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

Short answer

The area is 14.137units

The sketch can be given as

Step-by-step solution

Given information

We are given a integral$A={\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}{(2-\sin \theta )}^{2}d\theta$

Sketch the graph

Using the standard formula we get, $r=2-\sin \theta$

using graphing utility we get,

Evaluate

We have,

$$\begin{gathered} A={\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}{(2-\sin \theta )}^{2}d\theta \\ U\sin gCAScalculatorweget, \\ A=14.137units \end{gathered}$$