- #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?
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