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Q. 41

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 $A={\int}_{0}^{\frac{\pi}{2}}\mathrm{sin}2\theta d\theta $

The area is 1 units

The graph can be given as

We are given an integral representing area as $A={\int}_{0}^{\frac{\pi}{2}}\mathrm{sin}2\theta d\theta $

On comparing with standard equation we get,

$r=\sqrt{\mathrm{sin}2\theta}$

Using graphing utility we get,

We are given $A={\int}_{0}^{\frac{\pi}{2}}\mathrm{sin}2\theta d\theta $

Using CAS calculator we get,

$A=1unit$

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