Question

Find the area of the region outside r=8+8sinθ , but inside r=24sinθ.

Answer 1

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
\[\int_{\pi/6}^{5\pi/6}\int_{8+8\sin\theta}^{24\sin\theta}r\,dr\,d\theta=\frac{1}{2}\int_{\pi/6}^{5\pi/6}\left[(24\sin\theta)^2-(8+8\sin\theta)^2\right]\,d\theta \]

Answer 2

perfect!!