91影视

We are given vector field$\mathbf{F}(x,y)=y^2\mathbf{i}鈭�xy\mathbf{j}$.

The objective is to evaluate the integral ${鈭�}_C鈥�\mathbf{F}鈰�d\mathbf{r}$by using Green's Theorem, where $C$is the unit circle traversed counterclockwise.

"Let R be a region in the plane with smooth boundary curve C oriented counterclockwise by

$\mathbf{r}\left(t\right)=鉄�\left(x\right(t),y(t\left)\right)鉄�\text{for}a鈮�t鈮�b$

If a vector field is defined on R, then

${鈭�}_C鈥�\mathbf{F}鈰�d\mathbf{r}={鈭�}_R鈥�\left(\frac{\mathrm{鈭�}{F}_2}{\mathrm{鈭�}x}鈭�\frac{\mathrm{鈭�}{F}_1}{\mathrm{鈭�}y}\right)dA鈰�"....\left(1\right)$

For the vector field$\mathbf{F}(x,y)=y^2\mathbf{i}鈭�xy\mathbf{j}$,

Now, first, find localid="1654151631288" $\frac{\mathrm{鈭�}{F}_2}{\mathrm{鈭�}x}and\frac{\mathrm{鈭�}{F}_1}{\mathrm{鈭�}\mathrm{y}}$

localid="1651063303202" $\begin{array}{r}\frac{\mathrm{鈭�}{F}_2}{\mathrm{鈭�}x}=\frac{\mathrm{鈭�}}{\mathrm{鈭�}x}(鈭�xy)\\ =鈭�y\end{array}$

and,

localid="1654151713367" $\begin{array}{r}\frac{\mathrm{鈭�}{F}_1}{\mathrm{鈭�}y}=\frac{\mathrm{鈭�}}{\mathrm{鈭�}y}\left(y^2\right)\\ =2y\end{array}$

Using the Green Theorem the integral can be evaluated as

Here, the boundary curve C is a unit circle traversed counterclockwise, so the region R is bounded by a unit disk. In polar coordinates, the region of integration is described as follows,

$R=\left\{\right(r,胃)鈭�0鈮�r鈮�1,0鈮�胃鈮�2蟺\}.$

In this case,

$x=r\cos 鈦�胃,y=r\sin 鈦�胃,x^2+y^2=r^{2}anddA=rdrd胃$

Change this integral (2) to polar coordinates, and integrate it.

$\begin{array}{rcl}\begin{array}{r}{鈭�}_C鈥�\mathbf{F}鈰�d\mathbf{r}\end{array}& =& {鈭�}_R鈥�鈭�3ydA\\ & =& \begin{array}{r}{鈭�}_0^{2蟺}鈥�{鈭�}_0^1鈥�(鈭�3r\sin 鈦�胃)rdrd胃\end{array}\\ & =& \begin{array}{r}{鈭�}_0^{2蟺}鈥�{鈭�}_0^1鈥�\left(鈭�3r^2\sin 鈦�胃\right)drd胃\end{array}\\ & =& \begin{array}{r}{鈭�}_0^{2蟺}鈥�\left[鈭�\sin 鈦�胃{鈭�}_0^1鈥�\left(3r^2\right)dr\right]d胃\end{array}\\ & =& \begin{array}{r}{鈭�}_0^{2蟺}鈥�鈭�\sin 鈦�胃{\left[r^3\right]}_0^{1}d胃\end{array}\\ & =& \begin{array}{r}{鈭�}_0^{2蟺}鈥�鈭�\sin 鈦�胃\left[1^3鈭�0^3\right]d胃\end{array}\\ & =& \begin{array}{r}{鈭�}_0^{2蟺}鈥�鈭�\sin 鈦�胃d胃\end{array}\\ & =& \begin{array}{r}[\cos 鈦�胃{]}_0^{2蟺}\end{array}\\ & =& \begin{array}{r}\cos 鈦�2蟺鈭�\cos 鈦�0\end{array}\\ & =& \begin{array}{r}1鈭�1\end{array}\\ & =& 0\end{array}$

Therefore, the required integral is evaluated as${鈭�}_C鈥�\mathbf{F}鈰�d\mathbf{r}\mathbf{=}\mathbf{0}$.