16.6

site/posts/multivariable-thomas-16.5.html

@@ -45,8 +45,30 @@
     Finally: after parameterizing a surface, it becomes quite simple to find the area. Plug in your bounds for `u` and `v` into a double integral, and simply apply the area element!
     For a general parameterized function, the area element is actually just `abs {frac {dr} {du} times frac {dr} {dv}}` - the magnitude of the normal vector to the tangent plane.
     In our sphere example above, that would be `sqrt((-16 cos(u) sin^2 (v))^2 + (-16 sin(u) sin^2 (v))^2 + (-16 sin(v) cos(v))^2)` - which simplifies to `16sin(v)`, magically enough.
-    The integral is thus `int_0^{2pi} int_0^{pi} 16sin(v) dv du = 64pi`. This matches up perfectly with the formula for the surface area of a sphere, `4pi r^2`. Great!
+    The integral is thus `int_0^{2pi} int_0^{pi} 16sin(v) dv du = 64pi`. This matches up perfectly with the formula for the surface area of a sphere, `4pi r^2`. Great! Let's add that
+    to the element of integration table:
 </p>
+<h3>An Increasingly Exhaustive Table of Area Elements</h3>
+<ul>
+    <li>
+        `dV = dx dy dz` (or other orderings thereof)
+    </li>
+    <li>
+        `dV = r dz dr d theta` (or other orderings thereof)
+    </li>
+    <li>
+        `dV = rho^2 sin(phi) d rho d phi d theta` (or other orderings thereof)
+    </li>
+    <li>
+        `dV = Jac(r(u, v)) dx dy` (or other orderings thereof; same pattern in 3d)
+    </li>
+    <li>
+        `dV = |frac {dr}{dt}| dt`
+    </li>
+    <li>
+        `dA = abs {frac {dr} {du} times frac {dr} {dv}} du dv` for a surface parameterization `r(u, v)`.
+    </li>
+</ul>
 <p>
     I've intentionally ignored implicit surfaces because they don't seem to be showing up in this course. If you actually have the Thomas textbook, I'd recommend reading through
     that part, for interestingness value if nothing else. Until 16.6, <i>arrivederci</i> and good luck!