   Chapter 16.7, Problem 17E

Chapter
Section
Textbook Problem

Evaluate the surface integral.17. ∫∫s (x2z + y2z)dS, S is the hemisphere x2 + y2 + z2 = 4, z ≥ 0

To determine

To find: The value of S(x2z+y2z)dS .

Explanation

Given data:

x2+y2+z2=4 and z0 .

Formula used:

Sf(x,y,z)dS=Df(r(u,v))|ru×rv|dA (1)

rϕ=xϕi+yϕj+zϕk (2)

rθ=xθi+yθj+zθk (3)

By using the spherical coordinates to parameterize the sphere, consider r(ϕ,θ)=2sinϕcosθi+2sinϕsinθj+2cosϕk .

Find rϕ .

Substitute 2sinϕcosθ for x , 2sinϕsinθ for y and 2cosϕ for z in equation (2),

rϕ=ϕ(2sinϕcosθ)i+ϕ(2sinϕsinθ)j+ϕ(2cosϕ)k=(2cosθ)ϕ(sinϕ)i+(2sinθ)ϕ(sinϕ)j+(2)ϕ(cosϕ)k=2cosθcosϕi+2sinθcosϕj2sinϕk

Find rθ .

Substitute 2sinϕcosθ for x , 2sinϕsinθ for y and 2cosϕ for z in equation (3),

rθ=θ(2sinϕcosθ)i+θ(2sinϕsinθ)j+θ(2cosϕ)k=(2sinϕ)θ(cosθ)i+(2sinϕ)θ(sinθ)j+(2cosϕ)θ(1)k=(2sinϕ)(sinθ)i+(2sinϕ)(cosθ)j+(2cosϕ)(0)k=2sinϕsinθi+2sinϕcosθj

Find rϕ×rθ .

rϕ×rθ=(2cosθcosϕi+2sinθcosϕj2sinϕk)×(2sinϕsinθi+2sinϕcosθj)=|ijk2cosθcosϕ2sinθcosϕ2sinϕ2sinϕsinθ2sinϕcosθ0|={(0+4sin2ϕcosθ)i(0+4sin2ϕsinθ)j+(4cos2θcosϕsinϕ+4sin2θcosϕsinϕ)k}={4sin2ϕcosθi4sin2ϕsinθj+4cosϕsinϕ(cos2θ+sin2θ)k}={4sin2ϕcosθi4sin2ϕsinθj+4cosϕsinϕk} {cos2θ+sin2θ=1}

Find |rϕ×rθ| .

|rϕ×rθ|=|4sin2ϕcosθi4sin2ϕsinθj+4cosϕsinϕk|=(4sin2ϕcosθ)2+(4sin2ϕsinθ)2+(4cosϕsinϕ)2=16sin4ϕcos2θ+16sin4ϕsin2θ+16cos2ϕsin2ϕ=16sin4ϕ(cos2θ+sin2θ)+16cos2ϕsin2ϕ

Simplify the equation

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