16.8 + fix some formatting

site/posts/multivariable-exam-1-review.html

@@ -124,7 +124,7 @@ Cross products have a set of useful properties too:
         Simplify: <1, 1, 1>t + <1, 1, 2>s = <2, 2, 0>.
     </li>
     <li>
-        This can now be converted into an augmented matrix multiple of [t, s, 1]! <br>
+        This can now be converted into an augmented matrix multiple of `\[t, s, 1]`! <br>
         <table>
             <tr>
                 <td>1</td>
@@ -233,7 +233,7 @@ Cross products have a set of useful properties too:
 </p>
 <ul>
     <li>
-        `l'(t) = [3, 4, 0]`
+        `l'(t) = \[3, 4, 0]`
     </li>
     <li>
         `||l'(t)|| = sqrt(3^2 + 4^2 + 0^2) = 5`

site/posts/multivariable-thomas-15.6.html

@@ -31,7 +31,7 @@
     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 `[0, 0, frac {4} {3}]`. Nice!
+    `frac {67.0206} {50.2655}` gives us a fairly nice number: `frac {4} {3}`. So our center-of-mass is `\[0, 0, frac {4} {3}]`. Nice!
 </p>
 <p>
     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.

site/posts/multivariable-thomas-16.1.html

@@ -40,7 +40,7 @@
 </p>
 <p>
     Another example is to find the mass of a circular arc in the first quadrant, with radius 2 and density `x^2 + y^2`. Remember that the parameterization of
-    a circle is `vec c(t) = [rcos(t), rsin(t)]`, and the radial span of the first quadrant is from `0` to `frac pi 2`. Thus, `vec r(t) = [2cos(t), 2sin(t)]` for `0 <= t <= frac pi 2`.
+    a circle is `vec c(t) = \[rcos(t), rsin(t)]`, and the radial span of the first quadrant is from `0` to `frac pi 2`. Thus, `vec r(t) = \[2cos(t), 2sin(t)]` for `0 <= t <= frac pi 2`.
     The density function here is convenient because it's just `r^2`, based on the pythagorean identity. If you substitute in `2cos(t)` for `x` and `2sin(t)` for y, you will get
     `2^2 = 4`. Constant density is always nice! Finally, we need to find the element of integration. `vec r'(t) = \[-2sin(t), 2cos(t)]`, so `|r'|dt` = `sqrt(4sin^2(t) + 4cos^2(t))` -
     `2 dt`. Thus, we have the integral `int_0^{frac pi 2} 8 dt`; this obviously works out to `4pi`.

site/posts/multivariable-thomas-16.8.html

@@ -1,12 +1,40 @@
 [!]
 [=post-]
 <p>
-    Hello, everybody! This is the very last section in the Thomas textbook, and boy is it a doozy. We're finally applying divergence!
+    Hello, everybody! This is the very last section in the Thomas textbook, and it's (fortunately) not a very hard one. This section is on something called <i>divergence theorem</i>.
+</p>
+<p>
+    The critical idea behind divergence theorem can be summed up with this formula, where `S` is a closed smooth surface, `C` is the volume contained by it, and `F` is a differentiable vector field:
+    `int int_S F cdot n dA = int int_C int grad cdot F dV`. The flux of a vector field through a closed smooth surface is equal to the volume integral of the divergence of the vector field in
+    the volume contained by the surface.
+</p>
+<p>
+    This has obvious applications. In many cases, it's <i>much</i> easier to compute `grad cdot F` than `F cdot n`: finding a normal vector can be extremely difficult for complex surfaces.
+    It also means you don't have to worry about orientation, and in some cases you can greatly simplify the bounds of integration. Let's do an example! 
+</p>
+<p>
+    Straight from the textbook, we have nice spherical one: find the flux of the vector field `F = \[x, y, z]` over the sphere `x^2 + y^2 + z^2 = a^2`. Finding this the
+    conventional way is pretty easy, but I'll leave that as an exercise to the reader. This is very obviously a sphere with radius `a` centered at origin, so we set up
+    our substitution with `0 <= theta <= 2 pi`, `0 <= phi <= pi`, `0 <= rho <= a`, and `dV = rho^2 sin(phi)`. It's easy enough to find that `grad cdot F = 3`. Substituting
+    everything in gives us the integral `int_0^{2pi} int_0^{pi} int_0^a 3rho^2 sin(phi) d rho d phi d theta`. The inner integral is trivial - it works out to
+    `int_0^a 3 rho^2 sin(phi) d rho = a^3 sin(phi)`. This gives us an integral in `phi` of `int_0^{pi} a^3 sin(phi) d phi = |_0^{pi} -a^3 cos(phi) = 2a^3`.
+    Finally, because there is no `theta` term here, we just multiply by `2pi` to get `4pi a^3`.
+</p>
+<p>
+    This technique can also be used to make finding the flux in a nonsmooth surface much less terrible, if it can be reduced to smooth bounds. For instance:
+    to find the flux of `F(x, y, z) = \[cos^3(z), cos^3(z), z - e^{cos(x)sin(y)}]` through a cube described by the points `(0, 0, 0)` and `(1, 4, 9)`, we would normally have to calculate
+    the flux through each individual face, which would be absolutely terrible; but with divergence theorem, we can turn this into a much more manageable
+    triple integral over the volume `int_0^1 int_0^4 int_0^9 grad cdot F dz dy dx`. We're in the incredibly lucky position of having `grad cdot F = 1` - meaning
+    the flux is just the volume, `1 * 4 * 9 = 36`. If you don't believe that that was easier, I invite you to calculate `int e^{cos(x)sin(y)} dx`. It's not pleasant.
+</p>
+<p>
+    That's the end of the whole textbook - my copy, at least. See you in the summer with differential equations! Also keep a look-out in the physics section -
+    after finals, I'll be building railguns with the folks at <a href="https://scrappysgarage.org/">Scrappy's Garage</a>, and I'll definitely be posting the results
+    of our experimentation here. For the very last time (well, not really), <i>au revoir</i>, and good luck!
 </p>
 [/]
 [=title "Multivariable Exam 3 Review: Thomas 16.8"]
 [=author "Tyler Clarke"]
-[=date "2025-4-16"]
+[=date "2025-4-23"]
 [=subject "Calculus"]
-[=unpub]
 [#post.html]