Area between 2 polar equations

[Zack K]

Homework Statement

Find the area of the region that lies inside the first curve and outside the second curve.
$r=6$
$r=6-6sin(\theta)$

Homework Equations

$A=\frac {1} {2}r^2\theta$

The Attempt at a Solution

\[/B]
If I'm correct, the area should just be $\frac {1} {2}\int_{0}^{2\pi} 6^2 d\theta - \frac {1} {2}\int_{0}^{2\pi} (6-6sin(\theta))^2 d\theta$. If that's the case then I'm probably making simple error calculating it.

EDIT: I just realized by graphing that the range is between π to 2π. How would I figure that out? Would it be just by setting both equations equal and finding intercepts?

 

[LCKurtz]

Homework Statement

Find the area of the region that lies inside the first curve and outside the second curve.
$r=6$
$r=6-6sin(\theta)$

Homework Equations

$A=\frac {1} {2}r^2\theta$

The Attempt at a Solution

\[/B]
If I'm correct, the area should just be $\frac {1} {2}\int_{0}^{2\pi} 6^2 d\theta - \frac {1} {2}\int_{0}^{2\pi} (6-6sin(\theta))^2 d\theta$. If that's the case then I'm probably making simple error calculating it.

EDIT: I just realized by graphing that the range is between π to 2π. How would I figure that out? Would it be just by setting both equations equal and finding intercepts?

Plotting is the way to go, but I don't think your proposed limits are correct. Plot a few easy points and you will see. Remember you want outside the second curve.