Chapter H.2, Problem 5E

To determine

To calculate:

The area of the region given by the following curve equation:

  $r=\sqrt{\theta }$

Answer to Problem 5E

The area of the region $r=\sqrt{\theta }$ is $\frac{{\pi }^2}{4}$ square unit.

Explanation of Solution

Given information:

  

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 curve region is given as:

  

The region is of cardioids $\left(r=1+\cos \theta \right)$ , the shaded loop is swept out by a ray which rotates between the interval $\theta =0$ and $\theta =\pi$

The area of the given equation $\left(r=\sqrt{\theta }\right)$ of the polar region is:

  $\begin{array}{l}A={\displaystyle \underset{a}{\overset{b}{\int }}\frac{1}{2}}{r}^{2}d\theta \\ A={\displaystyle \underset{0}{\overset{\pi }{\int }}}\frac{1}{2}{\left(\sqrt{\theta }\right)}^{2}d\theta \\ A=\frac{1}{2}{\displaystyle \underset{0}{\overset{\pi }{\int }}}\theta d\theta \\ A=\frac{1}{2}{\left(\frac{{\theta }^2}{2}\right)}_0^{\pi }\left[\because {\displaystyle \underset{a}{\overset{b}{\int }}}{x}^{n}dx=\frac{x^{n+1}}{n+1}\right]\\ A=\frac{1}{2}\left(\frac{{\pi }^2}{2}-0\right)\\ A=\frac{{\pi }^2}{4}\end{array}$