# Find the area of the region that lies inside both curves $r = 5 \sin (2\theta)$, $r = 5 \sin (\theta)$

A friend of mine and I have this problem for homework, and he's my math tutor for all intents/purposes. He's spent a solid hour trying to figure this out, watching videos and testing different techniques, and he has no clue what to do. We know the base formula 1/2 Integral (a,b) of r^2 but this one is much different, and we have no idea what to do. It's our last assignment this semester and he's really pissed he can't do it, and I'm sure something like this will be on our final. Anyone out there have a simple process to follow to get it done?

• There are $2$ interpretations of a polar equation like $r=5\sin(2\theta)$. Roughly speaking they are (i) $r\lt 0$ not allowed and (ii) allowed but we reflect in the origin. It is the difference between a $2$-petal flower and a $4$-petal flower. The answer depends on which convention your course uses. – André Nicolas May 3 '14 at 5:16

A picture might help: $r=5\sin(2t)$ is shown in blue, $r=\sin t$ is shown in red, for $0\le t\le 2\pi$. By symmetry, you only need to find the area of the common part of the graphs shown on the right and double it. The intersection point is found by setting $5\sin(2t)=5\sin t$ and solving; it occurs at $t=\pi/3$, so you want $$2\left( \underbrace{{1\over 2}\int_0^{\pi/3} (5\sin t)^2\,dt}_\text{area of red region} + \underbrace{{1\over 2}\int_{\pi/3}^{\pi/2} (5\sin(2t))^2\,dt}_\text{area of blue region}\right)=\frac{25}{16} \left(4 \pi -3 \sqrt{3}\right)\approx 11.516.$$