Question

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

Answer 1

You integrate with respect to theta.

Answer 2

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

Answer 3

the limit of integration is what I have difficult with

Answer 4

I think there will be two integrals

Answer 5

You have to figure out which \(r\) is greater.

Answer 6

\[
7\sin(2\theta) = 14\sin(\theta)\cos(\theta)
\]

Answer 7

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

\]

Answer 8

Ok, I got one solution to be pi/3

Answer 9

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