# Math Help - Polar Area using a Double Integral

1. ## Polar Area using a Double Integral

Q: Obtain the area inside $r=1+cos\theta$ and outside $r=3cos\theta$

My first attempt was $A=2\int_{\frac{\pi}{3}}^{\pi}\int_{3cos\theta}^{1+ cos\theta}rdrd\theta$ which I quickly realized has incorrect angles, since $r=3cos\theta$ loops around twice as fast as the other graph. Thank you!

EDIT - I have just figured it out on my own! Sorry about that

2. You need two integrals.

The circle $r=3\cos\theta$ doesn't reach the second quadrant.

So you integrate from $\pi/3$ to $\pi/2$ with the bounds you used

BUT in the second integral you integrate from $\pi/2$ to $\pi$

where the lower bound is zero not $3\cos\theta$