Brain Gains (Rapid Recall, Jivin' Generation)

With appropriate assumptions place on the corresponding vector fields and objects involved, here are the three big theorems we've seen.

  • Green's Theorem: $\ds \iint_R N_x-M_y dA = \oint_{\partial R} M dx+Ndy $
  • Stokes' Theorem: $\ds \iint_S \vec \nabla \times \vec F \cdot \hat n dS = \int_{\partial S} Mdx+Ndy+Pdz$.
  • Divergence Theorem: $\ds \iiint_D \vec \nabla \cdot \vec F dV = \iint_{\partial D}\vec F\cdot\hat n dS$.

One (of many) use of each theorem is that sometimes one side of the equality is extremely simple to compute. Let's see this in a few settings.

  1. Let $\vec F = (2x-y^2z, 3y-4x^3, e^{x^2y}+z)$. Compute the outward flux of $\vec F$ across the surface $S$ of the cube $-1\leq x\leq 1$, $-1\leq y\leq 1$, $-1\leq z\leq 1$, so compute $\iint_{S}\vec F\cdot\hat n dS$.
  2. For the vector field $\vec F = (x^2+4z, 3y, -4x)$, compute $\int_C M dx+Ndy+Pdz$ along the curve $C$ parametrized by $\vec r(t) = (2+3\cos t, 5, 7+3\sin t)$ for $0\leq t\leq 2\pi$.
  3. Compute the counterclockwise circulation $\oint_C -\frac{y}{2}dx+\frac{x}{2}dy$ for the curve $C$ defined by $(x-3)^2 + (y+7)^2=4$.

Solutions

With each questions above, the key is to first identify the right theorem.

  1. Use the divergence theorem. Note that the divergence is $\vec\nabla \cdot \vec F = 2+3+1=6$. As such, the outward flux is $\iiint_D 6 dV = 6V = 6(2)(2)(2) = 48$.
  2. Use Stokes' Theorem. The curl is $\vec\nabla \times \vec F = (0,8,0)$. Because the curve is a circle, let's use the flat planar surface $S$ inside this circle. The only thing left is to figure out which orientation to use for the surface. The curve is a circle centered at $(2,5,7)$ of radius 3, in a plane parallel to the $xz$-plane, which means a unit normal vector can be chosen as $(0,1,0)$ or $(0,-1,0)$. The orientation of the curve is compatible with $(0,-1,0)$, which we can use in Stokes's theorem. We obtain $$\ds \iint_S \vec \nabla \times \vec F \cdot \hat n dS = \ds \iint_S (0,8,0) \cdot (0,-1,0) dS = \ds \iint_S -8 dS = -8\pi(3)^2 = -72\pi.$$ We reduced the problem to a surface area question.
  3. Using Green's theorem, we have $$\ds \iint_R N_x-M_y dA = \iint_R \frac{1}{2}-(-\frac{1}{2}) dA = A = \pi (2)^2 = 4\pi.$$ This particular integral is often used by a planimeter to compute the area inside odd shape regions.

Group Problems