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Calculus: Early Transcendentals

8th Edition
James Stewart
ISBN: 9781285741550

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BuyFindarrow_forward

Calculus: Early Transcendentals

8th Edition
James Stewart
ISBN: 9781285741550
Textbook Problem

Evaluate the iterated integral by converting to polar coordinates.

32. 0 2 0 2 x x 2 x 2 + y 2 d y d x

To determine

To evaluate: The iterated integral using polar coordinates.

Explanation

Given:

The function is, z=x2+y2 .

The variable x varies from 0 to 2 and y varies from 0 to 2xx2 .

Formula used:

If f is a polar rectangle R given by 0arb,αθβ, where 0βα2π , then, Rf(x,y)dA=αβabf(rcosθ,rsinθ)rdrdθ (1)

Calculation:

In order to convert the given function into polar coordinates, substitute x=rcosθ and y=rsinθ . Thus, z becomes,

z=x2+y2=r2=r

Moreover, from the given condition of x and y, the value of r varies from 0 to 2cosθ and the value of θ varies from 0 to π2 .

Therefore, by the equation (1), the value of the iterated integral becomes,

DzdA=0π202cosθr(r)drdθ=0π202cosθr2drdθ

Integrate with respect to r and apply the limit as shown below.

0π202cosθr2drdθ=0π2[r33]02cosθdθ=0π2[(2cosθ)33033]dθ=0π223cos3θ3dθ=830π2cos3θdθ

On further simplification, the integral value becomes,

0π202cosθr2drdθ=83

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Chapter 15 Solutions

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