Directly compute (i.e., without using Green’s Theorem) ${\iint }_R (3y-3x)dA$, where R is the portion of the disk of radius 2, centered at the origin, and lying above the x-axis.

We have been given${\iint }_R (3y-3x)dA$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,

$R=\left\{\left(r,\theta \right)|0\le r\le 2,0\le \theta \le \mathrm{\pi }\right\}$

In this case,

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

Substitute the values in the integral and evaluate it as:

$\begin{array}{rcl}\begin{array}{r}{\iint }_R (3y-3x)dA\end{array}& =& {\int }_0^{\pi } {\int }_0^2 (3r\sin \theta -3r\cos \theta )rdrd\theta \\ & =& {\int }_0^{\pi } {\int }_0^2 3r^2(\sin \theta -\cos \theta )drd\theta \\ & =& {\int }_0^{\pi } (\sin \theta -\cos \theta )\left[{\int }_0^2 3r^{2}dr\right]d\theta \\ & =& {\int }_0^{\pi } (\sin \theta -\cos \theta ){\left[r^3\right]}_0^{2}d\theta \\ & =& {\int }_0^{\pi } (\sin \theta -\cos \theta )\left[2^3-0^3\right]d\theta \\ & =& {\int }_0^{\pi } 8(\sin \theta -\cos \theta )d\theta \\ & =& 8\left[(-\cos \theta -\sin \theta {]}_0^{\pi }\right.\\ & =& 8\left[\right(-\cos \pi -\sin \pi )-(-\cos 0-\sin 0\left)\right]\\ & =& 8\left[\right(-(-1)-0)-(-1-0\left)\right]\\ & =& 8[1-(-1\left)\right]\\ & =& 8[1+1]\\ & =& 8·2\\ & =& 16\end{array}$

Therefore, the required integral is localid="1654104789028" ${\iint }_R (3y-3x)dA=16$.