Find the area of the region enclosed by the inner loop of the curve $r = 4 + 8 \sin \theta$ The loop is generated for $\theta \in \left[\frac76\pi, \frac{11}{6}\pi\right]$
(this is from setting $r = 0$). 

So, $$A = \int_{\frac76\pi}^{\frac{11}{6}\pi} \frac12(4 + 8 \sin \theta)^2 \,d\theta=\\
= \int_{\frac76\pi}^{\frac{11}{6}\pi} 8 + 32 \sin \theta + 32 \sin^2\theta\, d\theta =\\
= \int_{\frac76\pi}^{\frac{11}{6}\pi} 8 + 32 \sin \theta + 16(1 - \cos(2\theta))\, d\theta=\\
= \int_{7\pi/6}^{11\pi/6} 24 + 32 \sin \theta - 16 \cos(2\theta)\, d\theta =\\
= \bigg[24\theta - 32 \cos \theta - 8 \sin(2\theta)\bigg]_{\frac76\pi}^{\frac{11}{6}\pi} =\\
= (44\pi - 16\sqrt{3} + 4\sqrt{3}) - (28\pi + 16\sqrt{3} - 4\sqrt{3}) =\\
= 16\pi - 24\sqrt{3}$$

Is my answer right?
As much as Wolfram|Alpha is impressive, here is the answer again using the SageMath free/libre software:

sage: integral((1/2)*(4 + 8 * sin(x))^2, x, 7*pi/6, 11*pi/6).simplify()
16*pi - 24*sqrt(3)
Yes.

Another way of doing this would be:
$$\int_{7\pi/6}^{11\pi/6} \int_{0}^{4 + 8\sin \theta} rdrd\theta$$
$$ \ldots = 16\pi - 24\sqrt{3}$$
If you fancy multiple integrals.