Multivariable Exam 3 Review: Thomas 15.5

By Tyler Clarke in Calculus on 2025-3-27

Hey there, everyone! I come to you after a long and boring Spring break with unholy tidings of exams approaching. My first exam in multivariable will be covering everything in Thomas' Calculus (really good book, much recommend) between 15.5 and 16.8. To do those very complex sections justice, I'm writing a separate piece for each of them. It's dense stuff!

Basics of Triple Integrals

15.5 starts with a discussion of triple integrals. These really aren't all that complicated - if you know how to do basic integration, they shouldn't pose a challenge. Usually they're written as `int int_D int f dV`. In this case, `D` is the region, and `dV` is the area element - the `1 dx dy dz` part, if this was in terms of cartesian coordinates, but as we'll see there are many options.

The hard part of triple integration is determining bounds. Usually speaking, you'll have to either figure out the bounds yourself from a set of boundary functions, OR convert the bounds to a different form. There's a knack to it that you can pick up with practice, but the process amounts to picking an order of integration and finding bounds that don't leave dangling variables (for instance, if you try to integrate `int_0^y int_0^z int_1^3 sin(x) dz dy dx`, you'll end up with a function rather than a value). You also want to pick bounds of integration that arne't too complicated - in many, many cases, it will actually be easier to change variables to spherical or cylindrical coordinates rather than toughing it out in cartesian with nasty bounds.

Let's go through a few examples. A classic is the triangular area one: we have a region in the first quadrant below the plane x + y + z = 4 over which to integrate the function `f(x, y, z) = x`. This area is convenient because we don't have to think too much about it: it's obviously the lower triangular half of the (4, 4, 4) cube. We need to find the bounds of the triple integral that describes this area. We don't need to do anything special with the coordinate system here, so our `dV = dz dy dx`. How did I pick that order? This shape is symmetrical, so I picked at random - it'll be equally easy to produce bounds for any given order.

Because the x-bounds are on the outside, we want them to be constant (if they contain variables, our final result will contain variables). Because this is a region in the first quadrant, x must be no larger than 0; the largest possible value of x is the x-intercept, `x=4`. Thus we have our first constraint: `0 <= x <= 4`. Onto `y`! `y` is a bit more complex. We can't just say `0 <= y < = 4`, because for z=0 that would produce a square rather than a triangle on the xy plane. If you can visualize the intercept of the region and the xy plane, it's clearly a simple line `y = 4 - x`: we can use this! `0 <= y <= 4 - x` is valid.

Finally, the z-bounds. These are a bit tricky to intuit, but fortunately we've a nice equation! `x + y + z = 4` can be easily reduced to `z = 4 - x - y`, and this is in fact the upper bound of z over this entire region. `0 <= z <= 4 - x - y`!

We can insert those bounds and our element of integration directly into the integral, to get `int_0^4 int_0^{4-x} int_0^{4 - x - y} f(x, y, z) dz dy dx`. Beautiful! Because `f(x, y, z) = x`, this is equivalent to `int_0^4 int_0^{4-x} int_0^{4 - x - y} x dz dy dx`. Let's integrate! The innermost integral, `int_0^{4 - x - y} x dz`, evaluates obviously to `4x - x^2 - xy` (if you can't see how I got this, practice integration). This means our integral is now `int_0^4 int_0^{4-x} 4x - x^2 - xy dy dx`. The innermost integral is a bit trickier for this one, but works out to `4x(4 - x) - x^2(4 - x) - x(4 - x)^2 / 2`.

Yikes.

Before we can finish the integration, we have to simplify. A binomial expansion and some distribution leaves us with `16x - 4x^2 - 4x^2 + x^3 - x^3/2 + 4x^2 - 8x`, which simplifies to `8x - 4x^2 + x^3/2`. Not terrible! Integrating `int_0^4 8x - 4x^2 + x^3/2 dx` leaves us with a simple `|_0^4 4x^2 - 4/3x^3 + x^4/8`, which evaluates to `frac{32}{3}`. Not too hard!

An example of a harder problem, ripped straight from the textbook, is to find the volume of a region bounded below by `z = 3` and bounded above by a sphere of radius 5 (centered on the origin). I'll leave solving this one as an exercise to the reader.

A particularly common problem is finding the volume of a tetrahedron given its corner points. There's no real trick here - we just have to find a set of lines describing the extrema, and solve them properly. For instance (yoinked from the textbook): given a tetrahedron described by the points (0, 0, 0), (0, 1, 0), (1, 1, 0), and (0, 1, 1), what are the bounds?

Like the last one, it doesn't matter too much which dimension we pick; I'm reusing `dV = dz dy dx`. The bounds for x are obvious enough: the lowest x-value is 0, and the highest is 1, so `0 <= x <= 1`. The bounds for y are conveniently described by our points on the z=0 plane: (0, 0, 0), (0, 1, 0), and (1, 1, 0). These form a right triangle; the line we're concerned with is (coincidentally) the hypotenuse, described by `y = x`. In this case, `y` is above the line and below 1, so our y-bounds are `x <= y <= 1`. Finally, onto the z-bounds! To find these, consider that z increases with y and decreases with x, hinting not-so-subtly at a function quite like the one from before: `z = y - x`. `z` is below the function, here, so `0 <= z <= y - x`.

These are very difficult to derive without graphics. While sketching is out of the scope of this website (seeing as I'm so bad at it), I'd recommend anyone learning multivariable practice sketching complex functions in a way that conveys the area reasonably well. Oftentimes, the functions you need are blisteringly obvious once you can look at them drawn out.

Other basic calculus ideas like the average value of a function in a region apply here as well. The average value of `F` over a region is simply `frac {int int_D int F dV} {int int_D int 1 dV}` - the integral of `F` over the region, divided by the volume of the region. Easy!