Chapter H.2, Problem 3E

To determine

To calculate:

The area of the region given by the following curve equation and its specified limits:

  $r=\sin \theta ,\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$3$}\right.\le \theta \le \raisebox{1ex}{$2\pi $}\!\left/ \!\raisebox{-1ex}{$3$}\right.$

Answer to Problem 3E

The area of the region $r=\sin \theta$ is $\frac{\pi }{12}$ square unit.

Explanation of Solution

Given information:

  $\begin{array}{l}r=\sin \theta \\ \raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$3$}\right.\le \theta \le \raisebox{1ex}{$2\pi $}\!\left/ \!\raisebox{-1ex}{$3$}\right.\end{array}$

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 area of the given equation $\left(r=\sin \theta \right)$ of the polar region is:

  $\begin{array}{l}A={\displaystyle \underset{a}{\overset{b}{\int }}\frac{1}{2}}{r}^{2}d\theta \\ A={\displaystyle \underset{a}{\overset{b}{\int }}\frac{1}{2}}{\left(\sin \theta \right)}^{2}d\theta \end{array}$

  $\begin{array}{l}A=\frac{1}{2}{\displaystyle \underset{\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$3$}\right.}{\overset{\raisebox{1ex}{$2\pi $}\!\left/ \!\raisebox{-1ex}{$3$}\right.}{\int }}}\frac{1-\cos 2\theta }{2}d\theta \\ A=\frac{1}{4}\left({\displaystyle \underset{\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$3$}\right.}{\overset{\raisebox{1ex}{$2\pi $}\!\left/ \!\raisebox{-1ex}{$3$}\right.}{\int }}}1d\theta -{\displaystyle \underset{\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$3$}\right.}{\overset{\raisebox{1ex}{$2\pi $}\!\left/ \!\raisebox{-1ex}{$3$}\right.}{\int }}}\cos 2\theta d\theta \right)\end{array}$

  $A=\frac{1}{4}{\left(x-\frac{1}{2}\sin 2x\right)}_{\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$3$}\right.}^{2\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$3$}\right.}\left[\begin{array}{l}\because {\displaystyle \underset{a}{\overset{b}{\int }}}1dx=x\\ {\displaystyle \underset{a}{\overset{b}{\int }}}\cos 2xdx=\frac{1}{2}\sin 2x\end{array}\right]$

  $A=\frac{1}{4}\left[\left(\frac{2\pi }{3}-\frac{1}{2}\sin \frac{4\pi }{3}\right)-\left(\frac{\pi }{3}-\frac{1}{2}\sin \frac{2\pi }{3}\right)\right]$

  $A=\frac{1}{4}\left(-\frac{\pi }{3}\right)\left[\because \sin \frac{2\pi }{3}=\sin \frac{4\pi }{3}=0\right]$

  $A=\frac{\pi }{12}$