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Answers without the blur. Sign up and see all textbooks for free! Q. 47

Expert-verified Found in: Page 1132 ### Calculus

Book edition 1st
Author(s) Peter Kohn, Laura Taalman
Pages 1155 pages
ISBN 9781429241861 # Find the work done by the vector field$\mathbf{F}\left(x,y\right)=\left(\mathrm{cos}\left({x}^{2}\right)+4x{y}^{2}\right)\mathbf{i}+\left({2}^{y}-4{x}^{2}y\right)\mathbf{j}$in moving an object around the unit circle, starting and ending at $\left(1,0\right)$.

Ans: The required work done is $-2\pi$.

See the step by step solution

## Step 1. Given information:

$\mathbf{F}\left(x,y\right)=\left(\mathrm{cos}\left({x}^{2}\right)+4x{y}^{2}\right)\mathbf{i}+\left({2}^{y}-4{x}^{2}y\right)\mathbf{j}$

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

The work done by a vector field $\mathbf{F}$ 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,

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

## Step 4. Finding the vector:

For the vector field $\mathbf{F}\left(x,y\right)=\left(\mathrm{cos}\left({x}^{2}\right)+4x{y}^{2}\right)\mathbf{i}+\left({2}^{y}-4{x}^{2}y\right)\mathbf{j}$, ${F}_{1}\left(x,y\right)=\mathrm{cos}\left({x}^{2}\right)+4x{y}^{2}$ and ${F}_{2}\left(x,y\right)={2}^{y}-4{x}^{2}y.$

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

Then,

$\frac{\partial {F}_{2}}{\partial x}=\frac{\partial }{\partial x}\left({2}^{y}-4{x}^{2}y\right)\phantom{\rule{0ex}{0ex}}=-8xy$

and,

$\frac{\partial {F}_{1}}{\partial y}=\frac{\partial }{\partial y}\left(\mathrm{cos}\left({x}^{2}\right)+4x{y}^{2}\right)\phantom{\rule{0ex}{0ex}}=8xy$

## 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:

${\int }_{C}\mathbf{F}·d\mathbf{r}={\iint }_{R}\left(\frac{\partial {F}_{2}}{\partial x}-\frac{\partial {F}_{1}}{\partial y}\right)dA\phantom{\rule{0ex}{0ex}}={\iint }_{R}\left(-8xy-8xy\right)dA\phantom{\rule{0ex}{0ex}}={\iint }_{R}\left(-16xy\right)dA\dots \dots \left(2\right)$

## Step 6. Changing this integral (2) to polar coordinates:

Here, the boundary curve C is a unit circle, starting and ending at $\left(1,0\right)$. In polar coordinates, the region of integration is described as follows,

$R=\left\{\left(r,\theta \right)\mid 0\le r\le 1,0\le \theta \le 2\pi \right\}.$

In this case,

$x=r\mathrm{cos}\theta ,y=r\mathrm{sin}\theta ,{x}^{2}+{y}^{2}={r}^{2},\text{and}\phantom{\rule{0ex}{0ex}}dA=rdrd\theta$

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

Then, evaluate the integral (2) as follows:

${\int }_{C}\mathbf{F}·d\mathbf{r}={\iint }_{R}\left(-16xy\right)dA\phantom{\rule{0ex}{0ex}}={\int }_{0}^{2\pi }{\int }_{0}^{1}-16\left(r\mathrm{cos}\theta \right)\left(r\mathrm{sin}\theta \right)rdrd\theta \phantom{\rule{0ex}{0ex}}={\int }_{0}^{2\pi }{\int }_{0}^{1}-16{r}^{3}\mathrm{sin}\theta \mathrm{cos}\theta drd\theta \phantom{\rule{0ex}{0ex}}={\int }_{0}^{2\pi }\mathrm{sin}\theta \mathrm{cos}\theta \left[{\int }_{0}^{1}-16{r}^{3}dr\right]d\theta \phantom{\rule{0ex}{0ex}}={\int }_{0}^{2\pi }\mathrm{sin}\theta \mathrm{cos}\theta {\left[-4{r}^{4}\right]}_{0}^{1}d\theta \phantom{\rule{0ex}{0ex}}={\int }_{0}^{2\pi }\mathrm{sin}\theta \mathrm{cos}\theta \left[\left(-4·{1}^{4}\right)-\left(-4·{0}^{4}\right)\right]d\theta \phantom{\rule{0ex}{0ex}}={\int }_{0}^{2\pi }-4\mathrm{sin}\theta \mathrm{cos}\theta d\theta \phantom{\rule{0ex}{0ex}}={\left[2{\mathrm{cos}}^{2}\theta \right]}_{0}^{2\pi }\phantom{\rule{0ex}{0ex}}=2{\mathrm{cos}}^{2}2\pi -2{\mathrm{cos}}^{2}0\phantom{\rule{0ex}{0ex}}=2\left(1{\right)}^{2}-2\left(1{\right)}^{2}\phantom{\rule{0ex}{0ex}}=0.$
Therefore, the required work done is 0.

