Multivariable Exam 3 Review: Thomas 15.8

By Tyler Clarke in Calculus on 2025-4-1

Aaaand we're back. It was a pleasant Monday for all three people who actually like Georgia weather. How does it go from "unpleasantly chilly and rainy" to "gushing sweat" in the space of half an hour?

Today we're covering Thomas 15.8 - Substitutions in Multiple Integrals. This stuff is fun. If you remember u-substitution, this is that: the whole idea is to find some function `u` that turns `int f(x) dx` into `int u du`, but in multiple variables. One of the core pieces of trickery in u-substitution is a disgusting-yet-accurate piece of algebra: `frac du dx = g(x)` turns into `du = g(x) dx`, meaning if you have `g(x) dx` in your original integral, you can substitute in `du`. How do we do this in multiple variables? It's not much more complicated. Find a function `r` that maps your new variables to your old ones in some nice way, and substitute in the mappings; then, if your old variables are x and y, and your new variables are u and v, you have the convenient substitution `dx dy` = `|det[[frac dx {du}, frac dy {du}], [frac dx {dv}, frac dy {dv}]]| du dv`. Why this works is out of scope and frankly beyond my understanding, but it definitely does. Rather like with single variable u-substitution, you can choose to either substitute back in the old variables before evaluating the bounds, or change the bounds; we'll opt for the latter option. You'll have to use some algebra to find bounds in terms of u and v. It's not very hard, but it can be tricky if you're not used to it.

This technique has a surprising consequence: it predicts our cylindrical and spherical area elements! In cylindrical coordinates, we have `x = rcos(theta)`, `y = rsin(theta)`, and `z = z`. This gives us a Jacobian matrix of `[[cos(theta), sin(theta), 0], [-rsin(theta), rcos(theta), 0], [0, 0, 1]]`. The determinant of this is clearly `rcos^2(theta) + rsin^2(theta) = r` (if you don't know how I found that, I recommend practicing determinants!), which gives us the substitution `dx dy dz = r dz dr d theta` - exactly what we already knew. Nice! I'll leave deriving the one for spherical coordinates as an exercise to the reader.

Let's do an example from the book. Given that `u = frac {2x - y} 2` and `v = frac y 2`, evaluate `int_0^4 int_{frac y 2}^{frac y 2 + 1} frac {2x - y} 2 dx dy`. The first thing to do is solve the bounds. Knowing that `0 <= y <= 4`, we can use the function for `v` to get `0 <= 2v <= 4`. Dividing by two gives us `0 <= v <= 2`.

The one for `u` is a little more complicated, but still not too bad. Some algebra tells us that `x = u + frac y 2`, and we can substitute to get... `frac y 2 <= u + frac y 2 <= frac y 2 + 1`. This is messy, but it's also easy to reduce to `0 <= u <= 1` by subtracting `frac y 2`. Hooray for constant bounds!

Next up is the actual integrand. Substituting in gives us `frac {2(u + frac y 2) - 2v} 2`. There's a dangling `y`! Fortunately, we know what `frac y 2` is: it's `v`. `frac {2u + 2v - 2v} 2 = u`. Lastly, we need to find `dx dy`. Our `r` function is `r = [u, v] = [frac {2x - y} 2, frac y 2]`. Substituting in and differentiating gives us the Jacobian matrix `[[1, 1], [0, 2]]` - the determinant of which is just `2`. Nice! Putting this all together gives us the integral `int_0^2 int_0^1 2u du dv`. This is much easier to solve! Our final result is just `2`.

While I won't cover it here, it's also quite useful to know how to actually find those substitutions if they aren't provided.