91Ó°ÊÓ

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

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

${∫}_C\mathbf{F}·d\mathbf{r}\text{.}$

Hence, evaluate the line integral ${∫}_C\mathbf{F}·d\mathbf{r}$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 $\mathbf{r}\left(t\right)=⟨\left(x\right(t),y(t\left)\right)âŸ\copyright$for $a≤t≤b$.

If a vector field is defined on R, then,

For the vector field $\mathbf{F}(x,y)=\left(\cos \left(x^2\right)+4x{y}^2\right)\mathbf{i}+\left(2^y-4x^{2}y\right)\mathbf{j}$, $F_1(x,y)=\cos \left(x^2\right)+4x{y}^2$and $F_2(x,y)=2^y-4x^{2}y.$

Now, first, find $\frac{∂F_2}{∂x}$and $\frac{∂F_1}{∂y}$.

Then,

$$\begin{gathered} \frac{∂F_2}{∂x}=\frac{∂}{∂x}\left(2^y-4x^{2}y\right) \\ =-8xy \end{gathered}$$

and,

$$\begin{gathered} \frac{∂F_1}{∂y}=\frac{∂}{∂y}\left(\cos \left(x^2\right)+4x{y}^2\right) \\ =8xy \end{gathered}$$

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

$$\begin{gathered} {∫}_C\mathbf{F}·d\mathbf{r}={âˆ\neg }_R\left(\frac{∂F_2}{∂x}-\frac{∂F_1}{∂y}\right)dA \\ ={âˆ\neg }_R(-8xy-8xy)dA \\ ={âˆ\neg }_R(-16xy)dA……\left(2\right) \end{gathered}$$

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,

$R=\left\{\right(r,\mathrm{θ})âˆ\pounds 0≤r≤1,0≤\mathrm{θ}≤2\mathrm{Ï€}\}.$

In this case,

$$\begin{gathered} x=r\cos \mathrm{θ},y=r\sin \mathrm{θ},x^2+y^2=r^2,\text{and} \\ dA=rdrd\mathrm{θ} \end{gathered}$$

Then, evaluate the integral (2) as follows:

$$\begin{gathered} {∫}_C\mathbf{F}·d\mathbf{r}={âˆ\neg }_R(-16xy)dA \\ ={∫}_0^{2\mathrm{Ï€}}{∫}_0^1-16\left(r\cos \mathrm{θ}\right)\left(r\sin \mathrm{θ}\right)rdrd\mathrm{θ} \\ ={∫}_0^{2\mathrm{Ï€}}{∫}_0^1-16r^3\sin \mathrm{θ}\cos \mathrm{θ}drd\mathrm{θ} \\ ={∫}_0^{2\mathrm{Ï€}}\sin \mathrm{θ}\cos \mathrm{θ}\left[{∫}_0^1-16r^{3}dr\right]d\mathrm{θ} \\ ={∫}_0^{2\mathrm{Ï€}}\sin \mathrm{θ}\cos \mathrm{θ}{\left[-4r^4\right]}_0^{1}d\mathrm{θ} \\ ={∫}_0^{2\mathrm{Ï€}}\sin \mathrm{θ}\cos \mathrm{θ}\left[\left(-4·1^4\right)-\left(-4·0^4\right)\right]d\mathrm{θ} \\ ={∫}_0^{2\mathrm{Ï€}}-4\sin \mathrm{θ}\cos \mathrm{θ}d\mathrm{θ} \\ ={\left[2{\cos }^2\mathrm{θ}\right]}_0^{2\mathrm{Ï€}} \\ =2{\cos }^{2}2\mathrm{Ï€}-2{\cos }^{2}0 \\ =2(1{)}^2-2(1{)}^2 \\ =0. \end{gathered}$$

Therefore, the required work done is 0.

In Book, this problem is given WRONG.

Correct Problem:

Consider the following vector field:

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

$F_1(x,y)=\cos \left(x^2\right)+4x^{2}y$and $F_2(x,y)=2^y-4x{y}^2$

Now, first find $\frac{∂F_2}{∂x}$and $\frac{∂F_1}{∂y}$.

Then,

$$\begin{gathered} \frac{∂F_2}{∂x}=\frac{∂}{∂x}\left(2^y-4x{y}^2\right) \\ =-4y^2 \end{gathered}$$

and,

$$\begin{gathered} \frac{∂F_1}{∂y}=\frac{∂}{∂y}\left(\cos \left(x^2\right)+4x^{2}y\right) \\ =4x^2 \end{gathered}$$

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

$$\begin{gathered} {∫}_C\mathbf{F}·d\mathbf{r}={âˆ\neg }_R\left(\frac{∂F_2}{∂x}-\frac{∂F_1}{∂y}\right)dA \\ ={âˆ\neg }_R\left(-4y^2-4x^2\right)dA \\ ={âˆ\neg }_R-4\left(x^2+y^2\right)dA……\left(3\right) \end{gathered}$$

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,

$R=\left\{\right(r,\mathrm{θ})âˆ\pounds 0≤r≤1,0≤\mathrm{θ}≤2\mathrm{Ï€}\}.$

In this case,

$$\begin{gathered} x=r\cos \mathrm{θ},y=r\sin \mathrm{θ},x^2+y^2=r^2,\text{and} \\ dA=rdrd\mathrm{θ} \end{gathered}$$

Then, evaluate the integral (3) as follows:

$$\begin{gathered} {∫}_C\mathbf{F}·d\mathbf{r}={âˆ\neg }_R-4\left(x^2+y^2\right)dA \\ ={∫}_0^{2\mathrm{Ï€}}{∫}_0^1-4r^2·rdrd\mathrm{θ} \\ ={∫}_0^{2\mathrm{Ï€}}{∫}_0^1-4r^{3}drd\mathrm{θ} \\ ={∫}_0^{2\mathrm{Ï€}}\left[{∫}_0^1-4r^{3}dr\right]d\mathrm{θ} \\ ={∫}_0^{2\mathrm{Ï€}}{\left[-r^4\right]}_0^{1}d\mathrm{θ} \\ ={∫}_0^{2\mathrm{Ï€}}\left[\left(-1^4\right)-\left(-0^4\right)\right]d\mathrm{θ} \\ ={∫}_0^{2\mathrm{Ï€}}-1d\mathrm{θ} \\ =[-\mathrm{θ}{]}_0^{2\mathrm{Ï€}} \\ =(-2\mathrm{Ï€})-(-0) \\ =-2\mathrm{Ï€}. \end{gathered}$$

Therefore, the required work done is $-2\mathrm{Ï€}$.