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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.

30. 0 a a 2 y 2 a 2 y 2 ( 2 x + y ) d x d y

To determine

To evaluate: The iterated integral using polar coordinates.

Explanation

Given:

The function is, z=2x+y .

The variable x varies from 0 to a and y varies from a2x2 to a2x2 .

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)

If g(x) is the function of x and h(y) is the function of y then,

abcdg(x)h(y)dydx=abg(x)dxcdh(y)dy (2)

Calculation:

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

z=2x+y=2rcosθ+rsinθ=r(2cosθ+sinθ)

Moreover, from the given condition of x and y, the value of r varies from 0 to a and the value of θ varies from 0 to π .

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

DzdA=0π0ar(2cosθ+sinθ)(r)drdθ=0π0ar2(2cosθ+sinθ)drdθ

Integrate the function with respect to r and θ by using the equation (2)

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Chapter 15 Solutions

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