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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

Evaluate the integral by changing to spherical coordinates.

41. 0 1 0 1 x 2 x 2 + y 2 2 x 2 y 2 xy dz dy dx

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)=xy .

The rectangular coordinates of the given triple integral are {(x,y,z)|0x1,0y1x2,x2+y2z2x2y2} .

Calculation:

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

f(x,y,z)=xyf(ρ,θ,ϕ)=ρsinϕcosθρsinϕcosθf(ρ,θ,ϕ)=ρ2sin2ϕsinθcosθ

The limits become,

y=1x2x2+y2=1ρ2sin2ϕ(cos2θ+sin2θ)=1ρ2sin2ϕ=1

And,

z=x2+y2ρcosϕ=ρ2sin2ϕ(cos2θ+sin2θ)ρcosϕ=ρ2sin2ϕρcosϕ=ρsinϕ

Solve the equation and obtain ϕ=π4 .

z=2x2y2z2=2x2y2x2+y2+z2=2ρ2=2

From the equations above, it is observed that the region lies above the cone and below the sphere of radius 2 in the first octant

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Chapter 15 Solutions

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