Area between $r=4\sin(\theta)$ and $r=2$

Question

I'm trying to find the area between $r=4\sin(\theta)$ and $r=2$.

I found the points of intersections to be $\pi/6,5\pi/6$. Which implies the area is $$A=\frac{1}{2}\int_{\pi/6}^{5\pi/6}(4\sin(\theta))^2-2^2d\theta.$$

Is this correct? Or did I find the area for the following region

Answer 1

You are just intersecting two circles with the same radius, going through the center of each other.
The area of a circle sector with radius $R=2$ and amplitude $60^\circ$ is $\frac{1}{6}\pi R^2=\frac{2\pi}{3}$, while the area of an equilateral triangle with side length $2$ is given by $\sqrt{3}$, hence the area of the circle segment by the difference of these objects is $\frac{2\pi}{3}-\sqrt{3}$.

These results are enough to solve your question without integrals:

$$\color{red}{\mathcal{A}}=2\sqrt{3}+4\left(\frac{2\pi}{3}-\sqrt{3}\right)=\color{red}{\frac{8\pi}{3}-2\sqrt{3}}.$$

Answer 2

The red region can be split into three ranges.

For $\theta\in(0,2\pi/12)$, integrate along the blue $r = 4\sin\theta$ to get a segment of the blue circle.

For $\theta\in(2\pi/12, 2\pi\cdot5/12)$, integrate along the green $r=2$ to get a sector of the green circle.

The third region is in $\theta\in(2\pi\cdot5/12, 2\pi/2)$, and you can try to figure out what to integrate there.

Answer 3

The desired red region is just the area of a circle with radius $2$ minus the area of the blue region: $$ \pi(2)^2 - A = 4\pi - \frac{1}{2}\int_{\pi/6}^{5\pi/6}(4\sin(\theta))^2-2^2d\theta = 4\pi - \left( 2\sqrt 3 + \frac{4\pi}{3} \right) = \frac{8\pi}{3} - 2\sqrt 3 $$

Answer 4

The area that you have calculated is for the blue region, as you suspect. Thus, the area for the red would be that of the upper circle ($\pi\cdot2^2$) minus the blue area. See if you can express that directly as a polar integral.