91Ó°ÊÓ

- The current through a certain passage is the following vector field;

- Consider a disk R of radius 1 mile and centered on this region.

- The boundary of this disk is denoted by $∂R$.

${âˆ\neg }_R\left(\frac{∂F_2}{∂x}-\frac{∂F_1}{∂y}\right)dA$.

- ${∫}_{∂\mathcal{R}}F·nds={∫}_0^{2\mathrm{π}}\left(1.152-0.8{\cos }^2\mathrm{θ}\right)\sin \mathrm{θ}d\mathrm{θ}.$

(a). For the vector field ,$F_1(x,y)=0$and $F_2(x,y)=1.152-0.8x^2.$

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

Then,

$$\begin{gathered} \frac{∂F_2}{∂x}=\frac{∂}{∂x}\left(1.152-0.8x^2\right) \\ =-1.6x \end{gathered}$$

and,

$$\begin{gathered} \frac{∂F_1}{∂y}=\frac{∂}{∂y}\left(0\right) \\ =0. \end{gathered}$$

Here, the boundary $∂R$is a boundary of a disk R of radius 1 mile and centered on this region. 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,

$x=r\cos \mathrm{θ},y=r\sin \mathrm{θ},x^2+y^2=r^2$ and

$dA=rdrd\mathrm{θ}$

Then, evaluate the required integral as follows:

$$\begin{gathered} {âˆ\neg }_R\left(\frac{∂F_2}{∂x}-\frac{∂F_1}{∂y}\right)dA \\ ={âˆ\neg }_R(-1.6x-0)dA \\ ={âˆ\neg }_R-1.6xdA \\ ={∫}_0^{2\mathrm{Ï€}}{∫}_0^1-1.6\left(r\cos \mathrm{θ}\right)rdrd\mathrm{θ} \\ ={∫}_0^{2\mathrm{Ï€}}{∫}_0^1-1.6r^2\cos \mathrm{θ}drd\mathrm{θ} \\ ={∫}_0^{2\mathrm{Ï€}}-1.6\cos \mathrm{θ}\left[{∫}_0^1{r}^{2}dr\right]d\mathrm{θ} \\ ={∫}_0^{2\mathrm{Ï€}}-1.6\cos \mathrm{θ}{\left[\frac{r^3}{3}\right]}_0^{1}d\mathrm{θ} \\ ={∫}_0^{2\mathrm{Ï€}}-1.6\cos \mathrm{θ}\left[\frac{1^3}{3}-\frac{0^3}{3}\right]d\mathrm{θ} \\ ={∫}_0^{2\mathrm{Ï€}}-\frac{1.6}{3}\cos \mathrm{θ}d\mathrm{θ} \\ =-\frac{1.6}{3}[\sin \mathrm{θ}{]}_0^{2\mathrm{Ï€}} \\ =-\frac{1.6}{3}[\sin 2\mathrm{Ï€}-\sin 0] \\ =-\frac{1.6}{3}[0-0] \\ =0. \end{gathered}$$

Therefore, the required integral is ${âˆ\neg }_R\left(\frac{∂F_2}{∂x}-\frac{∂F_1}{∂y}\right)dA=0$.

(b). Proof:

Here, the boundary $∂R$is a boundary of a disk R of radius 1 mile and centered on this region, so it is parametrized by as follows:

$x\left(\mathrm{θ}\right)=\cos \mathrm{θ},y\left(\mathrm{θ}\right)=\sin \mathrm{θ}$, or

$\mathbf{r}\left(\mathrm{θ}\right)=⟨\cos \mathrm{θ},\sin \mathrm{θ}âŸ\copyright$

for $0≤\mathrm{θ}≤2\mathrm{π}$.

