Find the area of the specified region.
-Inside the circle $r = a \sin \theta$ and outside the cardioid $r = a ( 1 - \sin \theta ) , a 0$
A) $\frac { a ^ { 2 } } { 6 } ( 6 - \pi )$
B) $\frac { a ^ { 2 } } { 6 } ( 4 \pi - 3 \sqrt { 3 } )$
C) $\frac { a ^ { 2 } } { 3 } ( 3 \sqrt { 3 } - \pi )$
D) $\frac { a ^ { 2 } } { 12 } ( 2 \pi - 3 \sqrt { 3 } )$

Question

Find the area of the specified region.
-Inside the circle $r = a \sin \theta$ and outside the cardioid $r = a ( 1 - \sin \theta ) , a > 0$
A) $\frac { a ^ { 2 } } { 6 } ( 6 - \pi )$
B) $\frac { a ^ { 2 } } { 6 } ( 4 \pi - 3 \sqrt { 3 } )$
C) $\frac { a ^ { 2 } } { 3 } ( 3 \sqrt { 3 } - \pi )$
D) $\frac { a ^ { 2 } } { 12 } ( 2 \pi - 3 \sqrt { 3 } )$

Quizplus · Accepted Answer

First, we notice that the circle has equation $r=a\sin	heta$ and the cardioid has equation $r=a(1-\sin	heta)$. Rewriting the first equation gives us $r^2=a^2\sin^2	heta$, and the second as $r=a-a\sin	heta$. By squaring both sides we obtain $r^2=a^2-2a^2\sin	heta+a^2\sin^2	heta$. Collecting the terms with $r^2$ we can write the two equations as $r^2=a^2\sin^2	heta$ and $r^2=a^2-2a^2\sin	heta+a^2\sin^2	heta=a^2(1-\sin	heta)^2$. The intersection points of the two graphs satisfy the equations \begin{align*}
a^2\sin^2	heta&=a^2(1-\sin	heta)^2\
\Rightarrow\ \ a^2\sin^2	heta&=a^2(1-2\sin	heta+\sin^2	heta)\
\Rightarrow\ \ 0&=\sin^3	heta - 3\sin^2	heta + 2\sin	heta = \sin	heta(\sin	heta-2)(\sin	heta-1).
\end{align*} Thus we have $\sin	heta=0,1,2$, but $\sin	heta$ cannot be bigger than $1$, so we have $\sin	heta=0$ and $\sin	heta=1$ being the two solutions.

The area we are looking for is given by \begin{align*}
\frac{1}{2}\int_0^{2\pi}\left[a^2(1-\sin	heta)^2-a^2\sin^2	hetaight]d	heta&=\frac{1}{2}\left[\int_0^{2\pi}a^2(1-2\sin	heta+\sin^2	heta)d	heta-\int_0^{2\pi}a^2\sin^2	heta d	hetaight]\
&=\frac{1}{2}\left[a^2\int_0^{2\pi}(1-2\sin	heta+\sin^2	heta)d	heta-a^2\int_0^{2\pi}\sin^2	heta d	hetaight]\
&=\frac{1}{2}\left[a^2\left(\pi+\frac{\pi}{2}ight)-a^2\cdot\frac{\pi}{2}ight]=\frac{a^2}{2}\cdot\frac{3\pi}{2}\
&=\frac{a^2}{4}\cdot3\pi=\boxed{	extbf{(C) }\frac{a^2}{3}(3\sqrt{3}-\pi)}.
\end{align*}

Find the Area of the Specified Region r=asin⁡θr = a \sin \thetar=asinθ

Find the Area of the Specified Region $r = a \sin \theta$