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Calculus: Early Transcendentals

8th Edition
James Stewart
ISBN: 9781285741550

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Calculus: Early Transcendentals

8th Edition
James Stewart
ISBN: 9781285741550
Textbook Problem

Use spherical coordinates to evaluate 2 2 0 4 y 2 4 x 2 y 2 4 x 2 y 2 y 2 x 2 + y 2 + z 2 d z d x d y

To determine

To evaluate: The integral by changing to spherical coordinates.

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)=y2(x2+y2+z2)12.

The rectangular coordinates of the given triple integral are {(x,y,z)|2y2,0x4y2,4x2y2z4x2y2}.

Calculation:

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

f(x,y,z)=y2(x2+y2+z2)12f(ρ,θ,ϕ)=(ρsinϕsinθ)2(ρ2)12f(ρ,θ,ϕ)=ρ2sin2ϕsin2θ(ρ)f(ρ,θ,ϕ)=ρ3sin2ϕsin2θ

The limits become,

x=4y2x2+y2=4ρ2sin2ϕ(cos2θ+sin2θ)=4ρ2sin2ϕ=4

And

z=4x2y2z2=4x2y2x2+y2+z2=4ρ2sin2ϕ+(ρcosϕ)2=4

On simplifying further, this equation  becomes,

ρ2sin2ϕ+ρ2cos2ϕ=4ρ2(sin2ϕ+cos2ϕ)=4ρ2=4ρ=±2

From the equations above, it is observed that the region lies above the cylinder and below the sphere with center at origin. Thus, by the equation (1), the integral becomes,

2204y24x2y24x2y2y2(x2+y2+z2)12dzdxdy=π2π20π02(ρ3sin2ϕsin2θ)(ρ2sinϕ)dρdϕdθ=π2π20π02ρ5sin3ϕsin2θdρdϕdθ

Apply the equation (2) to separate the integrals

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Chapter 15 Solutions

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