Chapter H.2, Problem 1E

To determine

To find: the area of the region

Answer to Problem 1E

The area of the region is $\frac{{\pi }^5}{10240}$ .

Explanation of Solution

Given:

  $r={\theta }^2,0\le \theta \le =\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$4$}\right.$

Calculation:

Consider the curve $r={\theta }^2,0\le \theta \le =\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$4$}\right.$

Sketch the polar curve of the $r={\theta }^2,0\le \theta \le =\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$4$}\right.$ shown below

  

The formula for the area A of the polar region $\Re$ is

  $A={\displaystyle {\int }_a^b\frac{1}{2}{\left[f\left(\theta \right)\right]}^{2}d\theta }$

Here,

  $\{\begin{array}{l}f\left(\theta \right)={\theta }^{{}^2}\\ a=0\\ b=\frac{\pi }{4}\end{array}$

The area of the given curve $r={\theta }^2,0\le \theta \le =\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$4$}\right.$ is,

  $\begin{array}{l}A={\displaystyle \underset{0}{\overset{\frac{\pi }{4}}{\int }}\frac{1}{2}{\left[{\theta }^2\right]}^{2}d\theta }\\ ={\displaystyle \underset{0}{\overset{\frac{\pi }{4}}{\int }}\frac{1}{2}{\theta }^{4}d\theta }\\ =\frac{1}{2}{\left[\frac{{\theta }^5}{5}\right]}_0^{\frac{\pi }{4}}\\ =\frac{1}{10}\left(\frac{{\pi }^5}{4^5}-0\right)\\ =\frac{{\pi }^5}{10240}\end{array}$

Conclusion:

Therefore, the area of the region is $\frac{{\pi }^5}{10240}$ .