   Chapter 16.7, Problem 49E

Chapter
Section
Textbook Problem

Let F be an inverse square field, that is, |F(r) = cr/|r|3 for some constant c, where r = x i + y j + z k. Show that the flux of F across a sphere S with center the origin is independent of the radius of S.

To determine

To show: The flux of F across a sphere S with center the origin is independent of the radius of S.

Explanation

Given:

The inverse square field is given by F(r)=cr|r|3 (1)

The radius is given by r=xi+yj+zk

Formula used:

SFdS=DF(rϕ×rθ)dA (2)

rϕ=xϕi+yϕj+zϕk (3)

rθ=xθi+yθj+zθk (4)

Calculation:

By using the spherical coordinates parameterize the sphere, consider x=asinϕcosθ, y=asinϕsinθ and z=acosϕ with 0θ2π and 0ϕπ.

r(ϕ,θ)=asinϕcosθi+asinϕsinθj+acosϕk

Find |r| as,

|r|=(asinϕcosθ)2+(asinϕsinθ)2+(acosϕ)2=a2sin2ϕcos2θ+a2sin2ϕsin2θ+a2cos2ϕ=a2sin2ϕ(cos2θ+sin2θ)+a2cos2ϕ=a2sin2ϕ+a2cos2ϕ {cos2θ+sin2θ=1}

Simplify the equation.

|r|=a2(sin2ϕ+cos2ϕ)=a2=a

Thus, the inverse square field becomes,

F(r)=c(asinϕcosθi+asinϕsinθj+acosϕk)a3

To find rϕ, differentiate with respect to ϕ as follows.

Substitute asinϕcosθ for x, asinϕsinθ for y and acosϕ for z in equation (3),

rϕ=ϕ(asinϕcosθ)i+ϕ(asinϕsinθ)j+ϕ(acosϕ)k=(acosθ)ϕ(sinϕ)i+(asinθ)ϕ(sinϕ)j+(a)ϕ(cosϕ)k=acosθcosϕi+asinθcosϕjasinϕk

To find rθ, differentiate with respect to θ as follows.

Substitute asinϕcosθ for x, asinϕsinθ for y and acosϕ for z in equation (4),

rθ=θ(asinϕcosθ)i+θ(asinϕsinθ)j+θ(acosϕ)k=(asinϕ)θ(cosθ)i+(asinϕ)θ(sinθ)j+(acosϕ)θ(1)k=(asinϕ)(sinθ)i+(asinϕ)(cosθ)j+(acosϕ)(0)k=asinϕsinθi+asinϕcosθj

Find rϕ×rθ as,

rϕ×rθ=(acosθcosϕi+asinθcosϕjasinϕk)×(asinϕsinθi+asinϕcosθj)=|ijkacosθcosϕasinθcosϕasinϕ

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