16.7

site/posts/multivariable-thomas-16.7.html

@@ -8,16 +8,20 @@
     the circulation about any two open surfaces with the same boundary curve is the same.
 </p>
 <p>
-    Let's do an example. Take the hemisphere with radius 2 above the plane `x - z = 0`, centered on origin: what is the circulation of the vector field
-    `F(x, y, z) = \[-y, x, z]` about this surface? In 2 dimensions this is a somewhat complicated integral to perform, but Stokes' theorem allows us to reduce
-    it to a single dimension, if we can parameterize the boundary curve. The plane parameterization in 2 variables is obviously `p(u, v) = \[u, v, u]`,
-    and the sphere is `s(u, v) = \[2cos(u)sin(v), 2sin(u)sin(v), 2cos(v)]`. This gives us three equations for the intersection: `u = 2cos(u)sin(v)`, `v = 2sin(u)sin(v)`,
-    and `u = 2cos(v)`.
+    Let's do an example. Take the hemisphere with radius 2 above the xy plane: what is the circulation of the vector field `F(x, y, z) = \[-y, x, z]`
+    about this surface? This is a somewhat difficult problem in two dimensions, but Stokes' theorem means we can reduce it to a single dimension
+    integral. The parametrization of our boundary curve is the radius 2 circle on the xy plane: `r(u) = \[2cos(u), 2sin(u), 0]` for `0 <= u <= 2pi`. Taking the derivative
+    gives us `dr = \[-2sin(u), 2cos(u), 0] du`, and substituting our parametrization gives us `F(u) = \[-2sin(u), 2cos(u), 0]`. We know the circulation
+    is going to be `int_C F cdot dr`; `F cdot dr` is just `4sin^2(u) + 4cos^2(u)`. Substituting in gives us `int_0^{2pi} 4(sin^2(u) + cos^2(u)) du` - which
+    simplifies to `int_0^{2pi} 4 du = 8pi`. Easy!
+</p>
+<p>
+    This post was much shorter than I was hoping, largely due to time constraints. I recommend practicing all the complexities of Stokes' theorem in depth-
+    for instance, Stokes' can be applied to surfaces with holes, unlike Green's. See you in 16.8!
 </p>
 [/]
 [=author "Tyler Clarke"]
-[=date "2025-4-12"]
+[=date "2025-4-16"]
 [=subject "Calculus"]
 [=title "Multivariable Exam 3 Review: Thomas 16.7"]
-[=unpub]
 [#post.html]