Q 15. - Calculus

Question

The following integral expression may be used to find the area of a region in the polar coordinate plane: $\frac{1}{2}{\int }_0^{\frac{\pi }{4}}{\sin }^2\theta d\theta +\frac{1}{2}{\int }_{\frac{\pi }{4}}^{\frac{\pi }{2}}{\cos }^2\theta d\theta$

Sketch the region and then compute its area. (If you prefer, you may use a simpler integral to compute the same area.)

Step-by-Step Solution

Answer

$\left(\frac{\pi }{8}-\frac{1}{4}\right)$

Step 1: Given information

$\frac{1}{2}{\int }_0^{\frac{\pi }{4}}{\sin }^2\theta d\theta +\frac{1}{2}{\int }_{\frac{\pi }{4}}^{\frac{\pi }{2}}{\cos }^2\theta d\theta$

Step 2: Calculation

Consider the integral, $\frac{1}{2}{\int }_0^{\frac{\pi }{4}}{\sin }^2\theta d\theta +\frac{1}{2}{\int }_{\frac{\pi }{4}}^{\frac{\pi }{2}}{\cos }^2\theta d\theta$

The goal is to determine the integral's value.

Take the integral, $\frac{1}{2}{\int }_0^{\frac{\pi }{4}}{\sin }^2\theta d\theta +\frac{1}{2}{\int }_{\frac{\pi }{4}}^{\frac{\pi }{2}}{\cos }^2\theta d\theta$

Then,

$$\begin{gathered} \frac{1}{2}{\int }_0^{\frac{\pi }{4}}{\sin }^2\theta d\theta +\frac{1}{2}{\int }_{\frac{\pi }{4}}^{\frac{\pi }{2}}{\cos }^2\theta d\theta =\frac{1}{2}{\int }_0^{\frac{\pi }{4}}\frac{1-\cos 2\theta }{2}d\theta +\frac{1}{2}{\int }_{\frac{\pi }{4}}^{\frac{\pi }{2}}\frac{1+\cos 2\theta }{2}d\theta \\ \left[since{\cos }^2\theta =\frac{1+\cos 2\theta }{2},{\sin }^2\theta =\frac{1-\cos 2\theta }{2}\right] \\ \frac{1}{2}{\int }_0^{\frac{\pi }{4}}{\sin }^2\theta d\theta +\frac{1}{2}{\int }_{\frac{\pi }{4}}^{\frac{\pi }{2}}{\cos }^2\theta d\theta =\frac{1}{2}{\int }_0^{\frac{\pi }{4}}\frac{1}{2}-\frac{\cos 2\theta }{2}d\theta +\frac{1}{2}{\int }_{\frac{\pi }{4}}^{\frac{\pi }{2}}\frac{1}{2}+\frac{\cos 2\theta }{2}d\theta \end{gathered}$$

$$\begin{gathered} \left[since{\cos }^2\theta =\frac{1+\cos 2\theta }{2},{\sin }^2\theta =\frac{1-\cos 2\theta }{2}\right] \\ \frac{1}{2}{\int }_0^{\frac{\pi }{4}}{\sin }^2\theta d\theta +\frac{1}{2}{\int }_{\frac{\pi }{4}}^{\frac{\pi }{2}}{\cos }^2\theta d\theta =\frac{1}{2}{\int }_0^{\frac{\pi }{4}}\frac{1}{2}-\frac{\cos 2\theta }{2}d\theta +\frac{1}{2}{\int }_{\frac{\pi }{4}}^{\frac{\pi }{2}}\frac{1}{2}+\frac{\cos 2\theta }{2}d\theta \\ =\frac{1}{2}{\left[\frac{\theta }{2}-\frac{\sin 2\theta }{2·2}\right]}_0^{\frac{\pi }{4}}+\frac{1}{2}{\left[\frac{\theta }{2}+\frac{\sin 2\theta }{2·2}\right]}_{\frac{\pi }{4}}^{\frac{\pi }{2}} \end{gathered}$$

By applying the limits,

$$\begin{gathered} \frac{1}{2}{\int }_0^{\frac{\pi }{4}}{\sin }^2\theta d\theta +\frac{1}{2}{\int }_{\frac{\pi }{4}}^{\frac{\pi }{2}}{\cos }^2\theta d\theta =\frac{1}{2}\left[\frac{\pi }{8}-\frac{\sin 2·\frac{\pi }{4}}{2·2}-0-\frac{\sin 2·0}{2·2}\right]+\frac{1}{2}\left[\frac{\pi }{4}+\frac{\sin 2·\frac{\pi }{2}}{2·2}-\frac{\pi }{8}-\frac{\sin 2·\frac{\pi }{4}}{2·2}\right] \\ =\frac{1}{2}\left[\frac{\pi }{8}-\frac{1}{4}-0-0\right]+\frac{1}{2}\left[\frac{\pi }{4}+0-\frac{\pi }{8}-\frac{1}{4}\right] \\ =\frac{1}{2}\left[\frac{\pi }{8}-\frac{1}{4}+\frac{\pi }{4}-\frac{\pi }{8}-\frac{1}{4}\right] \end{gathered}$$

Step 3: Calculation

Thus,

$$\begin{gathered} \frac{1}{2}{\int }_0^{\frac{\pi }{4}}{\sin }^2\theta d\theta +\frac{1}{2}{\int }_{\frac{\pi }{4}}^{\frac{\pi }{2}}{\cos }^2\theta d\theta =\frac{1}{2}\left[\frac{\pi }{4}-\frac{2}{4}\right] \\ \frac{1}{2}{\int }_0^{\frac{\pi }{4}}{\sin }^2\theta d\theta +\frac{1}{2}{\int }_{\frac{\pi }{4}}^{\frac{\pi }{2}}{\cos }^2\theta d\theta =\left[\frac{\pi }{8}-\frac{1}{4}\right] \end{gathered}$$

Therefore, the value of the integral is $\left(\frac{\pi }{8}-\frac{1}{4}\right)$

Step 4: Calculation

The graphical representation is as follows. 

0

This is the graphical representation.