Double integration to find the area

• Jun 5th 2009, 07:20 AM
hilbuby
Double integration to find the area
Use a double integral to find the area of the region.
The region inside the cardioid r = 1 + cos(θ) and outside the circle r = 3cos(θ).

I found the boundaries to be from pi/3 to 5pi/3 and 3costheta to 1+costheta. But I'm not getting the right answer.. after integrating everything, I get the equation- 1/2 (3theta+2sin2theta-2sintheta). Plugging in the boundaries (5pi/3 and pi/3) I get 10pi/6, but that's not correct. Did I integrate wrong?
• Jun 5th 2009, 08:22 AM
Spec
It looks like you've integrated correctly, but inserting those limits does not give $\frac{10\pi}{6}$
• Jun 5th 2009, 05:39 PM
hilbuby
I keep getting -2pi, but that isn't right either. Am I doing something wrong here?

The boundaries for r: 3cos theta to 1+cos theta
The boundaries for theta: pi/3 to 5pi/3, found by setting the two r's equal to eachother.

Then, I set up the integral, and got:
the integral from pi/3 to 5pi/3 (1/2)(1+2cos theta-4theta-4cos 2theta.
Plugging in pi/3 and 5pi/3 I keep getting -2pi...

$[\frac{5\pi}{3} + 2 \sin (\frac{5\pi}{3}) - 4(\frac{5\pi}{3}) - 2 \sin(\frac{10\pi}{3})] - [\frac{\pi}{3} + 2 \sin (\frac{\pi}{3}) - 4(\frac{\pi}{3}) - 2 \sin(\frac{2\pi}{3})]$
• Jun 5th 2009, 05:46 PM
Spec
If you get a negative area that means that you've got the two r functions in the wrong order. Recall that

$\int_{a}^{b}f(x)dx=-\int_{b}^{a}f(x)dx$

Is the answer $\pi \over 4$ ?