Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 16 - Vector Calculus - 16.8 Stokes' Theorem - 16.8 Exercises: 17

Answer

$3$

Work Step by Step

1) Find Curl of $\vec{F}$ $$ \text{curl}\vec{F}=\vec{\nabla}\times\vec{F} $$ $$ =\begin{vmatrix} \hat{\text{i}} & \hat{\text{j}} & \hat{\text{k}}\\[1pt] \frac{\partial}{\partial{x}} & \frac{\partial}{\partial{y}} & \frac{\partial}{\partial{z}}\\[2pt] z^2 & 2xy & 4y^2 \end{vmatrix}= \begin{vmatrix} \frac{\partial}{\partial{y}} & \frac{\partial}{\partial{z}}\\[2pt] 2xy & 4y^2 \end{vmatrix} \hat{\text{i}}- \begin{vmatrix} \frac{\partial}{\partial{x}} & \frac{\partial}{\partial{z}}\\[2pt] z^2 & 4y^2 \end{vmatrix} \hat{\text{j}}+ \begin{vmatrix} \frac{\partial}{\partial{x}} & \frac{\partial}{\partial{y}}\\[2pt] z^2 & 2xy \end{vmatrix} \hat{\text{k}} $$ $$ =(8y-0)\hat{\text{i}}-(0-2z)\hat{\text{j}}+(2y-0)\hat{\text{k}} =8y\hat{\text{i}}-2z\hat{\text{j}}+2y\hat{\text{k}} $$ $$ \therefore \text{curl}\vec{F}=8y\hat{\text{i}}+2z\hat{\text{j}}+2y\hat{\text{k}} $$ 2) Define and Parametrize some Surface $S$ with Boundary of $C$ We are free to choose any surface with its boundaries defined by the given path $C$, the positively oriented closed curve from $O$ to $P, Q, R\ \text{and back to}\ O$. We will define a plane $S$ such that $S$ passes through all of the given points. The vectors $$\vec{OP}=\hat{\text{i}}+0\hat{\text{j}}+0\hat{\text{k}}\ \text{and}\ \vec{PQ}=0\hat{\text{i}} + 2\hat{\text{j}}+\hat{\text{k}}$$ both lie in $S$, so $$ \vec{OP}\times\vec{PQ}=\vec{n}. $$ Computing $\vec{OP}\times\vec{PQ}$, $$ \vec{OP}\times\vec{PQ}= \begin{vmatrix} \hat{\text{i}} & \hat{\text{j}} & \hat{\text{k}}\\[1pt] 1 & 0 & 0\\ 0 & 2 & 1\\ \end{vmatrix}= \begin{vmatrix} 0 & 0\\ 2 & 1 \end{vmatrix} \hat{\text{i}}- \begin{vmatrix} 1 & 0\\ 0 & 1 \end{vmatrix} \hat{\text{j}}+ \begin{vmatrix} 1 & 0\\ 0 & 2 \end{vmatrix} \hat{\text{k}} $$ $$ \therefore \vec{n}=-\hat{\text{j}}+2\hat{\text{k}} $$ Note that the equation of a plane $S$ with normal vector $-\hat{\text{j}}+2\hat{\text{k}}$ is $$ 0=-y+2z \Rightarrow S=z=\frac{1}{2}y $$ Now that $S$ is represented in terms of two variables $x$ and $y$, we can find a parametric representation and a vector representation of $S$, $\vec{r}$(x,y). $$ S=\Big\{(x,y,z)\ |\ x=x,\ y=y,\ z=\frac{1}{2}y\ |\ 0 \leq x \leq 1,\ 0 \leq y \leq 2 \Big\} $$ $$ \vec{r}(x,y)=x\hat{\text{i}}+y\hat{\text{j}}+\frac{1}{2}y\hat{\text{k}} $$ 3) Use Stokes' Theorem to Find Work Done Recall that $$ \text{work}=\oint_C \vec{F}\cdot d\vec{r} $$ Therefore, from Stokes' Theorem, $$ \text{work}=\iint \limits_S (\text{curl}\vec{F}(x,y,z)) \cdot\vec{n} dS= \iint \limits_S (\text{curl}\vec{F} \cdot \vec{n}) dS= $$ $$ \iint \limits_R (\text{curl} \vec{F}(x,y) \cdot (\vec{r}_x \times \vec{r}_y)) dydx $$ (where $R$ is the projection of $S$ in the $xy$-plane) Parametrize $\text{curl}\vec{F}$ according to the parametrization of $S$ to find $\text{curl}\vec{F}(x,y)$ $$ \text{curl}\vec{F}(x,y)=8y\hat{\text{i}}+y\hat{\text{j}}+2y\hat{\text{k}} $$ \noindent Compute $\text{curl} \vec{F}(x,y) \cdot (\vec{r}_x \times \vec{r}_y)$ $$ \vec{r}_x=\hat{\text{i}}+0\hat{\text{j}}+0\hat{\text{k}} $$ $$ \vec{r}_y=0\hat{\text{i}}+\hat{\text{j}}+\frac{1}{2}\hat{\text{k}} $$ $$ \vec{r}_x \times \vec{r}_y = \begin{vmatrix} \hat{\text{i}} & \hat{\text{j}} & \hat{\text{k}}\\ 1 & 0 & 0\\ 0 & 1 & \frac{1}{2} \end{vmatrix}= \begin{vmatrix} 0 & 0\\ 1 & \frac{1}{2} \end{vmatrix}\hat{\text{i}}- \begin{vmatrix} 1 & 0\\ 0 & \frac{1}{2} \end{vmatrix}\hat{\text{j}}+ \begin{vmatrix} 1 & 0\\ 0 & 1 \end{vmatrix}\hat{\text{k}} $$ $$ \therefore \vec{r}_x \times \vec{r}_y=0\hat{\text{i}}-\frac{1}{2}\hat{\text{j}}+\hat{\text{k}} $$ $$ \text{curl} \vec{F}(x,y) \cdot (\vec{r}_x \times \vec{r}_y)=(8y\hat{\text{i}}+y\hat{\text{j}}+2y\hat{\text{k}}) \cdot (0\hat{\text{i}}-\frac{1}{2}\hat{\text{j}}+\hat{\text{k}}) $$ $$ \therefore \text{curl} \vec{F}(x,y) \cdot (\vec{r}_x \times \vec{r}_y)=0-\frac{1}{2}y+2y=\frac{3}{2}y $$ Plug $\text{curl} \vec{F}(x,y) \cdot (\vec{r}_x \times \vec{r}_y)$ into our work equation from earlier: $$ \text{work}=\int_0^1 \int_0^2 \frac{3}{2}y\ dydx=\frac{3}{2}\int_0^2 y\ dy=\frac{3}{4}\Big[y^2\Big]_0^2 $$ $$ \therefore \text{work}=3 $$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.