91影视

We have been given${鈭�}_R鈥�(3y鈭�3x)dA$where $R$is the portion of the disk of radius $2$, centered at the origin, and lying above the x-axis.

Here, the region $R$is the portion of the disk of radius $2$, centered at the origin, and lying above the x-axis as shown in the following figure:

In polar coordinates, the region of integration is described as follows,

$R=\left\{\left(r,胃\right)|0鈮�r鈮�2,0鈮�胃鈮�\mathrm{蟺}\right\}$

In this case,

$$\begin{gathered} x=r\cos 胃 \\ y=r\sin 胃 \\ {x}^2+y^2=r^2 \\ dA=rdrd胃 \end{gathered}$$

Substitute the values in the integral and evaluate it as:

$\begin{array}{rcl}\begin{array}{r}{鈭�}_R鈥�(3y鈭�3x)dA\end{array}& =& {鈭�}_0^{蟺}鈥�{鈭�}_0^2鈥�(3r\sin 鈦�胃鈭�3r\cos 鈦�胃)rdrd胃\\ & =& {鈭�}_0^{蟺}鈥�{鈭�}_0^2鈥�3r^2(\sin 鈦�胃鈭�\cos 鈦�胃)drd胃\\ & =& {鈭�}_0^{蟺}鈥�(\sin 鈦�胃鈭�\cos 鈦�胃)\left[{鈭�}_0^2鈥�3r^{2}dr\right]d胃\\ & =& {鈭�}_0^{蟺}鈥�(\sin 鈦�胃鈭�\cos 鈦�胃){\left[r^3\right]}_0^{2}d胃\\ & =& {鈭�}_0^{蟺}鈥�(\sin 鈦�胃鈭�\cos 鈦�胃)\left[2^3鈭�0^3\right]d胃\\ & =& {鈭�}_0^{蟺}鈥�8(\sin 鈦�胃鈭�\cos 鈦�胃)d胃\\ & =& 8\left[(鈭�\cos 鈦�胃鈭�\sin 鈦�胃{]}_0^{蟺}\right.\\ & =& 8\left[\right(鈭�\cos 鈦�蟺鈭�\sin 鈦�蟺)鈭�(鈭�\cos 鈦�0鈭�\sin 鈦�0\left)\right]\\ & =& 8\left[\right(-(-1)-0)-(-1-0\left)\right]\\ & =& 8[1-(-1\left)\right]\\ & =& 8[1+1]\\ & =& 8路2\\ & =& 16\end{array}$

Therefore, the required integral is localid="1654104789028" ${鈭�}_R鈥�(3y鈭�3x)dA=16$.