## Step 8. Correcting the Problem of the textbook:

In Book, this problem is given WRONG.

Correct Problem:

Consider the following vector field:

$\mathbf{F}\left(x,y\right)=\left(\mathrm{cos}\left({x}^{2}\right)+4{x}^{2}y\right)\mathbf{i}+\left({2}^{y}-{4}{x}{{y}}^{{2}}\right)\mathbf{j}$

## Step 9. Rewriting the Step (4) with the correct problem by finding the vector:

For the vector field $\mathbf{F}\left(x,y\right)=\left(\mathrm{cos}\left({x}^{2}\right)+4{x}^{2}y\right)\mathbf{i}+\left({2}^{y}-4x{y}^{2}\right)\mathbf{j}$

${F}_{1}\left(x,y\right)=\mathrm{cos}\left({x}^{2}\right)+4{x}^{2}y$and ${F}_{2}\left(x,y\right)={2}^{y}-4x{y}^{2}$

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

Then,

$\frac{\partial {F}_{2}}{\partial x}=\frac{\partial }{\partial x}\left({2}^{y}-4x{y}^{2}\right)\phantom{\rule{0ex}{0ex}}=-4{y}^{2}$

and,

$\frac{\partial {F}_{1}}{\partial y}=\frac{\partial }{\partial y}\left(\mathrm{cos}\left({x}^{2}\right)+4{x}^{2}y\right)\phantom{\rule{0ex}{0ex}}=4{x}^{2}$

## Step 10. Rewriting the Step (5) with the correct problem by 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:

${\int }_{C}\mathbf{F}·d\mathbf{r}={\iint }_{R}\left(\frac{\partial {F}_{2}}{\partial x}-\frac{\partial {F}_{1}}{\partial y}\right)dA\phantom{\rule{0ex}{0ex}}={\iint }_{R}\left(-4{y}^{2}-4{x}^{2}\right)dA\phantom{\rule{0ex}{0ex}}={\iint }_{R}-4\left({x}^{2}+{y}^{2}\right)dA\dots \dots \left(3\right)$

## Step 11. Rewriting the Step (6) with the correct problem by  Changing this integral (3) to polar coordinates:

Change this integral (3) to polar coordinates, and integrate it.

Here, the boundary curve C is a unit circle, starting and ending at $\left(1,0\right)$. In polar coordinates, the region of integration is described as follows,

$R=\left\{\left(r,\theta \right)\mid 0\le r\le 1,0\le \theta \le 2\pi \right\}.$

In this case,

$x=r\mathrm{cos}\theta ,y=r\mathrm{sin}\theta ,{x}^{2}+{y}^{2}={r}^{2},\text{and}\phantom{\rule{0ex}{0ex}}dA=rdrd\theta$

## Step 12. Rewriting the Step (7) with the correct problem by evaluating the integral (3):

Then, evaluate the integral (3) as follows:

${\int }_{C}\mathbf{F}·d\mathbf{r}={\iint }_{R}-4\left({x}^{2}+{y}^{2}\right)dA\phantom{\rule{0ex}{0ex}}={\int }_{0}^{2\pi }{\int }_{0}^{1}-4{r}^{2}·rdrd\theta \phantom{\rule{0ex}{0ex}}={\int }_{0}^{2\pi }{\int }_{0}^{1}-4{r}^{3}drd\theta \phantom{\rule{0ex}{0ex}}={\int }_{0}^{2\pi }\left[{\int }_{0}^{1}-4{r}^{3}dr\right]d\theta \phantom{\rule{0ex}{0ex}}={\int }_{0}^{2\pi }{\left[-{r}^{4}\right]}_{0}^{1}d\theta \phantom{\rule{0ex}{0ex}}={\int }_{0}^{2\pi }\left[\left(-{1}^{4}\right)-\left(-{0}^{4}\right)\right]d\theta \phantom{\rule{0ex}{0ex}}={\int }_{0}^{2\pi }-1d\theta \phantom{\rule{0ex}{0ex}}=\left[-\theta {\right]}_{0}^{2\pi }\phantom{\rule{0ex}{0ex}}=\left(-2\pi \right)-\left(-0\right)\phantom{\rule{0ex}{0ex}}=-2\pi .$

Therefore, the required work done is $-2\pi$. ### Want to see more solutions like these? 