Chapter 10.3, Problem 53E

To determine

To find: The area of the common region enclosed by the circle $r=2$ and the cardioid $r=2\left(1-\cos \theta \right)$ .

Answer to Problem 53E

The area of the region enclosed by the given circle and the cardioid is $5\pi -8$ .

Explanation of Solution

Given information: Equation of circle is $r=2$ and the equation of cardioid is $r=2\left(1-\cos \theta \right)$ .

Formula used: The double-angle formula for cosine function is given by:

  ${\cos }^2\theta =\frac{1+\cos 2\theta }{2}$

Calculation:

Equate the given equation of the circle and the cardioid to find the value of $\theta$ .

  $\begin{array}{c}2-2\cos \theta =2\\ -2\cos \theta =0\\ \cos \theta =0\\ \theta \in \left\{\frac{\pi }{2},\frac{3\pi }{2}\right\}\end{array}$

According to the fact $\cos \alpha =\cos \left(\alpha -2\pi \right)$ for every angle $\alpha$ and $\cos \left(-\alpha +\beta \right)=-\cos \beta$ for $\theta \le \beta \le \frac{\pi }{2}$ . Find the value of $\cos \left(\frac{3\pi }{2}\right)$ .

  $\begin{array}{c}\cos \left(\frac{3\pi }{2}\right)=\cos \left(\frac{3\pi }{2}-2\pi \right)\\ =\cos \left(-\frac{\pi }{2}\right)\\ =\cos \left(-\pi +\frac{\pi }{2}\right)\end{array}$

Further simplify.

  $\begin{array}{c}\cos \left(\frac{3\pi }{2}\right)=-\cos \left(\frac{\pi }{2}\right)\\ =0\end{array}$

Draw the graph of the given circle and cardioid using the determined results.

  

Figure (1)

From the above graph it can be observed that the common region enclosed by the given circle and cardioid is same as the region enclosed by the polar curve $r=2\left(1-\cos \theta \right)$ for $0\le \theta \le \frac{\pi }{2}$ , the polar curve $r=2$ for $\frac{\pi }{2}\le \theta \le \frac{3\pi }{2}$ , and the polar curve $r=2\left(1-\cos \theta \right)$ for $\frac{3\pi }{2}\le \theta \le 2\pi$ .

Also, the common region enclosed by the given circle and cardioid is symmetrical with respect to the $x$ -axis. Consequently, it is clear that the area enclosed by the circle $r=1$ and the cardioid $r=2\left(1-\cos \theta \right)$ equals twice the areas of the region inside the polar curves $r=2\left(1-\cos \theta \right)$ for $0\le \theta \le \frac{\pi }{2}$ and $r=2$ for $\frac{\pi }{2}\le \theta \le \pi$ .

Solve the integral to find the required area.

  $\begin{array}{c}A=2\cdot \left({\displaystyle \underset{0}{\overset{\frac{\pi }{2}}{\int }}\frac{1}{2}{\left(2\left(1-\cos \theta \right)\right)}^{2}d\theta }+{\displaystyle \underset{\frac{\pi }{2}}{\overset{\pi }{\int }}\frac{1}{2}\cdot 2^{2}d\theta }\right)\\ ={\displaystyle \underset{0}{\overset{\frac{\pi }{2}}{\int }}4\left(1-2\cos \theta +{\cos }^2\theta \right)d\theta }+{\displaystyle \underset{\frac{\pi }{2}}{\overset{\pi }{\int }}4d\theta }\\ ={\displaystyle \underset{0}{\overset{\frac{\pi }{2}}{\int }}4d\theta }-{\displaystyle \underset{0}{\overset{\frac{\pi }{2}}{\int }}8\cos \theta d\theta }+{\displaystyle \underset{0}{\overset{\frac{\pi }{2}}{\int }}4{\cos }^2\theta d\theta }+4\cdot {\left[\theta \right]}_{\frac{\pi }{2}}^{\pi }\\ =4\cdot {\left[\theta \right]}_0^{\frac{\pi }{2}}-8{\left[\sin \theta \right]}_0^{\frac{\pi }{2}}+{\displaystyle \underset{0}{\overset{\frac{\pi }{2}}{\int }}4{\cos }^2\theta d\theta }+4\cdot \left(\pi -\frac{\pi }{2}\right)\end{array}$

Further solve the integral.

  $\begin{array}{c}A=4\cdot \left(\frac{\pi }{2}-0\right)-8\left(\sin \left(\frac{\pi }{2}\right)-\sin 0\right)+{\displaystyle \underset{0}{\overset{\frac{\pi }{2}}{\int }}4{\cos }^2\theta d\theta }+2\pi \\ =2\pi -8\left(1-0\right)+{\displaystyle \underset{0}{\overset{\frac{\pi }{2}}{\int }}4{\cos }^2\theta d\theta }+2\pi \\ =4\pi -8+{\displaystyle \underset{0}{\overset{\frac{\pi }{2}}{\int }}4{\cos }^2\theta d\theta }\end{array}$

Apply the trigonometric identity ${\cos }^2\theta =\frac{1+\cos 2\theta }{2}$ to solve the integral ${\displaystyle \underset{0}{\overset{\frac{\pi }{2}}{\int }}4{\cos }^2\theta d\theta }$ .

  $\begin{array}{c}{\displaystyle \underset{0}{\overset{\frac{\pi }{2}}{\int }}4{\cos }^2\theta d\theta }={\displaystyle \underset{0}{\overset{\frac{\pi }{2}}{\int }}4\cdot \frac{1+\cos 2\theta }{2}d\theta }\\ ={\displaystyle \underset{0}{\overset{\frac{\pi }{2}}{\int }}2\cdot \left(1+\cos 2\theta \right)d\theta }\\ ={\displaystyle \underset{0}{\overset{\frac{\pi }{2}}{\int }}2d\theta }+{\displaystyle \underset{0}{\overset{\frac{\pi }{2}}{\int }}2\cos 2\theta d\theta }\\ =2\cdot {\left[\theta \right]}_0^{\frac{\pi }{2}}+{\displaystyle \underset{0}{\overset{\frac{\pi }{2}}{\int }}2\cos 2\theta d\theta }\end{array}$

Further solve the integral.

  $\begin{array}{c}{\displaystyle \underset{0}{\overset{\frac{\pi }{2}}{\int }}4{\cos }^2\theta d\theta }=2\left(\frac{\pi }{2}-0\right)+{\displaystyle \underset{0}{\overset{\frac{\pi }{2}}{\int }}2\cos 2\theta d\theta }\\ =\pi +{\displaystyle \underset{0}{\overset{\frac{\pi }{2}}{\int }}2\cos 2\theta d\theta }\end{array}$

Evaluate the integral ${\displaystyle \underset{0}{\overset{\frac{\pi }{2}}{\int }}2\cos 2\theta d\theta }$ . Suppose $u=2\theta$ , so $du=2d\theta$ or $d\theta =\frac{du}{2}$ .

For $\theta =0$ , $u=0$ and for $\theta =\frac{\pi }{2}$ , $u=\pi$ .

  $\begin{array}{c}{\displaystyle \underset{0}{\overset{\frac{\pi }{2}}{\int }}2\cos 2\theta d\theta }={\displaystyle \underset{0}{\overset{\pi }{\int }}2\cos u\frac{du}{2}}\\ ={\left[\sin u\right]}_0^{\pi }\\ =\sin \left(\pi \right)-\sin 0\\ =0\end{array}$

Use the above results determine the area.

  $\begin{array}{c}A=4\pi -8+{\displaystyle \underset{0}{\overset{\frac{\pi }{2}}{\int }}4{\cos }^2\theta d\theta }\\ =4\pi -8+\pi +{\displaystyle \underset{0}{\overset{\frac{\pi }{2}}{\int }}2\cos 2\theta d\theta }\\ =5\pi -8+0\\ =5\pi -8\end{array}$

Therefore, the area of the region enclosed by the given circle and the cardioid is $5\pi -8$ .