<s>[INST] Find the area of the region that is enclosed by the cardioid $r=2+2\sin(\theta)$ We just learned polar integration, so I know that's how we're supposed to do it. I have a problem though: I'm getting a negative answer.
What I did:
Using the graph, which is:

I figured out that every $2\pi$ it repeated, so I did 
$$\int_0^{2\pi} {\frac{{(2+2\sin \theta)}^2}{2}d\theta}$$
Evaluating it I got $-6\pi$, which is negative and definitly not right. Can I have help please? Thanks!
 [/INST] Your integral is correct, but you must have made a mistake in integration.
\begin{array} \\
\frac{1}{2} \int_0^{2\pi} (2 + 2\sin \theta)^2 \, \mathrm{d}\theta &= \frac{1}{2} \int_0^{2\pi} (4 + 8 \sin \theta + 4 \sin^2 \theta) \, \mathrm{d}\theta \\
&= \frac{1}{2} \int_0^{2\pi} \left(4 + 8 \sin \theta + 4 \left(\frac{1 - \cos (2 \theta)}{2} \right)\right) \, \mathrm{d}\theta \\
&= \frac{1}{2} \int_0^{2\pi} (6 + 8 \sin \theta - 2 \cos (2 \theta)) \, \mathrm{d}\theta \\
&= \left. \frac{1}{2} (6 \theta - 8 \cos \theta - \sin(2\theta)) \right|_{\theta=0}^{2\pi} \\
&= \frac{1}{2} ((12 \pi - 8) - (-8)) \\
&= 6 \pi 
\end{array}
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