Chapter 10, Problem 36RE

To determine

To Find: The area of the region.

Answer to Problem 36RE

The area is $\frac{\pi }{12}$

Explanation of Solution

Given information: The equation is $r=\sin \left(3\theta \right)$

Calculation:

The graph of this equation is:

  

In the polar coordinate, r represents the origin and hence,

  $\begin{array}{l}\sin \left(3\theta \right)=0\\ 3\theta \in \{0,\pi \}\\ \theta \in \left\{o,\frac{\pi }{3}\right\}\end{array}$

Area of a polar region can be find out as:

  $\begin{array}{l}A={\displaystyle \underset{0}{\overset{\frac{\pi }{3}}{\int }}\frac{1}{2}{\left(\sin 3\theta \right)}^{2}d\theta }\\ A=\frac{1}{2}{\displaystyle \underset{0}{\overset{\frac{\pi }{3}}{\int }}{\sin }^{2}3\theta .d\theta }\\ A=\frac{1}{2}{\displaystyle \underset{0}{\overset{\frac{\pi }{3}}{\int }}\frac{1-\cos \left(2.3\theta \right)}{2}}d\theta \\ A=\frac{1}{4}{\displaystyle \underset{0}{\overset{\frac{\pi }{3}}{\int }}\left(1-\cos 6\theta \right)}d\theta \\ A=\frac{1}{4}\left[{\displaystyle \underset{0}{\overset{\frac{\pi }{3}}{\int }}1.d\theta }-{\displaystyle \underset{0}{\overset{\frac{\pi }{3}}{\int }}\cos 6\theta .d\theta }\right]\\ A=\frac{1}{4}.{\left[\theta \right]}_o^{\frac{\pi }{3}}-\frac{1}{4}.{\left[-\frac{\sin 6\theta }{6}\right]}_0^{\frac{\pi }{3}}\\ A=\frac{1}{4}.\frac{\pi }{3}+\frac{1}{24}\left[\sin 2\pi -\sin 0\right]\\ A=\frac{\pi }{12}+0\\ A=\frac{\pi }{12}\end{array}$