evaluate $iint_{r}3x da$ using polar coordinates, where $r$ is the region bounded by the $x$-axis and the…

Explanation:

Step1: Convert to polar coordinates

We know that $x = r\cos\theta$ and $dA=r\ dr\ d\theta$. The equation $x^{2}+(y - 2)^{2}=4$ can be rewritten as $x^{2}+y^{2}-4y=0$. In polar coordinates, $x^{2}+y^{2}=r^{2}$ and $y = r\sin\theta$, so $r^{2}-4r\sin\theta=0$, which gives $r = 4\sin\theta$. The line $x = 2$ in polar coordinates is $r\cos\theta=2$, or $r=\frac{2}{\cos\theta}$. The region $R$ is bounded by the $x -$axis ($\theta = 0$) and we need to find the intersection of $r = 4\sin\theta$ and $r=\frac{2}{\cos\theta}$. So $4\sin\theta\cos\theta=2$, $\sin2\theta = 1$, $2\theta=\frac{\pi}{2}$, $\theta=\frac{\pi}{4}$. The double - integral $\iint_{R}3x\ dA$ becomes $\int_{0}^{\frac{\pi}{4}}\int_{\frac{2}{\cos\theta}}^{4\sin\theta}3r\cos\theta\cdot r\ dr\ d\theta$.

Step2: Integrate with respect to $r$ first

[ \begin{align*} \int_{0}^{\frac{\pi}{4}}\cos\theta\left[\frac{3r^{3}}{3}\right]{\frac{2}{\cos\theta}}^{4\sin\theta}d\theta&=\int{0}^{\frac{\pi}{4}}\cos\theta\left(r^{3}\right)\big|{\frac{2}{\cos\theta}}^{4\sin\theta}d\theta\ &=\int{0}^{\frac{\pi}{4}}\cos\theta\left((4\sin\theta)^{3}-\left(\frac{2}{\cos\theta}\right)^{3}\right)d\theta\ &=\int_{0}^{\frac{\pi}{4}}\left(64\sin^{3}\theta\cos\theta-\frac{8}{\cos^{2}\theta}\right)d\theta \end{align*} ]

Step3: Integrate term - by - term

For the first integral $\int_{0}^{\frac{\pi}{4}}64\sin^{3}\theta\cos\theta\ d\theta$, let $u=\sin\theta$, $du=\cos\theta\ d\theta$. When $\theta = 0$, $u = 0$; when $\theta=\frac{\pi}{4}$, $u=\frac{\sqrt{2}}{2}$. So $\int_{0}^{\frac{\pi}{4}}64\sin^{3}\theta\cos\theta\ d\theta=64\int_{0}^{\frac{\sqrt{2}}{2}}u^{3}du=64\times\left[\frac{u^{4}}{4}\right]{0}^{\frac{\sqrt{2}}{2}}=64\times\frac{(\frac{\sqrt{2}}{2})^{4}}{4}=4$. For the second integral $\int{0}^{\frac{\pi}{4}}\frac{- 8}{\cos^{2}\theta}d\theta=-8\int_{0}^{\frac{\pi}{4}}\sec^{2}\theta\ d\theta=-8[\tan\theta]_{0}^{\frac{\pi}{4}}=-8(1 - 0)=-8$.

Step4: Find the final result

[ \begin{align*} \int_{0}^{\frac{\pi}{4}}\left(64\sin^{3}\theta\cos\theta-\frac{8}{\cos^{2}\theta}\right)d\theta&=4-8\ &=- 4 \end{align*} ]

Answer:

$-4$