Q. 40 - Calculus

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

Step-by-Step Solution

Answer

The area is 14.137units

The sketch can be given as

Step 1: Given information

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

Step 2: Sketch the graph

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

using graphing utility we get,

Step 3: Evaluate

We have,

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