Chapter H.2, Problem 16E

To determine

To find: area of the region

Answer to Problem 16E

The area of the region is $\frac{4\pi }{3}$

Explanation of Solution

Given:

Given that the curve is $r=4\sin 3\theta$

  

Calculation:

The area of the region enclosed by one loop is

  $\begin{array}{l}A={\displaystyle \underset{0}{\overset{\frac{\pi }{3}}{\int }}\frac{1}{2}}{\left[4\sin 3\theta \right]}^{2}d\theta \\ A={\displaystyle \underset{0}{\overset{\frac{\pi }{3}}{\int }}\frac{1}{2}}\left[16{\sin }^{2}3\theta \right]d\theta \\ A={\displaystyle \underset{0}{\overset{\frac{\pi }{3}}{\int }}\frac{16}{2}}\left[\frac{1-\cos 6\theta }{2}\right]d\theta \left[\because 2{\sin }^2\theta =1-\cos \theta \right]\\ A={\displaystyle \underset{0}{\overset{\frac{\pi }{3}}{\int }}4}\left[1-\cos 6\theta \right]d\theta \\ A=4{\left[\theta -\frac{\sin 6\theta }{6}\right]}_0^{\frac{\pi }{3}}\\ A=4\left[\frac{\pi }{3}-\frac{\sin 2\pi }{6}\right]-4\left[0-0\right]\\ A=\frac{4\pi }{3}-\frac{0}{6}\\ A=\frac{4\pi }{3}\end{array}$

Conclusion:

Therefore, the area of the region is $\frac{4\pi }{3}$