Find the area of the specified region.
-Inside the circle $r = \sqrt { 3 } \sin \theta$ and outside the cardioid $r = 1 + \cos \theta$
A) $\frac { \sqrt { 3 } } { 4 }$
B) $\frac { 3 \sqrt { 3 } } { 8 }$
C) $\frac { 3 \sqrt { 3 } } { 2 }$
D) $\frac { 3 \sqrt { 3 } } { 4 }$

Question

Quizplus · Accepted Answer

The graph of the circle is a circle centered at the origin with radius $\frac{\sqrt{3}}{2}.$ The graph of the cardioid is a shifted and scaled version of $r = 1 - \cos \theta.$
[asy] unitsize(3cm); import graph; size(8cm); real f(real x) { return(sqrt(3)*sin(x)); } real g(real x) { return(1+cos(x)); } filldraw(2*sqrt(3)*dir(210)--arc((0,0),sqrt(3)/2,210,330)--cycle,grey); draw(graph(f,-5*pi/6,5*pi/6),red); draw(graph(g,0,2*pi),red); draw((-sqrt(3)/2,0)--(sqrt(3)/2,0)); draw((0,-1.5)--(0,2)); dot((0,1)); dot((0,-1)); label("$\left(\frac{\sqrt{3}}{2},0\right)$",(sqrt(3)/2,0),E); label("$\left(-\frac{\sqrt{3}}{2},0\right)$",(-sqrt(3)/2,0),W);  [/asy]
The region is shaded in gray. By using symmetry, we see that the area we want is 4 times the area of the portion of the shaded region in the first quadrant, which is the region between the two curves from $\theta = 0$ to $\theta = \frac{\pi}{3}.$

The two curves intersect when $\sqrt{3} \sin \theta = 1 + \cos \theta,$ or $$\sin \theta = \frac{1}{\sqrt{3}} + \frac{1}{2} \cos \theta.$$ Squaring both sides, we have $$\sin^2 \theta = \frac{1}{3} + \frac{1}{2} \cos \theta + \frac{1}{4} \cos^2 \theta.$$ Multiplying by 4, shifting everything to the left, and factoring the quadratic, we have $$(2\cos \theta + 1)^2 = 7 - 3 \sin^2 \theta.$$ Thus, $$\cos \theta = \frac{-1 \pm \sqrt{7 - 3 \sin^2 \theta}}{2}.$$ We take the negative branch, since the other branch corresponds to the portion of the curve inside the circle, which we don't want.

Therefore, we can compute the area as \begin{align*}
4 \int_0^{\pi/3} \frac{1}{2} \left[ \left( 1 + \cos \theta \right)^2 - \left( \sqrt{3} \sin \theta \right)^2 \right] \, d \theta &= 2 \int_0^{\pi/3} (1 + 2\cos \theta + \cos^2 \theta - 3 \sin^2 \theta) \, d \theta \
&= 2 \left[ \theta + 2 \sin \theta - \frac{3}{2} \cdot \frac{\theta}{2} \right]_0^{\pi/3}.
\end{align*}

This evaluates to $\boxed{\frac{3\sqrt{3}}{4}}.$

Find the Area of the Specified Region r=3sin⁡θr = \sqrt { 3 } \sin \thetar=3​sinθ

Find the Area of the Specified Region $r = \sqrt { 3 } \sin \theta$