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Let $S$ be the set of points with polar coordinates $(r,\theta)$ such that $ 2 \le r \le 6$ and $\frac{\pi}{3} \le \theta \le \frac{5 \pi}{6} $. Find the area of $S$.

I haven't done many problems with Polar Coordinates, where should I start?

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    First interpret these inequalities geometrically. For example, $a \leq r$ means the region on and outside the circle with radius $a$ and centered at origin. – 2017-01-12

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I presumed that you know how to sketch the region $S$. Then, the area of $S$ is then given by $$\begin{align} A&=\int_{\frac{\pi}{3}}^{\frac{5\pi}{6}}\frac{1}{2}[6^2-2^2]d\theta\\ &=16\quad\Big[\theta\Big]_{\frac{\pi}{3}}^{\frac{5\pi}{6}}\\ &=16\left[\frac{5\pi}{6}-\frac{\pi}{3}\right]\\ &=8\pi. \end{align} $$

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    Thanks, this is what I noticed after drawing out the region. It is basically just 1/4 of the circle because the 2 sectors have radiuses of 2 and 6, and share a 90deg angle, I just did 1/4(6^2pi - 2^2pi) = 8pi – 2017-01-12
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    @user405892 Yes, you are precisely correct. But since then you mentioned Polar Coordinates, I opt to give an answer using the formula. – 2017-01-12
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A good starting point, as observed in the comments, is to picture the inequalities geometrically, so that you can convince yourself that you know how to set the boundaries of the area you are trying to find.

You should find that the region is a quarter of a thick ring. From that you could cheat and say that the full ring has area $36\pi - 4\pi = 32\pi$ so the region in question has area $8\pi$. But let's do this in polar coordinates, as asked.

The area is the integral of $1$, over the region you want to measure. So you will be doing a double integral in $dr$ and $d\theta$.

The key point is that the element of area in polar coordinates is not $dr\,d\theta$, it is $r\,dr\,d\theta$.

With that in mind: $$ \int_{r=2}^{6} \int_{\theta = \pi/3}^{5\pi/6} r\,dr\,d\theta = \int_{r=2}^{6} r \left. \theta\right|_{\theta = \pi/3}^{5\pi/6} = \int_{r=2}^{6} \frac{\pi}{2} r\,dr = \frac{\pi}{2} \left( \frac{6^2}{2} - \frac{2^2}{2}\right) = 8\pi $$