Find the area of the specified region.
-Shared by the circles $r = 3 \cos \theta$ and $r = 3 \sin \theta$
A) $\frac { 3 } { 2 } ( 2 - \sqrt { 2 } )$
B) $\frac { 9 } { 4 } \pi$
C) $\frac { 3 } { 8 } ( 4 \pi - 3 \sqrt { 3 } )$
D) $\frac { 9 } { 8 } ( \pi - 2 )$

Question

Quizplus · Accepted Answer

The area shared by the circles $r = 3 \cos \theta$ and $r = 3 \sin \theta$ can be found by integrating the difference in their radial functions over the range of $\theta$ where they overlap. The circles intersect where $3\cos\theta = 3\sin\theta$, which occurs at $\theta = \frac{\pi}{4}$ and $\theta = \frac{5\pi}{4}$. The shared area is symmetric about these points, so we can integrate from $\frac{\pi}{4}$ to $\frac{5\pi}{4}$ and find the area of one of the lobes formed by the intersection, then double it. The correct calculation leads to the area being $\frac{9}{8}(\pi - 2)$, which matches option D.

Find the Area of the Specified Region r=3cos⁡θr = 3 \cos \thetar=3cosθ

Find the Area of the Specified Region $r = 3 \cos \theta$