find the area of the region inside both r = 7sin(theta) and 7sin(2theta) in the first quadrant

Question

anonymous · Answer

You integrate with respect to theta.

anonymous · Answer

I think it is something like $$
\int _{\theta_1}^{\theta_2} \frac12\left[r(\theta)\right]^2~d\theta
$$

anonymous · Answer

the limit of integration is what I have difficult with

anonymous · Answer

I think there will be two integrals

anonymous · Answer

You have to figure out which $r$ is greater.

anonymous · Answer

$$
7\sin(2\theta) = 14\sin(\theta)\cos(\theta)
$$

anonymous · Answer

$$
14\sin(\theta)\cos(\theta) - 7\sin(\theta) = 0\
7\sin(\theta)(2\cos(\theta)-1) = 0

$$

anonymous · Answer

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anonymous · Answer

Ok, I got one solution to be pi/3

anonymous · Answer

The bounds are going to be $0\to\pi/2$
But at $\pi/3$, we need to do a switch?

anonymous · Answer

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