- #1

Zack K

- 166

- 6

## 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?

Last edited: