9-16 evaluate the given integral by changing to polar coordinates. 9. ∬_d x²y da, where d is the top half of…

Explanation:

Step1: Recall polar - coordinate transformation

In polar coordinates, $x = r\cos\theta$, $y = r\sin\theta$, and $dA=r\ dr\ d\theta$. Also, $x^{2}+y^{2}=r^{2}$.

Step2: Determine the limits of integration for problem 9

The region $D$ is the top - half of the disk with center at the origin and radius $5$. So, $0\leq r\leq5$ and $0\leq\theta\leq\pi$. The integrand $x^{2}y=(r\cos\theta)^{2}(r\sin\theta)=r^{3}\cos^{2}\theta\sin\theta$. The double - integral $\iint_{D}x^{2}y\ dA=\int_{0}^{\pi}\int_{0}^{5}r^{3}\cos^{2}\theta\sin\theta\cdot r\ dr\ d\theta=\int_{0}^{\pi}\cos^{2}\theta\sin\theta\ d\theta\int_{0}^{5}r^{4}\ dr$.

Integrate with respect to $r$ first

$\int_{0}^{5}r^{4}\ dr=\left[\frac{r^{5}}{5}\right]_{0}^{5}=\frac{5^{5}}{5}=5^{4}=625$.

Integrate with respect to $\theta$

Let $u = \cos\theta$, then $du=-\sin\theta\ d\theta$. When $\theta = 0$, $u = 1$; when $\theta=\pi$, $u=-1$. $\int_{0}^{\pi}\cos^{2}\theta\sin\theta\ d\theta=-\int_{1}^{-1}u^{2}\ du=\int_{-1}^{1}u^{2}\ du=\left[\frac{u^{3}}{3}\right]{-1}^{1}=\frac{1}{3}-\left(-\frac{1}{3}\right)=\frac{2}{3}$. So, $\iint{D}x^{2}y\ dA = 625\times\frac{2}{3}=\frac{1250}{3}$.

Step3: Determine the limits of integration for problem 10

The circle $x^{2}+y^{2}=4$ gives $r = 2$. The lines $x = 0$ and $y = x$ in polar coordinates: $x = 0$ corresponds to $\theta=\frac{\pi}{2}$ and $y = x$ corresponds to $\theta=\frac{\pi}{4}$. So, $0\leq r\leq2$ and $\frac{\pi}{4}\leq\theta\leq\frac{\pi}{2}$. The integrand $2x - y=2r\cos\theta-r\sin\theta=r(2\cos\theta-\sin\theta)$. The double - integral $\iint_{R}(2x - y)\ dA=\int_{\frac{\pi}{4}}^{\frac{\pi}{2}}\int_{0}^{2}r(2\cos\theta-\sin\theta)\cdot r\ dr\ d\theta=\int_{\frac{\pi}{4}}^{\frac{\pi}{2}}(2\cos\theta-\sin\theta)\ d\theta\int_{0}^{2}r^{2}\ dr$.

Integrate with respect to $r$

$\int_{0}^{2}r^{2}\ dr=\left[\frac{r^{3}}{3}\right]_{0}^{2}=\frac{8}{3}$.

Integrate with respect to $\theta$

$\int_{\frac{\pi}{4}}^{\frac{\pi}{2}}(2\cos\theta-\sin\theta)\ d\theta=[2\sin\theta+\cos\theta]{\frac{\pi}{4}}^{\frac{\pi}{2}}=(2\times1 + 0)-(2\times\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2})=2-\frac{3\sqrt{2}}{2}$. So, $\iint{R}(2x - y)\ dA=\frac{8}{3}(2 - \frac{3\sqrt{2}}{2})=\frac{16}{3}-4\sqrt{2}$.

Step4: Determine the limits of integration for problem 11

The region $R$ is in the first quadrant between the circles $r = 1$ and $r = 3$. So, $1\leq r\leq3$ and $0\leq\theta\leq\frac{\pi}{2}$. The integrand is $\sin(x^{2}+y^{2})=\sin(r^{2})$. The double - integral $\iint_{R}\sin(x^{2}+y^{2})\ dA=\int_{0}^{\frac{\pi}{2}}\int_{1}^{3}\sin(r^{2})\cdot r\ dr\ d\theta$. Let $u = r^{2}$, then $du = 2r\ dr$. When $r = 1$, $u = 1$; when $r = 3$, $u = 9$. $\int_{0}^{\frac{\pi}{2}}\int_{1}^{3}\sin(r^{2})\cdot r\ dr\ d\theta=\int_{0}^{\frac{\pi}{2}}d\theta\int_{1}^{9}\frac{1}{2}\sin u\ du$. $\int_{0}^{\frac{\pi}{2}}d\theta=\frac{\pi}{2}$, and $\int_{1}^{9}\frac{1}{2}\sin u\ du=\frac{1}{2}[-\cos u]{1}^{9}=\frac{1}{2}(\cos1-\cos9)$. So, $\iint{R}\sin(x^{2}+y^{2})\ dA=\frac{\pi}{4}(\cos1 - \cos9)$.

Step5: Determine the limits of integration for problem 12

The region $R$ is between the circles $x^{2}+y^{2}=a^{2}$ and $x^{2}+y^{2}=b^{2}$ with $0 < a < b$. So, $a\leq r\leq b$ and $0\leq\theta\leq2\pi$. The integrand $\frac{y^{2}}{x^{2}+y^{2}}=\frac{r^{2}\sin^{2}\theta}{r^{2}}=\sin^{2}\theta$. The double - integral $\iint_{R}\frac{y^{2}}{x^{2}+y^{2}}\ dA=\int_{0}^{2\pi}\int_{a}^{b}\sin^{2}\theta\cdot r\ dr\ d\theta$. Since $\sin^{2}\theta=\frac{1 - \cos(2\theta)}{2}$, $\int_{0}^{2\pi}\sin^{2}\theta\ d\theta=\int_{0}^{2\pi}\frac{1-\cos(2\theta)}{2}\ d\theta=\left[\frac{\theta}{2}-\frac{\sin(2\theta)}{4}\right]{0}^{2\pi}=\pi$. $\int{a}^{b}r\ dr=\left[\frac{r^{2}}{2}\right]{a}^{b}=\frac{b^{2}-a^{2}}{2}$. So, $\iint{R}\frac{y^{2}}{x^{2}+y^{2}}\ dA=\frac{\pi(b^{2}-a^{2})}{2}$.

Answer:

9. $\frac{1250}{3}$
10. $\frac{16}{3}-4\sqrt{2}$
11. $\frac{\pi}{4}(\cos1 - \cos9)$
12. $\frac{\pi(b^{2}-a^{2})}{2}$