Question

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 $s$.

Answer 1

The area defined by
2 ≤ r ≤ 6 and π/3 ≤ θ ≤ (5π)/6 is shown in the figure below.

An element of area is
dA = r dr dθ

Therefore the total area is
$A=\int _{ \frac{ \pi }{3}} ^{ \frac{5 \pi }{6} } \,d\theta \int_{2}^{6} \,rdr \\ A= [ \frac{5 \pi }{6}- \frac{ \pi }{3}]*[ \frac{6^{2}}{2} - \frac{2^{2}}{2}] = 8 \pi$

Answer: 8π