Polar Integral Confusion Yet again, a friend of mine asked for help with a polar integral, we both got the same answer, the book again gave a different answer.  
Question
Use a polar integral to find the area inside the circle $r=4 \sin \theta$ and outside $r=2$.
Proposed Solution
We see that the two circles intersect at $ \pm \frac{\pi}{6}$. Because of the symmetry of the region, we may write $$2 \int_{0}^{\frac{\pi}{6}} \int_{2}^{4 \sin \theta} r  \, \,dr d\theta $$ and equivalently $$2 \int_{0}^{\frac{\pi}{6}} 8 \sin^2 \theta - 2 \, \,d\theta$$ and finally $$-4 \int_{0}^{\frac{\pi}{6}} 2\cos 2\theta -1 \, \,d\theta$$ 
Which is $\frac{2}{3} ( \pi - 3\sqrt{3})$. The book gives an answer of $\frac{2}{3} ( 2\pi - 3\sqrt{3})$, the same as evaluating the integral from $0$ to $\frac{\pi}{3}$.
Which answer is correct?

The two circles intersect when:
$$\sin \theta = 1/2 \ \to \ \theta = \pi/6, \pi-\pi/6$$
So the integral should be from $\pi/6$ until $5\pi/6$, and not as you wrote. 
Note that you could have eliminated both answers by noting that the area should be larger than half of the circle, or $A>2\pi$, whereas your book's answer gives $\sim 1.3$, and your answer is negative.
The correct result should be:
$$\frac{2}{3}\left(2\pi\color{red}{+}3\sqrt{3}\right)$$
