Chapter H.2, Problem 17E

To determine

To calculate:

The area of the region enclosed by one loop only:

  $r=1+2\sin \theta$

Explanation of Solution

Given information:

  $r=1+2\sin \theta$

Formula Used:

The formula for the area of the polar region:

  $A={\displaystyle \underset{a}{\overset{b}{\int }}\frac{1}{2}}{r}^{2}d\theta$

Calculation:

The equation of the region is

  $r=1+2\sin 6\theta$

So, the curve region of $r=1+2\sin 6\theta$ is plotted as below:

  

Now, need to find the interval in which the inner loop exists:

  $\begin{array}{l}r=0\\ 1+2\sin \theta =0\\ \sin \theta =-\frac{1}{2}\\ \theta =\frac{7\pi }{6},\frac{11\pi }{6}\end{array}$

That means the area lies between the interval $\frac{7\pi }{6}$ and $\frac{11\pi }{6}$ .

Since, the curve of the equation is symmetric about the polar axis. Os the area of the loop is gets double between $\theta =\frac{7\pi }{6},\frac{3\pi }{2}$

The area of given equation of the polar region is:

  $\begin{array}{l}A={\displaystyle \underset{a}{\overset{b}{\int }}\frac{1}{2}}{r}^{2}d\theta \\ A={\displaystyle \underset{a}{\overset{b}{\int }}\frac{1}{2}}{\left(1+2\sin \theta \right)}^{2}d\theta \\ A=\frac{1}{2}{\displaystyle \underset{\frac{7\pi }{6}}{\overset{\frac{3\pi }{2}}{\int }}}\left(1+4\sin \theta +4{\sin }^2\theta \right)d\theta \\ A=\frac{1}{2}{\displaystyle \underset{\frac{7\pi }{6}}{\overset{\frac{3\pi }{2}}{\int }}}\left(1+4\sin \theta +4\times \frac{1-\cos 2\theta }{2}\right)d\theta \left[\because {\sin }^{2}x=\frac{1-\cos 2x}{2}\right]\end{array}$

  $\begin{array}{l}A=\frac{1}{2}{\displaystyle \underset{\frac{7\pi }{6}}{\overset{\frac{3\pi }{2}}{\int }}}\left(3+4\sin \theta -2\cos 2\theta \right)d\theta \\ A=\frac{1}{2}{\left(3+4\sin \theta -2\cos 2\theta \right)}_{\frac{7\pi }{6}}^{\frac{3\pi }{2}}\left[\begin{array}{l}\because {\displaystyle \underset{a}{\overset{b}{\int }}\cos 2x}dx=\frac{\sin 2x}{2}\\ {\displaystyle \underset{a}{\overset{b}{\int }}}\sin dx=-\cos x\\ {\displaystyle \underset{a}{\overset{b}{\int }}1}dx=x\end{array}\right]\\ A=\frac{1}{2}\left(3\left(\frac{3\pi }{2}\right)-0-0-3\left(\frac{7\pi }{2}\right)+\frac{\sqrt{3}}{2}-\frac{4\sqrt{3}}{2}\right)\\ A=\pi -\left(3\frac{\sqrt{3}}{2}\right)\end{array}$