Chapter 10.3, Problem 54E

To determine

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

Answer to Problem 54E

The area of the region enclosed by the given cardioids is $6\pi -16$ .

Explanation of Solution

Given information: Equation of the cardioids are $r=2\left(1+\cos \theta \right)$ and $r=2\left(1-\cos \theta \right)$ .

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 $\theta$ .

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

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 $\theta \le \beta \le \frac{\pi }{2}$ .

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

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.

  

Figure (1)

From the above graph it can be observed that the common region enclosed by the given cardioids 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 $r=2\left(1+\cos \theta \right)$ 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 cardioids is symmetrical with respect to the $x$ -axis and the $y$ -axis. Consequently, it is clear that the area enclosed by the cardioids $r=2\left(1+\cos \theta \right)$ and $r=2\left(1-\cos \theta \right)$ equals 4 times the areas of the region inside the polar curve $r=2\left(1-\cos \theta \right)$ for $0\le \theta \le \frac{\pi }{2}$ .

Calculate the required area enclosed by the given curves.

  $\begin{array}{c}A=4\cdot {\displaystyle \underset{0}{\overset{\frac{\pi }{2}}{\int }}\frac{1}{2}{\left(2\left(1-\cos \theta \right)\right)}^{2}d\theta }\\ ={\displaystyle \underset{0}{\overset{\frac{\pi }{2}}{\int }}2\cdot 4\cdot \left(1-2\cos \theta +{\cos }^2\theta \right)d\theta }\\ ={\displaystyle \underset{0}{\overset{\frac{\pi }{2}}{\int }}8d\theta }-{\displaystyle \underset{0}{\overset{\frac{\pi }{2}}{\int }}16\cos \theta d\theta }+{\displaystyle \underset{0}{\overset{\frac{\pi }{2}}{\int }}8{\cos }^2\theta d\theta }\\ =8\cdot {\left[\theta \right]}_0^{\frac{\pi }{2}}-16{\left[\sin \theta \right]}_0^{\frac{\pi }{2}}+{\displaystyle \underset{0}{\overset{\frac{\pi }{2}}{\int }}8{\cos }^2\theta d\theta }\end{array}$

Further simplify the above integral.

  $\begin{array}{c}A=8\cdot \left(\frac{\pi }{2}-0\right)-16\left(\sin \left(\frac{\pi }{2}\right)-\sin 0\right)+{\displaystyle \underset{0}{\overset{\frac{\pi }{2}}{\int }}8{\cos }^2\theta d\theta }\\ =4\pi -16\left(1-0\right)+{\displaystyle \underset{0}{\overset{\frac{\pi }{2}}{\int }}8{\cos }^2\theta d\theta }\\ =4\pi -16+{\displaystyle \underset{0}{\overset{\frac{\pi }{2}}{\int }}8{\cos }^2\theta d\theta }\end{array}$

Apply the trigonometric identity ${\cos }^2\theta =\frac{1+\cos 2\theta }{2}$ to find the integral ${\displaystyle \underset{0}{\overset{\frac{\pi }{2}}{\int }}8{\cos }^2\theta d\theta }$ .

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

Further simplify.

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

Evaluate the integral ${\displaystyle \underset{0}{\overset{\frac{\pi }{2}}{\int }}4\cos 2\theta d\theta }$ . Suppose $u=2\theta$ , so $du=2d\theta$ or $d\theta =\frac{du}{2}$ .

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

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

Using the above results determine the integral $A=4\pi -16+{\displaystyle \underset{0}{\overset{\frac{\pi }{2}}{\int }}8{\cos }^2\theta d\theta }$ .

  $\begin{array}{c}A=4\pi -16+{\displaystyle \underset{0}{\overset{\frac{\pi }{2}}{\int }}8{\cos }^2\theta d\theta }\\ =4\pi -16+2\pi +{\displaystyle \underset{0}{\overset{\frac{\pi }{2}}{\int }}4\cos 2\theta d\theta }\\ =6\pi -16+0\\ =6\pi -16\end{array}$

Therefore, the area of the region enclosed by the given cardioids is $6\pi -16$ .