Polar Area using a Double Integral

Nov 2009
94
6
Q: Obtain the area inside \(\displaystyle r=1+cos\theta\) and outside \(\displaystyle r=3cos\theta\)

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

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

matheagle

MHF Hall of Honor
Feb 2009
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1,146
You need two integrals.

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

So you integrate from \(\displaystyle \pi/3\) to \(\displaystyle \pi/2\) with the bounds you used

BUT in the second integral you integrate from \(\displaystyle \pi/2\) to \(\displaystyle \pi\)

where the lower bound is zero not \(\displaystyle 3\cos\theta\)