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Q. 38

Expert-verified
Found in: Page 756

### Calculus

Book edition 1st
Author(s) Peter Kohn, Laura Taalman
Pages 1155 pages
ISBN 9781429241861

# 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 expressionThe integral is $\frac{1}{2}{\int }_{0}^{2\pi }{\left(1+\mathrm{sin}\theta \right)}^{2}d\theta$

The value of integral is 4.71 units

The graph can be given as

See the step by step solution

## Step 1: Given information

We are given an integral $\frac{1}{2}{\int }_{0}^{2\pi }{\left(1+\mathrm{sin}\theta \right)}^{2}d\theta$

## Step 2: Sketch the function

Comparing with standard equation we get $r=1+\mathrm{sin}\theta$

Using graphing utility we get

## Step 3: Evaluate the integral

We get,

$A=\frac{1}{2}{\int }_{0}^{2\pi }{\left(1+\mathrm{sin}\theta \right)}^{2}d\theta \phantom{\rule{0ex}{0ex}}A=\frac{1}{2}{\int }_{0}^{2\pi }\left(1+2\mathrm{sin}\theta +{\mathrm{sin}}^{2}\theta \right)d\theta \phantom{\rule{0ex}{0ex}}A=4.71units$