   Chapter 15.3, Problem 29E

Chapter
Section
Textbook Problem

Evaluate the iterated integral by converting to polar coordinates.29. ∫ 0 2 ∫ 0 4 − x 2 e − x 2 − y 2 d y   d x

To determine

To evaluate: The iterated integral by using polar coordinates.

Explanation

Given:

The function is, z=ex2y2 .

The variable x varies from 0 to 2 and y varies from 0 to 4x2 .

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=e(x2+y2)=er2

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

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

DzdA=0π202er2(r)drdθ=0π202rer2drdθ

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

0π202rer2drdθ=0π2dθ02rer2dr

Let t=r2

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