Chapter H.2, Problem 4E

To determine

To calculate:

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

  $r=\sqrt{\sin \theta },0\le \theta \le \pi$

Answer to Problem 4E

The area of the region $r=\sqrt{\sin \theta }$ is $1$ square unit.

Explanation of Solution

Given information:

  $r=\sqrt{\sin \theta },0\le \theta \le \pi$

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 $\left(r=\sqrt{\sin \theta }\right)$ of the polar region is:

  $\begin{array}{l}A={\displaystyle \underset{a}{\overset{b}{\int }}\frac{1}{2}}{r}^{2}d\theta \\ A={\displaystyle \underset{0}{\overset{\pi }{\int }}}\frac{1}{2}{\left(\sqrt{\sin \theta }\right)}^{2}d\theta \\ A=\frac{1}{2}{\displaystyle \underset{0}{\overset{\pi }{\int }}}\sin \theta d\theta \\ A=\frac{1}{2}{\left(-\cos \theta \right)}_0^{\pi }\left[\because {\displaystyle \underset{a}{\overset{b}{\int }}}\sin xdx=-\cos x\right]\\ A=\frac{1}{2}\left(-\cos \left(\pi \right)+\cos \left(0\right)\right)\\ A=\frac{1}{2}\left(1+1\right)\left[\begin{array}{l}\because \cos \left(\pi \right)=-1\\ \cos \left(0\right)=1\end{array}\right]\\ A=1\end{array}$