Rapid Recall
- Consider the vector field $\vec F(x,y,z) = (M,N,P)$, where each of $M$, $N$, and $P$ are functions of $x$, $y$, and $z$. Compute $D\vec F(x,y,z)$ symbolically, filling the 3 by 3 matrix with appropriate partials of $M$, $N$, and $P$.
Solution
The solution is $$D\vec F = \begin{bmatrix} \begin{matrix}M_x\\N_x\\P_x\end{matrix} &\begin{matrix}M_y\\N_y\\P_y\end{matrix} &\begin{matrix}M_z\\N_z\\P_z\end{matrix} \end{bmatrix}$$
- Consider the surface $\vec r(u,v) = (2u+3v,4u+5v,6u+7v)$. The differential is $$d\vec r=\begin{pmatrix}2\\4\\6\end{pmatrix}du+\begin{pmatrix}3\\5\\7\end{pmatrix}dv.$$ Find the area of the parallelogram with edges $(2,4,6)du$ and $(3,5,7)dv$.
Solution
The magnitude of the cross product of these two vectors is $$\begin{align} |(2,4,6)du\times(3,5,7)dv| &=\left|(2,4,6)\times(3,5,7)\right|dudv\\ &=\left|(4(7)-5(6),3(6)-2(7),2(5)-3(4))\right|dudv\\ &=\left|(-2,4,-2)\right|dudv\\ &=\sqrt{4+16+4}dudv\\ &=\sqrt{24}dudv \end{align}$$
- For a vector field $\vec F(x,y)=(M,N)$ that is continuously differentiable everywhere, Green's theorem states $$\oint_C Mdx+Ndy = \iint_R N_x-M_ydA,$$ where $R$ is region inside a simple closed piecewise smooth curve $C$ that traverses around the boundary of $R$ in a counterclockwise fashion. Let $\vec F = (3x+4y,6x+7y)$ and let $C$ be the curve which starts at $(1,-3)$, travels right to $(5,-3)$, up to $(5,7)$, left to $(1,7)$, and then back down to $(1,-3)$. Find the work done by $\vec F$ along $C$.
Solution
Note that $N_x-M_y=6-4=2$. By Green's theorem, the requested integral is the same as $\iint_R2dA=2A$, twice the area of the rectangle $R$ inside $C$. The width of this rectangle is 4, and the height is 10, so the answer is $W=2A = 2(40)=80$.
Group problems
- Consider the vector field $\vec F = (M,N,P)$. Compute each of the following symbolically, or explain why it is impossible.
- $\vec \nabla \cdot \vec F$
- $\vec \nabla\times \vec F$
- $\vec \nabla\times(\vec \nabla \cdot \vec F)$
- $\vec \nabla\cdot (\vec \nabla \times \vec F)$
- $\vec \nabla\cdot (\vec \nabla \cdot \vec F)$
- $\vec \nabla\times(\vec \nabla \times \vec F)$
- The conclusion of Green's Theorem is that $ \oint_{C} M dx+Ndy=\iint_R N_x-M_y dA$. In which of the above computations do you find this quantity?
- Compute the work done by $\vec F = (-3y,3x)$ to move an object counterclockwise once along the circle $\vec r(t) = (5\cos t, 5\sin t).$
- Compute this work using $ \oint_{C} M dx+Ndy$.
- Compute this work using $\iint_R N_x-M_y dA$.
- Compute the work done by $\vec F = (2x-y,2x+4y)$ to move an object counterclockwise once along the triangle with corners $(0,0)$, $(2,0)$, and $(0,3)$.
- Set up the single double integral $\iint_R N_x-M_y dA$.
- Set up three integrals needed using $ \oint_{C} M dx+Ndy$.
- Compute one of these integrals.
- Consider the surface parametrized by $\vec r(u,v) = (u\cos v, u\sin v, u^2+v^2)$ for $-3\leq u\leq 3$ and $0\leq v\leq 3$.
- Draw the surface.
- Compute $d\sigma = \left |\dfrac{\partial \vec r}{\partial u}\times\dfrac{\partial \vec r}{\partial u}\right|dudv$.
- Set up an integral formula to compute $\bar z$ for this surface.
- Draw each curve or surface given below.
- $\vec r(u,v) = (4\cos u,v, 3\sin u)$ for $0\leq u\leq \pi$ and $0\leq v\leq 7$.
- $\vec r(t) = (3\cos t,3\sin t,t)$ for $0\leq t\leq 6\pi$.
- $\vec r(u,v) = (u\cos v,u\sin v,v)$ for $0\leq v\leq 6\pi$ and $2\leq u\leq 4$.
- $\vec r(t) = (0,t,9-t^2)$ for $0\leq t\leq 3$.
- $\vec r(x,y) = (x,y,9-x^2-y^2)$ for for $0\leq x \leq 3$ and $-3\leq y\leq 3$.
- $\vec r(u,v) = (u\cos v,u\sin v,9-u^2)$ for $0\leq u\leq 3$ and $0\leq v\leq 2\pi$.
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