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Q. 38

Expert-verifiedFound in: Page 756

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 expression

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

The value of integral is 4.71 units

The graph can be given as

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

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

Using graphing utility we get

We get,

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

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