11. the area between the two loops of the limacon r = 1 + 2sin θ.\n12. the area inside the small loop of the…

Explanation:

Step1: Recall the formula for the area in polar coordinates

The area $A$ in polar coordinates is given by $A=\frac{1}{2}\int_{\alpha}^{\beta}r^{2}d\theta$.

Step2: For problem 11 (area between two - loops of $r = 1+2\sin\theta$)

First, find the values of $\theta$ for which $r = 0$. So, $1 + 2\sin\theta=0$, then $\sin\theta=-\frac{1}{2}$, and $\theta=\frac{7\pi}{6},\frac{11\pi}{6}$. The large - loop is traced for $\theta$ from $\frac{\pi}{2}$ to $\frac{5\pi}{2}$ and the small - loop from $\frac{7\pi}{6}$ to $\frac{11\pi}{6}$. The area between the two loops is $A=\frac{1}{2}\int_{\frac{\pi}{2}}^{\frac{5\pi}{2}}(1 + 2\sin\theta)^{2}d\theta-\frac{1}{2}\int_{\frac{7\pi}{6}}^{\frac{11\pi}{6}}(1 + 2\sin\theta)^{2}d\theta$. Expand $(1 + 2\sin\theta)^{2}=1 + 4\sin\theta+4\sin^{2}\theta=1 + 4\sin\theta+4\times\frac{1 - \cos2\theta}{2}=3 + 4\sin\theta-2\cos2\theta$. $\int(3 + 4\sin\theta-2\cos2\theta)d\theta=3\theta-4\cos\theta-\sin2\theta+C$. $\frac{1}{2}\int_{\frac{\pi}{2}}^{\frac{5\pi}{2}}(3 + 4\sin\theta-2\cos2\theta)d\theta=\frac{1}{2}\left[3\theta-4\cos\theta-\sin2\theta\right]{\frac{\pi}{2}}^{\frac{5\pi}{2}}=\frac{1}{2}\left((\frac{15\pi}{2}-0 - 0)-(\frac{3\pi}{2}+0 - 0)\right)=3\pi$. $\frac{1}{2}\int{\frac{7\pi}{6}}^{\frac{11\pi}{6}}(3 + 4\sin\theta-2\cos2\theta)d\theta=\frac{1}{2}\left[3\theta-4\cos\theta-\sin2\theta\right]_{\frac{7\pi}{6}}^{\frac{11\pi}{6}}=\frac{1}{2}\left((\frac{11\pi}{2}+2\sqrt{3}+\frac{\sqrt{3}}{2})-(\frac{7\pi}{2}-2\sqrt{3}-\frac{\sqrt{3}}{2})\right)=\pi - 2\sqrt{3}$. The area between the two loops is $2\pi + 2\sqrt{3}$.

Step3: For problem 12 (area inside the small - loop of $r = 1+2\sin\theta$)

We already know that for the small - loop $r = 1+2\sin\theta$ and $\theta$ ranges from $\frac{7\pi}{6}$ to $\frac{11\pi}{6}$. $A=\frac{1}{2}\int_{\frac{7\pi}{6}}^{\frac{11\pi}{6}}(1 + 2\sin\theta)^{2}d\theta$. Since $(1 + 2\sin\theta)^{2}=3 + 4\sin\theta-2\cos2\theta$, then $\int(3 + 4\sin\theta-2\cos2\theta)d\theta=3\theta-4\cos\theta-\sin2\theta+C$. $A=\frac{1}{2}\left[3\theta-4\cos\theta-\sin2\theta\right]_{\frac{7\pi}{6}}^{\frac{11\pi}{6}}=\pi - 2\sqrt{3}$.

