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• I-Learn, Class Pictures, Learning Targets, Text Book Practice
• Prep Tasks: Unit 1 - Motion, Unit 2 - Derivatives, Unit 3 - Integration, Unit 4 - Vector Calculus

This is day 7 of Unit 4.

Brain Gains (Rapid Recall, Jivin' Generation)

• Let $\vec F(x,y) = (x^2, 3z, -3y)$. Let $C$ be the curve $\vec r(t) = (3, 4\cos t, 4\sin t)$ for $0\leq t\leq 2\pi$. Let $S$ be the surface parametrized by $\vec r(u,v) = (3, u\cos v, u\sin v)$.
  • Set up and compute the $\ds \int_C Mdx+Ndy+Pdz$.
  • Compute the curl of $\vec F$, so compute $(P_y-N_z,M_z-P_x, N_x-M_y)$.
  • Compute $\ds \iint_S \vec \nabla\times\vec F\cdot \hat n dS$ for the surface $S$ using the orientation $\hat n$ that is compatible with the orientation of $C$.
• Let $R$ be the region in space that lies inside the square with $-2\leq x\leq 2$ and $-2\leq y\leq 2$ and outside the circle $x^2+y^2=1$. Let $C_1$ be the curve that wraps around the square in a counter-clockwise fashion. Let $C_2$ be the curve that wraps around the circle in a counter-clockwise fashion. Let $\vec F (x,y) = \frac{(-y,x)}{x^2+y^2}$. Note that $\vec F$ is not defined at $(0,0)$.
  • Draw the region $R$ and curves $C_1$ and $C_2$.
  • Show that $N_x-M_y=0$ for $(x,y)\neq 0$.
  • Let $D$ be the disc inside $C_2$. Why can we not use Green's theorem to compute $\ds\int_{C_2}Mdx+Ndy$ using $\iint_D N_x-M_ydA$?
  • Compute $\ds\int_{C_2}Mdx+Ndy$ directly.
  • Green's theorem states that $\ds\iint_R N_x-M_y dA = \int_{\partial R}Mdx+Ndy$, provided that $\vec F$ is continuously differentiable along all of the region $R$, and all curves in the boundary $\partial R$ are oriented so that the region $R$ is on the left. Use Green's theorem to explain why $\ds\int_{C_1}Mdx+Ndy = \int_{C_2}Mdx+Ndy$.
  • Let $C_3$ be any simple closed piecewise smooth curve which does not pass through the origin. Explain why $\ds\int_{C_3}Mdx+Ndy = 2\pi$.

Solutions

I'll let you draw the square and circle.

We compute $$\begin{align*} N_x - M_y &= \frac{\partial}{\partial x}(\frac{y}{x^2+y^2}) - \frac{\partial}{\partial y}(\frac{x}{x^2+y^2}) \\ &= \frac{(x^2+y^2)(0)-(-y)(2x)}{(x^2+y^2)^2} - \frac{(x^2+y^2)(0)-(x)(2y)}{(x^2+y^2)^2} \\ &= 0. \end{align*}$$

Note that $\vec F$ is not defined at $(0,0)$, let alone differentiable there. We can't employ Green's theorem, as we would have to integrate $N_x-M_y$ over a region $D$ where $N_x-M_y$ is undefined. The region $R$ conveniently dodges the origin, but has two curves that make up its boundary.

We are on a circle of radius 1, so we know that $x^2+y^2=1$ and hence $F = (-y,x)$ when on this circle. Using the parametrization $\vec r(t) = (\cos t, \sin t)$ for $0\leq t\leq 2\pi$, we have $$\begin{align*} \int_{C_2}Mdx+Ndy &= \int_{C_2} (-y)dx+xdy \\ &= \int_{0}^{2\pi} (-(\sin t))(-\sin t)+(\cos t)(\cos t)dt \\ &= \int_{0}^{2\pi} 1 dt \\ &= 2\pi. \end{align*}$$

