
Polar Curve Area
Find the area inside both r=5sin(2theta) and r=5cos(2theta).
I have NO idea how to do it. My book has no examples and I couldn't find anything online.
I ended up trying to do the integral of (5sin(2theta))^2, evaluated from 0 to Pi/8 + the integral of (5cos(2theta))^2 evaluated from 0 to Pi/8 and then multiplying the whole thing by 8 because of symmetry. I don't believe that to be correct.
Any explanations would be great!

One point the curves intersect in at Pi/8. If we find the area inside
$\displaystyle 5sin(2{\theta})$ from 0 to Pi/8 and multiply by 16, we can find the area of all
the regions inside the two functions.
$\displaystyle 8\int_{0}^{\frac{\pi}{8}}\left[5sin(2{\theta})\right]^{2}d{\theta}$

First, I don't understand why you disregard the 5cos(2theta) completely. Why do you only use the sin equation?
I also don't understand why you multiply by 16. There are only 8 sections that are in BOTH graphs, so wouldn't you multiply by 8?
Edit: Ah, I see you fixed it. I was wondering why it wasn't squared. Thanks for adding a graph, too.