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Expert-verified Found in: Page 1133 ### Calculus

Book edition 1st
Author(s) Peter Kohn, Laura Taalman
Pages 1155 pages
ISBN 9781429241861 # Find the work done by the vector field$\mathbf{F}\left(x,y\right)=\left(\mathrm{cos}x-3y{e}^{x}\right)\mathbf{i}+\mathrm{sin}x\mathrm{sin}y\mathbf{j}$in moving an object around the periphery of the rectangle with vertices $\left(0,0\right),\left(2,0\right),\left(2,\pi \right)$, and $\left(0,\pi \right)$, starting and ending at $\left(2,\pi \right)$.

Ans: The required work done is $2\mathrm{sin}2+3\pi \left({e}^{2}-1\right)$

See the step by step solution

## Step 1. Given information:

• $\mathbf{F}\left(x,y\right)=\left(\mathrm{cos}x-3y{e}^{x}\right)\mathbf{i}+\mathrm{sin}x\mathrm{sin}y\mathbf{j}$
Moving an object around the periphery of the rectangle with vertices $\left(0,0\right),\left(2,0\right),\left(2,\pi \right)$ and $\left(0,\pi \right)$
• starting and ending at $\left(2,\pi \right)$

## Step 2. Finding the work done by the vector filed:

The work done by a vector field F moving a particle along the curve C is computed by the following line integral:

${\int }_{C}\mathbf{F}·d\mathbf{r}\text{.}$

Hence, evaluate the line integral ${\int }_{C}\mathbf{F}·d\mathbf{r}$ to evaluate the required work done.

## Step 3. Using Green's Theorem to evaluate the required the line integral:

Green's Theorem states that,

"Let R be a region in the plane with smooth boundary curve C oriented counterclockwise by $\mathbf{r}\left(t\right)=⟨\left(x\left(t\right),y\left(t\right)\right)⟩$ for $a\le t\le b$.

If a vector field is defined on R, then,

${\int }_{C}\mathbf{F}·d\mathbf{r}={\iint }_{R}\left(\frac{\partial {F}_{2}}{\partial x}-\frac{\partial {F}_{1}}{\partial y}\right)dA."\dots \dots \left(1\right)$

## Step 4. Finding the vector:

For the vector field $\mathbf{F}\left(x,y\right)=\left(\mathrm{cos}x-3y{e}^{x}\right)\mathbf{i}+\mathrm{sin}x\mathrm{sin}y\mathbf{j}$,

${F}_{1}\left(x,y\right)=\mathrm{cos}x-3y{e}^{x}$ and ${F}_{2}\left(x,y\right)=\mathrm{sin}x\mathrm{sin}y$.

Now, first find $\frac{\partial {F}_{2}}{\partial x}$ and $\frac{\partial {F}_{1}}{\partial y}$.

Then,

$\frac{\partial {F}_{2}}{\partial x}=\frac{\partial }{\partial x}\left(\mathrm{sin}x\mathrm{sin}y\right)\phantom{\rule{0ex}{0ex}}=\mathrm{cos}x\mathrm{sin}y$

and,

$\frac{\partial {F}_{1}}{\partial y}=\frac{\partial }{\partial y}\left(\mathrm{cos}x-3y{e}^{x}\right)\phantom{\rule{0ex}{0ex}}=-3{e}^{x}$

## Step 5. Using Green's Theorem (1) to evaluate the integral:

Now, use Green's Theorem (1) to evaluate the integral ${\int }_{C}\mathbf{F}·d\mathbf{r}$ as follows:

${\int }_{C}\mathbf{F}·d\mathbf{r}={\iint }_{R}\left(\frac{\partial {F}_{2}}{\partial x}-\frac{\partial {F}_{1}}{\partial y}\right)dA\phantom{\rule{0ex}{0ex}}={\iint }_{R}\left(\mathrm{cos}x\mathrm{sin}y-\left(-3{e}^{x}\right)\right)dA\phantom{\rule{0ex}{0ex}}={\iint }_{R}\left(\mathrm{cos}x\mathrm{sin}y+3{e}^{x}\right)dA\dots \dots .\left(2\right)$

## Step 6. Finding the region of integration:

Here, the boundary curve C is a rectangle with vertices $\left(0,0\right),\left(2,0\right),\left(2,\pi \right),\left(0,\pi \right)$,

starting and ending at $\left(2,\pi \right)$.

So the region R bounded by this rectangle is shown in the following figure. Viewed it as an x - or y-simple region, then the region of integration will be,

$R=\left\{\left(x,y\right)\mid 0\le x\le 2,0\le y\le \pi \right\}$.

## Step 7. Evaluating the integral (2):

Then, evaluate the integral (2) as follows:

${\int }_{C}\mathbf{F}·d\mathbf{r}={\int }_{R}\left(\mathrm{cos}x\mathrm{sin}y+3{e}^{x}\right)dA\phantom{\rule{0ex}{0ex}}={\int }_{0}^{\pi }{\int }_{0}^{2}\left(\mathrm{cos}x\mathrm{sin}y+3{e}^{x}\right)dxdy\phantom{\rule{0ex}{0ex}}={\int }_{0}^{\pi }\left[{\int }_{0}^{2}\left(\mathrm{cos}x\mathrm{sin}y+3{e}^{x}\right)dx\right]dy\phantom{\rule{0ex}{0ex}}={\int }_{0}^{\pi }{\left[\mathrm{sin}x\mathrm{sin}y+3{e}^{x}\right]}_{0}^{2}dy\phantom{\rule{0ex}{0ex}}={\int }_{0}^{\pi }\left[\left(\mathrm{sin}2\mathrm{sin}y+3{e}^{2}\right)-\left(\mathrm{sin}0\mathrm{sin}y+3{e}^{0}\right)\right]dy={\int }_{0}^{\pi }\left[\left(\mathrm{sin}2\mathrm{sin}y+3{e}^{2}\right)-\left(0·\mathrm{sin}y+3·1\right)\right]dy\phantom{\rule{0ex}{0ex}}={\int }_{0}^{\pi }\left(\mathrm{sin}2\mathrm{sin}y+3{e}^{2}-3\right)dy\phantom{\rule{0ex}{0ex}}={\left[-\mathrm{sin}2\mathrm{cos}y+3{e}^{2}y-3y\right]}_{0}^{\pi }\phantom{\rule{0ex}{0ex}}=\left(-\mathrm{sin}2·\left(-1\right)+3{e}^{2}\pi -3\pi \right)-\left(-\mathrm{sin}2\mathrm{cos}0+3{e}^{2}·0-3·0\right)\phantom{\rule{0ex}{0ex}}=\left(\mathrm{sin}2+3\pi \left({e}^{2}-1\right)\right)-\left(-\mathrm{sin}2·1+0\right)\phantom{\rule{0ex}{0ex}}=\mathrm{sin}2+3\pi \left({e}^{2}-1\right)+\mathrm{sin}2\phantom{\rule{0ex}{0ex}}=2\mathrm{sin}2+3\pi \left({e}^{2}-1\right)$

Therefore, the required work done is $2\mathrm{sin}2+3\pi \left({e}^{2}-1\right)$. ### Want to see more solutions like these? 