   Chapter 16.7, Problem 16E

Chapter
Section
Textbook Problem

Evaluate the surface integral.16 ∫∫s y2 dS, S is the part of the sphere x2 + y2 + z2 = 1 that lies above the cone z   =   x 2 + y 2

To determine

To find: The value of Sy2dS .

Explanation

Given data:

x2+y2+z2=1 and z=x2+y2 .

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)

The sphere intersects the cone in the circle x2+y2=12 , z=12 , As S is the portion of the sphere where z12 .

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

Find rϕ .

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

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

Find rθ .

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

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

Find rϕ×rθ .

rϕ×rθ=(cosθcosϕi+sinθcosϕjsinϕk)×(sinϕsinθi+sinϕcosθj)=|ijkcosθcosϕsinθcosϕsinϕsinϕsinθsinϕcosθ0|=[(0+sin2ϕcosθ)i(0+sin2ϕsinθ)j+(cos2θcosϕsinϕ+sin2θcosϕsinϕ)k]=sin2ϕcosθisin2ϕsinθj+cosϕsinϕ(cos2θ+sin2θ)k=sin2ϕcosθisin2ϕsinθj+cosϕsinϕk {cos2θ+sin2θ=1}

Find |rϕ×rθ|

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