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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 integral by changing to spherical coordinates.

43.   2 2 4 x 2 4 x 2 2 4 x 2 y 2 2 + 4 x 2 y 2 (xz + yz + z2)3/2 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)=(x2+y2+z2)32 .

The rectangular coordinates of the given triple integral are {(x,y,z)|2x2,4x2y4x2,24x2y2z2+4x2y2} .

Calculation:

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

f(x,y,z)=(x2+y2+z2)32f(ρ,θ,ϕ)=(ρ2)32f(ρ,θ,ϕ)=ρ3

The limits become,

y=4x2x2+y2=4ρ2sin2ϕ(cos2θ+sin2θ)=4ρ2sin2ϕ=4

And

z=2+4x2y2(z2)2=4x2y2x2+y2+(z2)2=4ρ2sin2ϕ+(ρcosϕ2)2=4

On simplifying further, this equation  becomes,

ρ2sin2ϕ+ρ2cos2ϕ+44ρcosϕ=4ρ24ρcosϕ=0ρ4cosϕ=0ρ=4cosϕ

From the equations above, it is observed that the region lies above the cylinder and below the sphere with center (0,0,2)

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Chapter 15 Solutions

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