluate. We'll see more discussion of this problem in the post on 15.7, but for now, just believe me when I say that it evaluates to 67.0206. Dividing `frac {67.0206} {50.2655}` gives us a fairly nice number: `frac {4} {3}`. So our center-of-mass is ``. Nice!

The next thing to talk about is moment of inertia. This is actually a physics concept, but it's an interesting illustration of another property of triple integrals. Moment of inertia can be thought of as rotational mass - it's the property that resists torque, much like how mass is the property that resists force. The moment of inertia is also known as the second moment. You can find the second moment about any given line or axis just by multiplying the density by the function for squared distance from said line or axis - `int int_D int r^2(x, y, z)F dV`. For instance, the moment of inertia about the x axis is `int int_D int (y^2 + z^2)F dV`. The process here is otherwise exactly the same as finding the first moment.