A community for students.
Here's the question you clicked on:
 0 viewing
anonymous
 one year ago
Find the area of the region outside r=8+8sinθ , but inside r=24sinθ.
anonymous
 one year ago
Find the area of the region outside r=8+8sinθ , but inside r=24sinθ.

This Question is Closed

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Start 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 \[\int_{\pi/6}^{5\pi/6}\int_{8+8\sin\theta}^{24\sin\theta}r\,dr\,d\theta=\frac{1}{2}\int_{\pi/6}^{5\pi/6}\left[(24\sin\theta)^2(8+8\sin\theta)^2\right]\,d\theta \]
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.