anonymous 4 years ago a. find the area enclosed in the inner loop of the limacon, r=1+2costheta. b. find hte area inside the limacon but outside the inner loop

1. anonymous

b) $2 \int_0^{\frac{2 \pi }{3}} \frac{1}{2} (2 \cos (\theta )+1)^2 \, d\theta=\frac{3 \sqrt{3}}{2}+2 \pi$

2. anonymous

a)$2 \int_{\frac{2 \pi }{3}}^{\pi } \frac{1}{2} (2 \cos (\theta )+1)^2 \, d\theta=\pi -\frac{3 \sqrt{3}}{2}$

3. anonymous

i'm not supposed to use a double integral?

4. anonymous

the way I see it the first step is to find the bounds on theta I think having the picture ion front of you helps

5. anonymous

Both of them are single integrals

6. anonymous
7. anonymous

sorry those bounds are wrong $r=0=1+2\cos\theta\implies \theta=\frac{2\pi}3,\frac{4\pi}3$which means the bounds on theta are$\frac{2\pi}3\le\theta\le\frac{4\pi}3$

8. anonymous

The bounds on r are just the polar function itself$0\le r\le1+2\cos\theta$and the area differential in polar coordinates is$dA=rdrd\theta$hence the intergal for the area of the inner loop should be$\large \int\int dA=\int_{\frac{2\pi}3}^{\frac{4\pi}3}\int_{0}^{1+2r\cos\theta}rdr\theta$

9. anonymous

...I assumed we are working in the interval $$[0,2\pi]$$

10. anonymous

and @colorful i'm at work so i can't be checking this all the time :( but it was a homework problem last night and it blew my mind cuz i've never even seen a limacon before last night.. -_-"

11. anonymous

The bounds for my solutions above are fine