find the area of the region that lies inside both curves $r=3+2\cos\theta ; r=3+2\sin\theta$
The points of intersection should be $\frac {\pi}{4} and \frac {5\pi}{4} $
I don't think these graphs are symmetrical and I am lost setting up the problem. any help would be greatly appreciated.
If you prefer seeing things visually, draw a graph to help you see the picture more clearly.
First thing to do is to set the two equations equal to one another and solve for theta.
So, as we can see $sin\theta = cos\theta$, so the intersection points would be $\frac {\pi}{4} and \frac {5\pi}{4} $.
Now, let's draw the graph:
http://www.wolframalpha.com/input/?i=r+%3D+3+%2B+2*sin%28x%29%2C+r+%3D+3+%2B+2*cos%28x%29+polar
As we can see, the area would be the area for $3 + 2cos\theta$ from $\pi/4$ to $5\pi/4.$
Therefore, the integral would be $(1/2)*(3 + 2cos\theta)^2$ from $\pi/4$ to $5\pi/4$