Find the area of the region that is enclosed by the cardioid $r=2+2\sin(\theta)$

Question

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!

Answer 1

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}