Chapter 15.8, Problem 48E

### Calculus: Early Transcendentals

8th Edition
James Stewart
ISBN: 9781285741550

Chapter
Section

### Calculus: Early Transcendentals

8th Edition
James Stewart
ISBN: 9781285741550
Textbook Problem

# Show that ∫ − ∞ ∞ ∫ − ∞ ∞ ∫ − ∞ ∞ x 2 + y 2 + z 2 e-(x2+y2+z2) dx dy dz = 2π (The improper triple integral is defined as the limit of a triple integral over a solid sphere as the radius of the sphere increases indefinitely.)

To determine

To Show: The value of the integral is equal to 2π .

Explanation

Formula used:

If f is a spherical region E given by aρb,αθβ,cϕd , then, Ef(x,y,z)dV=αβabcdf(ρsinϕcosθ,ρsinϕsinθ,ρcosϕ)ρ2sinϕdϕdρdθ (1)

If g(x) is the function of x and h(y) is the function of y and k(z) is the function of z  then, abcdefg(x)h(y)k(z)dzdydx=abg(x)dxcdh(y)dyefk(z)dz (2)

The spherical coordinates (ρ,θ,ϕ) corresponding to the rectangular coordinates (x,y,z) is,

ρ=x2+y2+z2ϕ=cos1(zρ)θ=cos1(xρsinϕ)

Given:

The function is f(x,y,z)=(x2+y2+z2)e(x2+y2+z2) .

Calculation:

Substitute x=ρsinϕcosθ,y=ρsinϕsinθ,z=ρcosϕ in the given function f(x,y,z) .

f(x,y,z)=(x2+y2+z2)e(x2+y2+z2)f(ρ,θ,ϕ)=(ρ2)eρ2f(ρ,θ,ϕ)=ρeρ2

Since triple integral is given, the region lies in the three dimensional space. Hence, θ varies from 0 to 2π , ϕ varies from 0 to π and ρ can vary upto . Thus, by the equation (1), the integral becomes,

(x2+y2+z2)e(x2+y2+z2)dzdydx=02π0π0(ρeρ2)(ρ2sinϕ)dρdθdϕ=02π0π0ρ3eρ2sinϕdρdθdϕ=lima02π0π0aρ3eρ2sinϕdρdθdϕ

Apply the equation (2) to separate the integrals and integrate it

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