For the parametrization $\mathbf{r}\left(\mathrm{θ}\right)=⟨\cos \mathrm{θ},\sin \mathrm{θ}âŸ\copyright$, rewrite the field as follows:

Then, the required Line Integral will be,

Next, to simplify the above integral, apply the following Theorem states that, "If $∂R$is a curve with a smooth parametrization $\mathbf{r}\left(\mathrm{θ}\right)=⟨x\left(\mathrm{θ}\right),y\left(\mathrm{θ}\right)âŸ\copyright$for $a≤\mathrm{θ}≤b$, then, $ds=||{\mathbf{r}}^{\text{'}}\left(\mathrm{θ}\right)||d\mathrm{θ}{.}^{\text{'}\text{'}}$

Then, the last integral (1) becomes,

role="math" localid="1650457081409" $$\begin{gathered} {∫}_{∂R}F·nds={∫}_0^{2\mathrm{π}}\left(1.152-0.8{\cos }^2\mathrm{θ}\right)\sin \mathrm{θ}ds \\ ={∫}_0^{2\mathrm{π}}\left(1.152-0.8{\cos }^2\mathrm{θ}\right)\sin \mathrm{θ}||{\mathbf{r}}^{\text{'}}\left(\mathrm{θ}\right)||d\mathrm{θ}……\left(2\right) \end{gathered}$$

The derivative of $\mathbf{r}\left(\mathrm{θ}\right)$will be,

Then,

$$\begin{gathered} ||{\mathbf{r}}^{\text{'}}\left(\mathrm{θ}\right)||=‖⟨-\sin \mathrm{θ},\cos \mathrm{θ}âŸ\copyright ‖ \\ =\sqrt{(-\sin \mathrm{θ}{)}^2+(\cos \mathrm{θ}{)}^2} \\ =\sqrt{{\sin }^2\mathrm{θ}+{\cos }^2\mathrm{θ}} \\ =\sqrt{1} \\ =1 \end{gathered}$$

Then, the required line integral (2) will be,

$$\begin{gathered} {∫}_{ZR}F·nds={∫}_0^{2\mathrm{π}}\left(1.152-0.8{\cos }^2\mathrm{θ}\right)\sin \mathrm{θ}||{\mathbf{r}}^{\text{'}}\left(\mathrm{θ}\right)||d\mathrm{θ} \\ ={∫}_0^{2\mathrm{π}}\left(1.152-0.8{\cos }^2\mathrm{θ}\right)\sin \mathrm{θ}·1d\mathrm{θ} \\ ={∫}_0^{2\mathrm{π}}\left(1.152-0.8{\cos }^2\mathrm{θ}\right)\sin \mathrm{θ}d\mathrm{θ}. \end{gathered}$$

Thus, it is proved that,

${∫}_{∂R}F·nds={∫}_0^{2\mathrm{π}}\left(1.152-0.8{\cos }^2\mathrm{θ}\right)\sin \mathrm{θ}d\mathrm{θ}\text{.}$

Next, evaluate the last integral as follows:

$$\begin{gathered} {∫}_{2R}F·nds={∫}_0^{2\mathrm{π}}\left(1.152-0.8{\cos }^2\mathrm{θ}\right)\sin \mathrm{θ}d\mathrm{θ} \\ ={∫}_0^{2\mathrm{π}}\left(1.152\sin \mathrm{θ}-0.8{\cos }^2\mathrm{θ}\sin \mathrm{θ}\right)d\mathrm{θ} \\ ={\left[-1.152\cos \mathrm{θ}+\frac{0.8}{3}{\cos }^3\mathrm{θ}\right]}_0^{2\mathrm{π}} \\ =\left(-1.152\cos 2\mathrm{π}+\frac{0.8}{3}{\cos }^{3}2\mathrm{π}\right)-\left(-1.152\cos 0+\frac{0.8}{3}{\cos }^{3}0\right) \\ =\left(-1.152\left(1\right)+\frac{0.8}{3}(1{)}^3\right)-\left(-1.152\left(1\right)+\frac{0.8}{3}(1{)}^3\right) \\ =0. \end{gathered}$$

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,

${∫}_C\mathbf{F}·d\mathbf{r}={âˆ\neg }_R\left(\frac{∂F_2}{∂x}-\frac{∂F_1}{∂y}\right)dA·"$

By result of last step,

role="math" localid="1650457467982" ${∫}_{∂R}F·nds=0\text{.}$

As a result of part (a),

${âˆ\neg }_R\left(\frac{∂F_2}{∂x}-\frac{∂F_1}{∂y}\right)dA=0.$

This implies that,

${∫}_{\tilde{R}}F·nds={âˆ\neg }_R\left(\frac{∂F_2}{∂x}-\frac{∂F_1}{∂y}\right)dA\text{.}$

Thus, it is concluded that Green's Theorem is valid for the current in this area.

(c)

The integrals from Green's Theorem tell that there is no creation of water inside the passage. The water flowing in and out is balanced.