Find the area of the region specified in polar coordinates.
-the region inside both $r = 6 \sin \theta$ and $r = 6 \cos \theta$
A) $\frac { 9 } { 2 } ( \pi - 1 )$
B) $9 ( \pi - 1 )$
C) $\frac { 9 } { 2 } ( \pi - 2 )$
D) $9 ( \pi - 2 )$

Question

Quizplus · Accepted Answer

We need to find the area enclosed by the curves $r=6\sin	heta$ and $r=6\cos	heta$, which intersect at $	heta=\frac{\pi}{4}$. To find the bounds of integration for $	heta$, we set $6\sin	heta=6\cos	heta$ and solve for $	heta$, giving us $	heta=\frac{\pi}{4}$. The area can be found by integrating $\frac{1}{2}r^2$ from $	heta=0$ to $	heta=\frac{\pi}{4}$ and multiplying by 8 since the region is symmetric about the lines $	heta=0$, $	heta=\frac{\pi}{4}$, $	heta=\frac{\pi}{2}$, and $	heta=\frac{3\pi}{4}$. Thus, the area is equal to $8\int_0^{\frac{\pi}{4}}\frac{1}{2}(6\sin	heta)^2\ d	heta=\frac{9}{2}(2\pi-4)=\frac{9}{2}(\pi-2)$. Therefore, the answer is $\boxed{	extbf{(C)}}$.

Find the Area of the Region Specified in Polar Coordinates r=6sin⁡θr = 6 \sin \thetar=6sinθ

Find the Area of the Region Specified in Polar Coordinates $r = 6 \sin \theta$