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The region inside one loop of the lemniscater2=sin2θ

Short Answer

Expert verified

In the first quadrant, the region bordered by one loop of lemniscates isA=12

Step by step solution

01

Given information

r2=sin2θ

02

Assessing the integral

The goal of this issue is to find and evaluate an iterated integral in polar coordinates that reflects the area of a given region in the polar plane.

The lemniscates encircle the region r2=sin2θ

03

The first quadrant of lemniscates

The area of the region in the first quadrant surrounded by one loop of lemniscates can be represented as A=0π/20r+sin2θrdrdθ

Integrate first with regard to r

A=0π/2r220sin2θA=0π/2sin2θ02A=cos2θ40π/2

substituting the limits,

A=cosπcos04A=114A=12

The area of the region in the first quadrant limited by one loop of lemniscates isA=12

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