Tyler Clarke
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3.7 KiB
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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"
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in the space of half an hour?
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</p>
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<p>
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Today we're covering Thomas 15.8 - Substitutions in Multiple Integrals. This stuff is fun. If you remember u-substitution, <i>this is that</i>: the whole idea is to find
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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
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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?
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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
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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
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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
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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,
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but it can be tricky if you're not used to it.
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</p>
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<p>
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This technique has a surprising consequence: it predicts our cylindrical and spherical area elements! In cylindrical coordinates, we have
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`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]]`.
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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
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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.
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</p>
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<p>
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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`.
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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`.
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</p>
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<p>
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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`.
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This is messy, but it's also easy to reduce to `0 <= u <= 1` by subtracting `frac y 2`. Hooray for constant bounds!
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</p>
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<p>
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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`.
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`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
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the Jacobian matrix `\[\[1, 1], \[0, 2]]` - the determinant of which is just `2`.
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Nice! Putting this all together gives us the integral
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`int_0^2 int_0^1 2u du dv`. This is <i>much</i> easier to solve! Our final result is just `2`.
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</p>
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<p>
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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.
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</p>
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[/]
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[=title "Multivariable Exam 3 Review: Thomas 15.8"]
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[=author "Tyler Clarke"]
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[=date "2025-4-1"]
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[=subject "Calculus"]
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