Here's the question you clicked on:
itzmashy
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
b) \[2 \int_0^{\frac{2 \pi }{3}} \frac{1}{2} (2 \cos (\theta )+1)^2 \, d\theta=\frac{3 \sqrt{3}}{2}+2 \pi \]
a)\[2 \int_{\frac{2 \pi }{3}}^{\pi } \frac{1}{2} (2 \cos (\theta )+1)^2 \, d\theta=\pi -\frac{3 \sqrt{3}}{2} \]
i'm not supposed to use a double integral?
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
Both of them are single integrals
http://www.wolframalpha.com/input/?i=plot%20r%3D1%2B2costheta.%20%20&t=crmtb01
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\[\frac{2\pi}3\le\theta\le\frac{4\pi}3\]
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\[\large \int\int dA=\int_{\frac{2\pi}3}^{\frac{4\pi}3}\int_{0}^{1+2r\cos\theta}rdr\theta\]
...I assumed we are working in the interval \([0,2\pi]\)
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.. -_-"
The bounds for my solutions above are fine