## anonymous one year ago Find the area of the region outside r=8+8sinθ , but inside r=24sinθ.

1. anonymous

Start by finding any intersection points, if they exist. These occur for $$\theta$$ such that $$r_1=r_2$$: \begin{align*}8+8\sin\theta&=24\sin\theta\\[1ex] 8&=16\sin\theta\\[1ex] \sin\theta&=\frac{1}{2}&\implies\theta=\frac{\pi}{6}+2\pi k,\,\theta=\frac{5\pi}{6}+2\pi k \end{align*}where $$k\in\mathbb{Z}$$. Assume $$k=0$$. To see which curve is "greater", take some $$\theta$$ between $$\dfrac{\pi}{6}$$ and $$\dfrac{5\pi}{6}$$. Suppose $$\theta=\dfrac{\pi}{2}$$. Then $r_1=8+8\sin\frac{\pi}{2}=16\\[1ex] r_2=24\sin\frac{\pi}{2}=24$Since $$r_1>r_2$$ for $$\dfrac{\pi}{6}<\theta<\dfrac{5\pi}{6}$$, the integral representing the region's area is $\int_{\pi/6}^{5\pi/6}\int_{8+8\sin\theta}^{24\sin\theta}r\,dr\,d\theta=\frac{1}{2}\int_{\pi/6}^{5\pi/6}\left[(24\sin\theta)^2-(8+8\sin\theta)^2\right]\,d\theta$

2. Loser66

perfect!!