use a graphing utility to graph the polar equation. between the loops of $r = 6(1 + 2sin(\theta))$ find the…

Explanation:

Step1: Recall area formula in polar coordinates

The area $A$ of a polar - curve $r = f(\theta)$ is given by $A=\frac{1}{2}\int_{\alpha}^{\beta}r^{2}d\theta$. First, find the values of $\theta$ for which $r = 0$. Set $r=6(1 + 2\sin\theta)=0$. Then $1 + 2\sin\theta=0$, so $\sin\theta=-\frac{1}{2}$. The solutions in the range $[0,2\pi]$ are $\theta=\frac{7\pi}{6}$ and $\theta=\frac{11\pi}{6}$.

Step2: Set up the integral for the area between the loops

The area between the loops of the polar curve $r = 6(1 + 2\sin\theta)$ is given by $A=\frac{1}{2}\int_{\frac{7\pi}{6}}^{\frac{11\pi}{6}}[6(1 + 2\sin\theta)]^{2}d\theta$. Expand $[6(1 + 2\sin\theta)]^{2}=36(1 + 4\sin\theta+4\sin^{2}\theta)$. We know that $\sin^{2}\theta=\frac{1 - \cos(2\theta)}{2}$. So $36(1 + 4\sin\theta+4\sin^{2}\theta)=36\left(1 + 4\sin\theta+4\times\frac{1 - \cos(2\theta)}{2}\right)=36(1 + 4\sin\theta + 2-2\cos(2\theta))=36(3 + 4\sin\theta-2\cos(2\theta))$.

Step3: Integrate the function

$\frac{1}{2}\int_{\frac{7\pi}{6}}^{\frac{11\pi}{6}}36(3 + 4\sin\theta-2\cos(2\theta))d\theta = 18\int_{\frac{7\pi}{6}}^{\frac{11\pi}{6}}(3 + 4\sin\theta-2\cos(2\theta))d\theta$. Integrating term - by - term: $\int(3)d\theta=3\theta$, $\int(4\sin\theta)d\theta=-4\cos\theta$, $\int(-2\cos(2\theta))d\theta=-\sin(2\theta)$. $18\left[3\theta-4\cos\theta-\sin(2\theta)\right]_{\frac{7\pi}{6}}^{\frac{11\pi}{6}}$.

Step4: Evaluate the definite integral

First, substitute $\theta=\frac{11\pi}{6}$: $3\times\frac{11\pi}{6}-4\cos\frac{11\pi}{6}-\sin(2\times\frac{11\pi}{6})=\frac{11\pi}{2}-4\times\frac{\sqrt{3}}{2}+\frac{\sqrt{3}}{2}=\frac{11\pi}{2}-\frac{3\sqrt{3}}{2}$. Then substitute $\theta=\frac{7\pi}{6}$: $3\times\frac{7\pi}{6}-4\cos\frac{7\pi}{6}-\sin(2\times\frac{7\pi}{6})=\frac{7\pi}{2}+4\times\frac{\sqrt{3}}{2}+\frac{\sqrt{3}}{2}=\frac{7\pi}{2}+\frac{5\sqrt{3}}{2}$. $18\left[\left(\frac{11\pi}{2}-\frac{3\sqrt{3}}{2}\right)-\left(\frac{7\pi}{2}+\frac{5\sqrt{3}}{2}\right)\right]=18\left(\frac{11\pi - 7\pi}{2}-\frac{3\sqrt{3}+5\sqrt{3}}{2}\right)=18\left(2\pi - 4\sqrt{3}\right)=36\pi-72\sqrt{3}$.

Answer:

$36\pi - 72\sqrt{3}$