Thread: Polar Integration

Ok, find the area of the region that lies inside the first curve (r = 3cosTheta) and outside the second curve (r = 1 + cosTheta). Now the way I went about it is finding a point of intersection in the first quadrant which happened to be pi/3. I went and found the area of the whole circle (r = 3cosTheta) and wanted to subtract double the area of the top half of the second curve (r = 1 + cosTheta) by integrating from 0 to pi/3. However, after some thinking, I realized that that would still leave some area between the left side of the circle and the line at pi/3. What can I do about this?

Nevermind, I figured out what I had to do.

Originally Posted by LeoBloom.

Nevermind, I figured out what I had to do.

good