Because $N_x-M_y=0$ inside all of the region $R$ (the region $R$ does not contain $(0,0)$), we have that $$\ds\iint_R N_x-M_y dA = \int_{\partial R}Mdx+Ndy.$$ The boundary $\partial R$ consists of the two curves $C_1$ and $C_2$, but to make sure the region $R$ stays on the left side of each curve, we must traverse $C_2$ in a clockwise fashion. This gives $$\ds\iint_R N_x-M_y dA = [\int_{C_1}Mdx+Ndy] + [-\int_{C_2}Mdx+Ndy]. $$ Because $N_x-M_y=0$, then we have $$0 = [\int_{C_1}Mdx+Ndy] + [-\int_{C_2}Mdx+Ndy]. $$ We can rearrange the above to obtain $\int_{C_1}Mdx+Ndy = \int_{C_2}Mdx+Ndy.$

Let $C$ be any circle centered at the origin. The same argument above will show that $\int_{C}Mdx+Ndy = \int_{C_2}Mdx+Ndy=2\pi$. Now let $C_3$ be a simple closed piecewise smooth curve that does not pass through the origin. Pick a curve $C_4$ that is a circle centered around the origin with a radius small enough that the entire curve $C_4$ lies inside $C_3$. Then let $R$ be the region between $C_3$ and $C_4$. Because $N_x-M_y = 0$, Green's theorem again gives $0 = [\int_{C_3}Mdx+Ndy] + [-\int_{C_4}Mdx+Ndy],$ and because $\int_{C_4}Mdx+Ndy=2\pi$, we have $\int_{C_3}Mdx+Ndy]=2\pi$.

Carefully chosing a region $R$ on which $N_x-M_y=0$ was crucial to all the work above. We had to dodge the place where $\vec F$ was undefined. Stokes' theorem and the divergence theorem have similar parallels to this problem, and picking a region where the curl or divergence of a vector field are zero will let you swap out one region for another.

If we're lucky, we can assume a spherical cow (see https://en.wikipedia.org/wiki/Spherical_cow).

Group Problems

1. Consider the vector field $\ds \vec F = \frac{1}{2\pi}\left(\frac{-y}{x^2+y^2},\frac{x}{x^2+y^2}\right)$.
   1. Let $C$ be the curve parametrized by $\vec r(t) = (3\cos t, 3\sin t)$ for $0\leq t\leq 2\pi$. Compute the work done by $\vec F$ to move an object along $C$. [Hint: The answer is NOT zero.]
   2. Let $C$ be the curve parametrized by $\vec r(t) = (7\cos t, 7\sin t)$ for $0\leq t\leq 4\pi$. Compute the work done by $\vec F$ to move an object along $C$.
   3. Let $C$ be the curve parametrized by $\vec r(t) = (a\cos t, a\sin t)$ for $0\leq t\leq 2k\pi$. Compute the work done by $\vec F$ to move an object along $C$.
   4. Show that $D\vec F(x,y)$ is symmetric.
   5. Does $\vec F$ have a potential?

ClearAll[F, r]
F[x_, y_] := 1/(2 Pi (x^2 + y^2)) {-y, x}
r[t_] := {a Cos[t], a Sin[t]}

(*Is the derivative symmetric*)
D[F[x, y], {{x, y}}] // Simplify // MatrixForm

(*Try to find a potential*)
Integrate[F[x, y][[1]], x]
Integrate[F[x, y][[2]], y]

(*Compute work done by wrapping k times around a circle of radius a*)
Dot[F @@ r@t, D[r[t], t]] // Simplify
Integrate[Dot[F @@ r@t, D[r[t], t]], {t, 0, 2 k Pi}]

1. Consider the vector field $\vec F = (x,x-z,y+z)$, the surface $S$ parametrized by $\vec r(u,v)=(u^2, u\cos v, u\sin v)$ for $0\leq u\leq 2$ and $0\leq v\leq 2\pi$, and the curve $C$ parametrized by $\vec r(t) = (4,2\cos t, 2\sin t)$ for $0\leq t\leq 2\pi$.
   1. Draw the surface $S$ and curve $C$. How are these two objects related?
   2. Compute $\vec N = \vec r_u\times \vec r_v$ and determine if $\vec N$ points inward toward the $x$-axis, or outwards away from the $x$-axis.
   3. Set up and compute the integral $\ds \int_C Mdx+Ndy+Pdz$, computing the work done by $\vec F$ along $C$.
   4. Set up and compute $\ds \iint_S \vec \nabla \times \vec F\cdot \hat n dS$, computing the flux of the curl of $\vec F$ across $S$ in the direction $\hat n$ outwards away from the $x$-axis.

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