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Q. 47

Expert-verifiedFound in: Page 1132

Book edition
1st

Author(s)
Peter Kohn, Laura Taalman

Pages
1155 pages

ISBN
9781429241861

Find the work done by the vector field

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

in moving an object around the unit circle, starting and ending at $(1,0)$.

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

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

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

${\int}_{C}\mathbf{F}\xb7d\mathbf{r}\text{.}$

Hence, evaluate the line integral ${\int}_{C}\mathbf{F}\xb7d\mathbf{r}$ to evaluate the required work done.

Green's Theorem states that,

"Let R be a region in the plane with smooth boundary curve C oriented counterclockwise by $\mathbf{r}\left(t\right)=\u27e8\left(x\right(t),y(t\left)\right)\u27e9$ for $a\le t\le b$.

If a vector field $\mathbf{F}(x,y)=>">{F}_{1}(x,y),{F}_{2}(x,y)$ is defined on R, then,

${\int}_{C}\mathbf{F}\xb7d\mathbf{r}={\iint}_{R}\left(\frac{\partial {F}_{2}}{\partial x}-\frac{\partial {F}_{1}}{\partial y}\right)dA."\dots ..\left(1\right)$For the vector field $\mathbf{F}(x,y)=\left(\mathrm{cos}\left({x}^{2}\right)+4x{y}^{2}\right)\mathbf{i}+\left({2}^{y}-4{x}^{2}y\right)\mathbf{j}$, ${F}_{1}(x,y)=\mathrm{cos}\left({x}^{2}\right)+4x{y}^{2}$ and ${F}_{2}(x,y)={2}^{y}-4{x}^{2}y.$

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,

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

Now, use Green's Theorem (1) to evaluate the integral ${\int}_{C}\mathbf{F}\xb7d\mathbf{r}$ as follows:

${\int}_{C}\mathbf{F}\xb7d\mathbf{r}={\iint}_{R}\left(\frac{\partial {F}_{2}}{\partial x}-\frac{\partial {F}_{1}}{\partial y}\right)dA\phantom{\rule{0ex}{0ex}}={\iint}_{R}(-8xy-8xy)dA\phantom{\rule{0ex}{0ex}}={\iint}_{R}(-16xy)dA\dots \dots \left(2\right)$

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

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

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 $

Then, evaluate the integral (2) as follows:

${\int}_{C}\mathbf{F}\xb7d\mathbf{r}={\iint}_{R}(-16xy)dA\phantom{\rule{0ex}{0ex}}={\int}_{0}^{2\pi}{\int}_{0}^{1}-16\left(r\mathrm{cos}\theta \right)\left(r\mathrm{sin}\theta \right)rdrd\theta \phantom{\rule{0ex}{0ex}}={\int}_{0}^{2\pi}{\int}_{0}^{1}-16{r}^{3}\mathrm{sin}\theta \mathrm{cos}\theta drd\theta \phantom{\rule{0ex}{0ex}}={\int}_{0}^{2\pi}\mathrm{sin}\theta \mathrm{cos}\theta \left[{\int}_{0}^{1}-16{r}^{3}dr\right]d\theta \phantom{\rule{0ex}{0ex}}={\int}_{0}^{2\pi}\mathrm{sin}\theta \mathrm{cos}\theta {\left[-4{r}^{4}\right]}_{0}^{1}d\theta \phantom{\rule{0ex}{0ex}}={\int}_{0}^{2\pi}\mathrm{sin}\theta \mathrm{cos}\theta \left[\left(-4\xb7{1}^{4}\right)-\left(-4\xb7{0}^{4}\right)\right]d\theta \phantom{\rule{0ex}{0ex}}={\int}_{0}^{2\pi}-4\mathrm{sin}\theta \mathrm{cos}\theta d\theta \phantom{\rule{0ex}{0ex}}={\left[2{\mathrm{cos}}^{2}\theta \right]}_{0}^{2\pi}\phantom{\rule{0ex}{0ex}}=2{\mathrm{cos}}^{2}2\pi -2{\mathrm{cos}}^{2}0\phantom{\rule{0ex}{0ex}}=2(1{)}^{2}-2(1{)}^{2}\phantom{\rule{0ex}{0ex}}=0.$

Therefore, the required work done is 0.

In Book, this problem is given WRONG.

Correct Problem:

Consider the following vector field:

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

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

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,

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

Now, use Green's Theorem (1) to evaluate the integral ${\int}_{C}\mathbf{F}\xb7d\mathbf{r}$ as follows:

${\int}_{C}\mathbf{F}\xb7d\mathbf{r}={\iint}_{R}\left(\frac{\partial {F}_{2}}{\partial x}-\frac{\partial {F}_{1}}{\partial y}\right)dA\phantom{\rule{0ex}{0ex}}={\iint}_{R}\left(-4{y}^{2}-4{x}^{2}\right)dA\phantom{\rule{0ex}{0ex}}={\iint}_{R}-4\left({x}^{2}+{y}^{2}\right)dA\dots \dots \left(3\right)$

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

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

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

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 $

Then, evaluate the integral (3) as follows:

${\int}_{C}\mathbf{F}\xb7d\mathbf{r}={\iint}_{R}-4\left({x}^{2}+{y}^{2}\right)dA\phantom{\rule{0ex}{0ex}}={\int}_{0}^{2\pi}{\int}_{0}^{1}-4{r}^{2}\xb7rdrd\theta \phantom{\rule{0ex}{0ex}}={\int}_{0}^{2\pi}{\int}_{0}^{1}-4{r}^{3}drd\theta \phantom{\rule{0ex}{0ex}}={\int}_{0}^{2\pi}\left[{\int}_{0}^{1}-4{r}^{3}dr\right]d\theta \phantom{\rule{0ex}{0ex}}={\int}_{0}^{2\pi}{\left[-{r}^{4}\right]}_{0}^{1}d\theta \phantom{\rule{0ex}{0ex}}={\int}_{0}^{2\pi}\left[\left(-{1}^{4}\right)-\left(-{0}^{4}\right)\right]d\theta \phantom{\rule{0ex}{0ex}}={\int}_{0}^{2\pi}-1d\theta \phantom{\rule{0ex}{0ex}}=[-\theta {]}_{0}^{2\pi}\phantom{\rule{0ex}{0ex}}=(-2\pi )-(-0)\phantom{\rule{0ex}{0ex}}=-2\pi .$

Therefore, the required work done is $-2\pi $.

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