Question

Each of the integral in exercise 38-44 represents the area of a region in a plane use polar coordinates to sketch the region and evaluate the expression

The integral is

$A={\int }_{-\frac{\pi }{4}}^{\frac{\pi }{2}}{(\frac{\sqrt{2}}{2}+\sin \theta )}^{2}d\theta -{\int }_{-\frac{\pi }{2}}^{-\frac{\pi }{4}}{(\frac{\sqrt{2}}{2}+\sin \theta )}^{2}d\theta$

Short answer

The value of the integral is 1.0707 units

Step-by-step solution

Given information

We are given the integral as$A={\int }_{-\frac{\pi }{4}}^{\frac{\pi }{2}}{(\frac{\sqrt{2}}{2}+\sin \theta )}^{2}d\theta -{\int }_{-\frac{\pi }{2}}^{-\frac{\pi }{4}}{(\frac{\sqrt{2}}{2}+\sin \theta )}^{2}d\theta$

Plot the graph of the function

Using graphing utility we have,

Evaluate the integral

We have,

$$\begin{gathered} A={\int }_{-\frac{\pi }{4}}^{\frac{\pi }{2}}{(\frac{\sqrt{2}}{2}+\sin \theta )}^{2}d\theta -{\int }_{-\frac{\pi }{2}}^{-\frac{\pi }{4}}{(\frac{\sqrt{2}}{2}+\sin \theta )}^{2}d\theta \\ A={\int }_{-\frac{\pi }{4}}^{\frac{\pi }{2}}(\frac{1}{2}+\sqrt{2}\sin \theta +{\sin }^2\theta )d\theta \\ -{\int }_{-\frac{\pi }{2}}^{-\frac{\pi }{4}}(\frac{1}{2}+\sqrt{2}\sin \theta +{\sin }^2\theta )d\theta \\ Nowu\sin gCAScalculatorweget \\ A=1.0707units \end{gathered}$$