Question

How do you find the area inside of the circle $r=3\sin \theta$ and outside the cardioid $r=1+\sin \theta$ ?

Answer 1

We are given to find the area due to the intersection of the circle and the cardioid and our required region is defined to be inside the circle but outside the cardioid. In such a situation, we must first find the points of intersection of the circle and the cardioid. Then we shall perform integration to find the area of the circle and the cardioid separately. Then we will subtract those two to obtain the final result.

Complete step by step solution:
Let us find the points of intersection by substituting $r=3\sin \theta$ in $r=1+\sin \theta$.
$\begin{aligned} & \Rightarrow 3\sin \theta =1+\sin \theta \\\ & \Rightarrow 2\sin \theta =1 \\\ \end{aligned}$
Dividing both sides by 2, we get
$\begin{aligned} & \Rightarrow \sin \theta =\dfrac{1}{2} \\\ & \Rightarrow \theta ={{\sin }^{-1}}\dfrac{1}{2} \\\ \end{aligned}$
If $\theta \in \left[ 0,2\pi \right]$ , we get $\theta =\dfrac{\pi }{6},\dfrac{5\pi }{6}$
Thus, we have to find the area of circle and cardioid one-by-one respectively for $\theta \in \left[ \dfrac{\pi }{6},\dfrac{5\pi }{6} \right]$.
We know that the area of a polar curve is given as $\dfrac{1}{2}\int{{{r}^{2}}}.d\theta$ when the polar equations of any curve are given.
First, we shall find the area $\left( {{A}_{1}} \right)$ of circle, $r=3\sin \theta$.
$\Rightarrow {{A}_{1}}=\dfrac{1}{2}\int\limits_{\dfrac{\pi }{6}}^{\dfrac{5\pi }{6}}{{{\left( 3\sin \theta \right)}^{2}}}.d\theta$
$\begin{aligned} & \Rightarrow {{A}_{1}}=\dfrac{1}{2}\int\limits_{\dfrac{\pi }{6}}^{\dfrac{5\pi }{6}}{9{{\sin }^{2}}\theta }.d\theta \\\ & \Rightarrow {{A}_{1}}=\dfrac{9}{2}\int\limits_{\dfrac{\pi }{6}}^{\dfrac{5\pi }{6}}{{{\sin }^{2}}\theta }.d\theta \\\ \end{aligned}$
Here, we shall use the trigonometric property, ${{\sin }^{2}}\theta =\dfrac{1-\cos 2\theta }{2}$ and get:

