Find the area of the specified region.
-Inside the lemniscate $r ^ { 2 } = 4 \sin 2 \theta$ and outside the circle $r = \sqrt { 2 }$
A) $\frac { 2 } { 3 } ( 3 \sqrt { 3 } - \pi + 6 )$
B) $\frac { 4 \pi } { 3 }$
C) $\frac { 2 } { 3 } ( 3 \sqrt { 3 } - \pi )$
D) $\frac { 1 } { 2 } ( 2 \pi + 3 \sqrt { 3 } )$

Question

Quizplus · Accepted Answer

The lemniscate is symmetric about the origin and has four symmetric parts. We will only find the area of one part and then multiply by 4 to get the total area.
Set up the integral to find the area inside the lemniscate and outside the circle:
$$A = 4\int_{0}^{\frac{\pi}{4}} \frac{1}{2}[(2\sqrt{2})^2-(r^2)]d	heta - \int_{0}^{\frac{\pi}{4}}\frac{1}{2}[(\sqrt{2})^2-r^2]d	heta $$
Simplify and convert to polar coordinates: 
$$A = 4\int_{0}^{\frac{\pi}{4}} 4\sin{2	heta}d	heta - \int_{0}^{\frac{\pi}{4}}(\frac{1}{2}-\frac{1}{2}r^2)d	heta$$
$$A = 8\int_{0}^{\frac{\pi}{4}} \sin{2	heta}d	heta - \frac{1}{2}\int_{0}^{\frac{\pi}{4}}(1-r^2)d	heta$$
$$A = 8\int_{0}^{\frac{\pi}{4}} \sin{2	heta}d	heta - \frac{1}{2}(\frac{\pi}{4}-\frac{2}{3})$$
Evaluate the integral and simplify:
$$A = \frac{16}{3} - \frac{\pi}{8} - \frac{1}{3} = \frac{2}{3}(3\sqrt{3}-\pi)$$
Therefore, the answer is C.

Find the Area of the Specified Region r2=4sin⁡2θr ^ { 2 } = 4 \sin 2 \thetar2=4sin2θ

Find the Area of the Specified Region $r ^ { 2 } = 4 \sin 2 \theta$