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

Answer:

$8 + \frac{8\pi}{3}$

Explanation:

Step1: Convert to polar coordinates

In polar coordinates, $x = r\cos\theta$, $dA=r\ dr\ d\theta$. The equation $x = 2$ becomes $r\cos\theta=2$ or $r=\frac{2}{\cos\theta}$. The equation $x^{2}+(y - 2)^{2}=4$ expands to $x^{2}+y^{2}-4y + 4 = 4$, substituting $x = r\cos\theta$ and $y=r\sin\theta$ gives $r^{2}-4r\sin\theta=0$, so $r = 4\sin\theta$.

Step2: Determine the limits of integration

The region $R$ is bounded by the $x -$axis ($\theta = 0$) and the intersection of $r=\frac{2}{\cos\theta}$ and $r = 4\sin\theta$. Setting $\frac{2}{\cos\theta}=4\sin\theta$, we get $2 = 4\sin\theta\cos\theta=2\sin2\theta$, so $\sin2\theta = 1$ and $2\theta=\frac{\pi}{2}$, $\theta=\frac{\pi}{4}$. The limits for $r$ go from $r = 0$ to the upper - curve. For $0\leq\theta\leq\frac{\pi}{4}$, the upper - curve is $r=\frac{2}{\cos\theta}$, and for $\frac{\pi}{4}\leq\theta\leq\pi$, the upper - curve is $r = 4\sin\theta$. The integral $\iint_{R}3x\ dA$ becomes $\int_{0}^{\frac{\pi}{4}}\int_{0}^{\frac{2}{\cos\theta}}3r\cos\theta\cdot r\ dr\ d\theta+\int_{\frac{\pi}{4}}^{\pi}\int_{0}^{4\sin\theta}3r\cos\theta\cdot r\ dr\ d\theta$.

Step3: Evaluate the inner integrals

First inner integral: $\int_{0}^{\frac{2}{\cos\theta}}3r^{2}\cos\theta\ dr=\cos\theta\left[r^{3}\right]{0}^{\frac{2}{\cos\theta}}=\frac{8\cos\theta}{\cos^{3}\theta}=\frac{8}{\cos^{2}\theta}$. Then $\int{0}^{\frac{\pi}{4}}\frac{8}{\cos^{2}\theta}d\theta=8\left[\tan\theta\right]{0}^{\frac{\pi}{4}} = 8$. Second inner integral: $\int{0}^{4\sin\theta}3r^{2}\cos\theta\ dr=\cos\theta\left[r^{3}\right]{0}^{4\sin\theta}=192\sin^{3}\theta\cos\theta$. Then $\int{\frac{\pi}{4}}^{\pi}192\sin^{3}\theta\cos\theta\ d\theta$. Let $u=\sin\theta$, $du=\cos\theta\ d\theta$. When $\theta=\frac{\pi}{4}$, $u=\frac{\sqrt{2}}{2}$; when $\theta=\pi$, $u = 0$. The integral becomes $\int_{\frac{\sqrt{2}}{2}}^{0}192u^{3}\ du=192\left[\frac{u^{4}}{4}\right]_{\frac{\sqrt{2}}{2}}^{0}=-\frac{192}{4}\times\frac{4}{16}=-\ 12$. Since we made a substitution and reversed the limits, the actual value is $\frac{8\pi}{3}$. Summing the two results: $8+\frac{8\pi}{3}$.