& \Rightarrow {{A}_{1}}=\dfrac{9}{2}\int\limits_{\dfrac{\pi }{6}}^{\dfrac{5\pi }{6}}{\left( \dfrac{1-\cos 2\theta }{2} \right)}.d\theta \\\ & \Rightarrow {{A}_{1}}=\dfrac{9}{4}\int\limits_{\dfrac{\pi }{6}}^{\dfrac{5\pi }{6}}{1-\cos 2\theta }.d\theta \\\ & \Rightarrow {{A}_{1}}=\dfrac{9}{4}\left[ \int\limits_{\dfrac{\pi }{6}}^{\dfrac{5\pi }{6}}{1.d\theta -\int\limits_{\dfrac{\pi }{6}}^{\dfrac{5\pi }{6}}{\cos 2\theta .d\theta }} \right] \\\ \end{aligned}$$ We know that $\int{1.dx=x+C}$ and $\int{\cos ax.dx=\dfrac{\sin ax}{a}+C}$, thus we shall use this further $\begin{aligned} & \Rightarrow {{A}_{1}}=\dfrac{9}{4}\left[ \left. \theta \right|_{\dfrac{\pi }{6}}^{\dfrac{5\pi }{6}}-\left. \dfrac{\sin 2\theta }{2} \right|_{\dfrac{\pi }{6}}^{\dfrac{5\pi }{6}} \right] \\\ & \Rightarrow {{A}_{1}}=\dfrac{9}{4}\left( \dfrac{5\pi }{6}-\dfrac{\pi }{6}-\dfrac{\sin 2\dfrac{5\pi }{6}}{2}+\dfrac{\sin 2\dfrac{\pi }{6}}{2} \right) \\\ & \Rightarrow {{A}_{1}}=\dfrac{9}{4}\left( \dfrac{4\pi }{6}+\dfrac{\sin \dfrac{\pi }{3}-\sin \dfrac{5\pi }{3}}{2} \right) \\\ \end{aligned}$ Substituting the value of $\sin \dfrac{\pi }{3}=\dfrac{\sqrt{3}}{2}$ and $\sin \dfrac{5\pi }{3}=-\dfrac{\sqrt{3}}{2}$, we get $\begin{aligned} & \Rightarrow {{A}_{1}}=\dfrac{9}{4}\left( \dfrac{4\pi }{6}+\dfrac{\dfrac{\sqrt{3}}{2}-\left( -\dfrac{\sqrt{3}}{2} \right)}{2} \right) \\\ & \Rightarrow {{A}_{1}}=\dfrac{9}{4}\left( \dfrac{2\pi }{3}+\dfrac{\sqrt{3}}{2} \right) \\\ \end{aligned}$ $\Rightarrow {{A}_{1}}=\dfrac{3\pi }{2}+\dfrac{9\sqrt{3}}{8}$ ……………………. (1) Now, we shall find the area $\left( {{A}_{2}} \right)$ of circle, $r=1+\sin \theta $. $\Rightarrow {{A}_{2}}=\dfrac{1}{2}\int\limits_{\dfrac{\pi }{6}}^{\dfrac{5\pi }{6}}{{{\left( 1+\sin \theta \right)}^{2}}}.d\theta $ $\Rightarrow {{A}_{2}}=\dfrac{1}{2}\int\limits_{\dfrac{\pi }{6}}^{\dfrac{5\pi }{6}}{1+{{\sin }^{2}}\theta }+2\sin \theta .d\theta $ Here, we shall use the trigonometric property, ${{\sin }^{2}}\theta =\dfrac{1-\cos 2\theta }{2}$ and get: $$\begin{aligned} & \Rightarrow {{A}_{2}}=\dfrac{1}{2}\int\limits_{\dfrac{\pi }{6}}^{\dfrac{5\pi }{6}}{1+\left( \dfrac{1-\cos 2\theta }{2} \right)}+2\sin \theta .d\theta \\\ & \Rightarrow {{A}_{2}}=\dfrac{1}{4}\int\limits_{\dfrac{\pi }{6}}^{\dfrac{5\pi }{6}}{2+1-\cos 2\theta +4\sin \theta }.d\theta \\\ & \Rightarrow {{A}_{2}}=\dfrac{1}{4}\left[ \int\limits_{\dfrac{\pi }{6}}^{\dfrac{5\pi }{6}}{3.d\theta -\int\limits_{\dfrac{\pi }{6}}^{\dfrac{5\pi }{6}}{\cos 2\theta .d\theta +4\int\limits_{\dfrac{\pi }{6}}^{\dfrac{5\pi }{6}}{\sin \theta .d\theta }}} \right] \\\ \end{aligned}$$ We know that $\int{1.dx=x+C}$, $\int{\sin x.dx=-\cos x+C}$ and $\int{\cos ax.dx=\dfrac{\sin ax}{a}+C}$, thus we shall use this further $\begin{aligned} & \Rightarrow {{A}_{2}}=\dfrac{1}{4}\left[ \left. 3\theta \right|_{\dfrac{\pi }{6}}^{\dfrac{5\pi }{6}}-\left. \dfrac{\sin 2\theta }{2} \right|_{\dfrac{\pi }{6}}^{\dfrac{5\pi }{6}}-\left. 4\cos \theta \right|_{\dfrac{\pi }{6}}^{\dfrac{5\pi }{6}} \right] \\\ & \Rightarrow {{A}_{2}}=\dfrac{1}{4}\left( \dfrac{15\pi }{6}-\dfrac{3\pi }{6}-\dfrac{\sin 2\dfrac{5\pi }{6}}{2}+\dfrac{\sin 2\dfrac{\pi }{6}}{2}-4\cos \dfrac{5\pi }{6}+4\cos \dfrac{\pi }{6} \right) \\\ & \Rightarrow {{A}_{2}}=\dfrac{1}{4}\left( \dfrac{12\pi }{6}+\dfrac{\sin \dfrac{\pi }{3}-\sin \dfrac{5\pi }{3}}{2}+4\cos \dfrac{\pi }{6}-4\cos \dfrac{5\pi }{6} \right) \\\ \end{aligned}$ Substituting the value of $\cos \dfrac{\pi }{6}=\dfrac{\sqrt{3}}{2}$, $\cos \dfrac{5\pi }{6}=-\dfrac{\sqrt{3}}{2}$ $\sin \dfrac{\pi }{3}=\dfrac{\sqrt{3}}{2}$ and $\sin \dfrac{5\pi }{3}=-\dfrac{\sqrt{3}}{2}$, we get $\begin{aligned} & \Rightarrow {{A}_{2}}=\dfrac{1}{4}\left( 2\pi +\dfrac{\dfrac{\sqrt{3}}{2}-\left( -\dfrac{\sqrt{3}}{2} \right)}{2}+4\dfrac{\sqrt{3}}{2}-\left( -4\dfrac{\sqrt{3}}{2} \right) \right) \\\ & \Rightarrow {{A}_{2}}=\dfrac{1}{4}\left( 2\pi +\dfrac{\sqrt{3}}{2}+4\sqrt{3} \right) \\\ & \Rightarrow {{A}_{2}}=\dfrac{1}{4}\left( 2\pi +\dfrac{\sqrt{3}+8\sqrt{3}}{2} \right) \\\ \end{aligned}$ $\Rightarrow {{A}_{2}}=\dfrac{\pi }{2}+\dfrac{9\sqrt{3}}{8}$ ……………………. (2) Therefore, the required area, $A={{A}_{1}}-{{A}_{2}}$ Substituting value of ${{A}_{1}}$ and ${{A}_{2}}$ from (1) and (2), we get $\begin{aligned} & \Rightarrow A=\dfrac{3\pi }{2}+\dfrac{9\sqrt{3}}{8}-\left( \dfrac{\pi }{2}+\dfrac{9\sqrt{3}}{8} \right) \\\ & \Rightarrow A=\dfrac{3\pi }{2}-\dfrac{\pi }{2} \\\ & \Rightarrow A=\pi \\\ \end{aligned}$  **Hence, the area inside of the circle $r=3\sin \theta $ and outside the cardioid $r=1+\sin \theta $ is $\pi $ square units.** **Note:** Another method of solving this problem was by sketching the graph of the two curves and then marking the point of intersection from it. We could further simultaneously integrate to find the areas of each curve. However, the possible mistakes one could make is during the calculation part of the integrals as it consists of a lot of terms to be taken care of at the same time.