Answer to Question #176682 in Calculus for Joshua

Question

Find the area that is inside r=4-2cos∅ and the outside r=6+2cos∅.

Answer 1

$r=4-2\cos \theta$ and $r=6+2\cos \theta$

Let us find points of intersection:

$4-2\cos \theta =6+2\cos \theta$

$4\cos \theta=-2$

$\cos \theta =-\frac{1}{2}$

$\theta _1=\frac{2\pi}{3}$ and $\theta _2=\frac{4\pi}{3}$

Formula of area of the region bounded by the graph of $r=f(\theta )$ between the radial lines $\theta =\alpha$ and $\theta =\beta$ is $A=\frac{1}{2}\int\limits_{\alpha }^{\beta}f(\theta)^2d\theta$ . Using this formula, we obtain

 $A=\frac{1}{2}\int\limits _{\frac{2\pi}{3}}^{\frac{4\pi}{3}}\left((4-2\cos \theta)^2-(6+2\cos \theta)^2 \right)d\theta=\frac{1}{2}\int\limits _{\frac{2\pi}{3}}^{\frac{4\pi}{3}}\left(-20-40\cos \theta \right)d\theta =$

$=-10\int\limits _{\frac{2\pi}{3}}^{\frac{4\pi}{3}}\left(1+2\cos \theta \right)d\theta =-10\left(\theta+2\sin \theta\right)\bigg|^\frac{4\pi}{3}_{\frac{2\pi}{3}}=$

$=-10\left(\frac{4\pi}{3}-2\cdot \frac{\sqrt{3}}{2}\right)+10\left(\frac{2\pi}{3}+2\cdot \frac{\sqrt{3}}{2}\right)=20\sqrt{3}-\frac{20\pi}{3}$

Answer: $A= 20\sqrt{3}-\frac{20\pi}{3}$ .