Chapter 10.3, Problem 55E

To determine

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

Answer to Problem 55E

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

Explanation of Solution

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

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

  ${\sin }^2\theta =\frac{1-\cos 2\theta }{2}$

Calculation:

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

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

According to the fact $\sin \left(\frac{\pi }{2}+\alpha \right)=\cos \alpha$ for $\theta \le \alpha \le \frac{\pi }{2}$ .

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

Draw the graph of the given circle and cardioid.

  

Figure (1)

From the above graph it can be observed that $\theta \in \left[0,\pi \right]$ for the region inside the given circle and outside the given cardioid.

Calculate the required area enclosed by the curves.

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

Further simplify the integral.

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

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

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

Further simplify.

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

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

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

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

Use the above results determine the integral $A=8-{\displaystyle \underset{0}{\overset{\pi }{\int }}2{\sin }^2\theta d\theta }$ .

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

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