Step4: For problem 13 (area inside $r = 10\cos\theta$ and to the right of $r=\sec\theta$)

The line $r = \sec\theta$ is equivalent to $r\cos\theta=1$ or $x = 1$. The circle $r = 10\cos\theta$ implies $r^{2}=10r\cos\theta$ or $x^{2}+y^{2}=10x$, which can be rewritten as $(x - 5)^{2}+y^{2}=25$. To find the intersection points of $r = 10\cos\theta$ and $r=\sec\theta$, we set $10\cos\theta=\sec\theta$, so $10\cos^{2}\theta=1$, $\cos\theta=\pm\frac{1}{\sqrt{10}}$. The area $A=\frac{1}{2}\int_{-\arccos\frac{1}{\sqrt{10}}}^{\arccos\frac{1}{\sqrt{10}}}(10\cos\theta)^{2}d\theta-\int_{1}^{5}\sqrt{25-(x - 5)^{2}}dx$. Using the double - angle formula $\cos^{2}\theta=\frac{1+\cos2\theta}{2}$, $\frac{1}{2}\int_{-\arccos\frac{1}{\sqrt{10}}}^{\arccos\frac{1}{\sqrt{10}}}(10\cos\theta)^{2}d\theta = 25\int_{-\arccos\frac{1}{\sqrt{10}}}^{\arccos\frac{1}{\sqrt{10}}}(1+\cos2\theta)d\theta=25\left[\theta+\frac{\sin2\theta}{2}\right]{-\arccos\frac{1}{\sqrt{10}}}^{\arccos\frac{1}{\sqrt{10}}}$. The second integral $\int{1}^{5}\sqrt{25-(x - 5)^{2}}dx$ can be evaluated using the substitution $x-5 = 5\sin t$, $dx=5\cos tdt$. The area $A = 25\arccos\frac{1}{\sqrt{10}}+5$.

Step5: For problem 14 (area inside $r = 8\cos\theta$ and to the right of $r\cos\theta=2$)

The circle $r = 8\cos\theta$ implies $r^{2}=8r\cos\theta$ or $x^{2}+y^{2}=8x$, $(x - 4)^{2}+y^{2}=16$. The line $r\cos\theta=2$ is $x = 2$. Find the intersection points: $8\cos\theta\cos\theta=2$, $\cos^{2}\theta=\frac{1}{4}$, $\cos\theta=\pm\frac{1}{2}$. $A=\frac{1}{2}\int_{-\frac{\pi}{3}}^{\frac{\pi}{3}}(8\cos\theta)^{2}d\theta-\int_{2}^{4}\sqrt{16-(x - 4)^{2}}dx$. Using $\cos^{2}\theta=\frac{1+\cos2\theta}{2}$, $\frac{1}{2}\int_{-\frac{\pi}{3}}^{\frac{\pi}{3}}(8\cos\theta)^{2}d\theta=16\int_{-\frac{\pi}{3}}^{\frac{\pi}{3}}(1+\cos2\theta)d\theta=16\left[\theta+\frac{\sin2\theta}{2}\right]{-\frac{\pi}{3}}^{\frac{\pi}{3}}=\frac{16\pi}{3}+4\sqrt{3}$. The second integral $\int{2}^{4}\sqrt{16-(x - 4)^{2}}dx$ using substitution $x - 4=4\sin t$, $dx = 4\cos tdt$. The area $A=\frac{16\pi}{3}+4\sqrt{3}-\frac{4\pi}{3}=4\pi + 4\sqrt{3}$.

Step6: For problem 15 (area inside $r = 4\cos\theta$ and outside $r = 2\cos\theta$)

The circles $r = 4\cos\theta$ implies $x^{2}+y^{2}=4x$, $(x - 2)^{2}+y^{2}=4$ and $r = 2\cos\theta$ implies $x^{2}+y^{2}=2x$, $(x - 1)^{2}+y^{2}=1$. The area $A=\frac{1}{2}\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}(4\cos\theta)^{2}d\theta-\frac{1}{2}\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}(2\cos\theta)^{2}d\theta$. Using $\cos^{2}\theta=\frac{1+\cos2\theta}{2}$, $\frac{1}{2}\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}(4\cos\theta)^{2}d\theta=4\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}(1+\cos2\theta)d\theta=4\left[\theta+\frac{\sin2\theta}{2}\right]{-\frac{\pi}{2}}^{\frac{\pi}{2}}=4\pi$. $\frac{1}{2}\int{-\frac{\pi}{2}}^{\frac{\pi}{2}}(2\cos\theta)^{2}d\theta=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}(1+\cos2\theta)d\theta=\left[\theta+\frac{\sin2\theta}{2}\right]_{-\frac{\pi}{2}}^{\frac{\pi}{2}}=\pi$. The area $A = 3\pi$.

