# Question e8206

Nov 22, 2017

A=(4pi)/12-4[1/24 (2 π - 3 sqrt(3))]~~sqrt(3)/2

#### Explanation:

The graph of $r = 1$ intersects the graph of $r = 2 \sin \left(\theta\right)$ when

$2 \sin \left(\theta\right) = 1$

$\sin \left(\theta\right) = \frac{1}{2}$

$\theta = 2 \pi n + \frac{\pi}{6}$ for $n \in \mathbb{Z}$

Looking at the graph, we can see that those angles correspond the intersection in quadrant I and quadrant II.

Analyzing the graph, we can develop a strategy for finding the area of the overlapping regions. For example, we could get the area of $r = 1$ (in red) from $\theta = \pi / 6$ to $\theta = 5 \pi / 6$. Then we would need to add on both remaining slivers of $r = 2 \sin \left(\theta\right)$ (in blue). Those two slivers are identical in area, so we could add the area of $r = 2 \sin \left(\theta\right)$ between $\theta = 0$ and $\theta = \pi / 6$, then double it.

This gives us

$A = \frac{1}{2} {\int}_{\frac{\pi}{6}}^{\frac{5 \pi}{6}} {\left(1\right)}^{2} d \theta + 2 \cdot \frac{1}{2} {\int}_{0}^{\frac{\pi}{6}} {\left(2 \sin \left(\theta\right)\right)}^{2} d \theta$

A=[1/2 theta]_(pi/6)^((5pi)/6)-4[1/2 (θ - sin(θ) cos(θ))]_0^(pi/6)

A=(4pi)/12-4[1/24 (2 π - 3 sqrt(3))]~~sqrt(3)/2#