
find area of polar graph
the area shared by the circle r=2 and the cardiod r=2(1cos(x))
so it looks like the limits of integration are from pi/2 to pi/2.
squaring both gets 4 and 48cos(x)+4cos(x)^2
i'm not sure which is supposed to subtracted from which... but i'll go 4(48cos(x)+4cos(X)^2.
which gives 8cos(x)  4cos(x)^2
half of that is 4cos(x) 2cos(x)^2
then i can get 4cos(x)  1  cos(2x)/2
integrating gives
4sin(x) x  sin(2x)/4
and plugging in pi/2 and pi/2 i get 8pi. but that's not the correct answer that i've been given. where am i going wrong?

first off, your limits of integration are wrong. you really need to start looking at the graph of the situation.
let $\displaystyle r_1(\theta) = 2$
$\displaystyle r_2(\theta) = 2(1\cos{\theta})$
using symmetry ...
$\displaystyle A = 2 \int_0^{\frac{\pi}{2}} \frac{[r_2(\theta)]^2}{2} \, d\theta + 2\int_{\frac{\pi}{2}}^{\pi} \frac{[r_1(\theta)]^2}{2} \, d\theta$