Step7: For problem 16 (area inside $r = 5\sin\theta$ and outside $r = 3-\sin\theta$)

Find the intersection points: $5\sin\theta=3-\sin\theta$, $6\sin\theta=3$, $\sin\theta=\frac{1}{2}$, $\theta=\frac{\pi}{6},\frac{5\pi}{6}$. $A=\frac{1}{2}\int_{\frac{\pi}{6}}^{\frac{5\pi}{6}}(5\sin\theta)^{2}d\theta-\frac{1}{2}\int_{\frac{\pi}{6}}^{\frac{5\pi}{6}}(3 - \sin\theta)^{2}d\theta$. Expand $(5\sin\theta)^{2}=25\sin^{2}\theta=\frac{25(1 - \cos2\theta)}{2}$ and $(3 - \sin\theta)^{2}=9-6\sin\theta+\sin^{2}\theta=9-6\sin\theta+\frac{1 - \cos2\theta}{2}=\frac{19}{2}-6\sin\theta-\frac{\cos2\theta}{2}$. $\int(25\sin^{2}\theta)d\theta=\frac{25}{2}\int(1 - \cos2\theta)d\theta=\frac{25}{2}\left(\theta-\frac{\sin2\theta}{2}\right)+C$. $\int\left(\frac{19}{2}-6\sin\theta-\frac{\cos2\theta}{2}\right)d\theta=\frac{19}{2}\theta + 6\cos\theta-\frac{\sin2\theta}{4}+C$. $A=\frac{1}{2}\left[\frac{25}{2}\left(\theta-\frac{\sin2\theta}{2}\right)-\left(\frac{19}{2}\theta + 6\cos\theta-\frac{\sin2\theta}{4}\right)\right]_{\frac{\pi}{6}}^{\frac{5\pi}{6}}=3\pi + 3\sqrt{3}$.

Step8: For problem 17 (area inside $r^{2}=8\sin2\theta$ and outside $r = 2$)

Find the intersection points: $8\sin2\theta=4$, $\sin2\theta=\frac{1}{2}$, $2\theta=\frac{\pi}{6},\frac{5\pi}{6},\frac{13\pi}{6},\frac{17\pi}{6}$, $\theta=\frac{\pi}{12},\frac{5\pi}{12},\frac{13\pi}{12},\frac{17\pi}{12}$. $A=\frac{1}{2}\int_{\frac{\pi}{12}}^{\frac{5\pi}{12}}8\sin2\theta d\theta+\frac{1}{2}\int_{\frac{13\pi}{12}}^{\frac{17\pi}{12}}8\sin2\theta d\theta-\frac{1}{2}\int_{\frac{\pi}{12}}^{\frac{5\pi}{12}}4d\theta-\frac{1}{2}\int_{\frac{13\pi}{12}}^{\frac{17\pi}{12}}4d\theta$. $\int8\sin2\theta d\theta=-4\cos2\theta+C$ and $\int4d\theta = 4\theta+C$. $A=\frac{4\pi}{3}-2\sqrt{3}$.

Answer:

11. $2\pi + 2\sqrt{3}$
12. $\pi - 2\sqrt{3}$
13. $25\arccos\frac{1}{\sqrt{10}}+5$
14. $4\pi + 4\sqrt{3}$
15. $3\pi$
16. $3\pi + 3\sqrt{3}$
17. $\frac{4\pi}{3}-2\sqrt{3}$