   Chapter 16.7, Problem 25E

Chapter
Section
Textbook Problem

Evaluate the surface integral ∫∫s F · dS for the given vector field F and the oriented surface S. In other words, find the flux of F across S. For closed surfaces, use the positive (outward) orientation.25. F(x, y, z) = x i + y j + z2 k, S is the sphere with radius 1 and center the origin

To determine

To find: The value of SFdS for F(x,y,z)=xi+yj+z2k .

Explanation

Given data:

F(x,y,z)=xi+yj+z2k (1)

Formula used:

SFdS=DF(rϕ×rθ)dA (2)

rϕ=xϕi+yϕj+zϕk (3)

rθ=xθi+yθj+zθk (4)

By using the spherical coordinates to parameterize the sphere, consider r(ϕ,θ)=sinϕcosθi+sinϕsinθj+cosϕk with 0θ2π and 0ϕπ .

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+(sin2ϕsinθ+0)j+(cos2θcosϕsinϕ+sin2θcosϕsinϕ)k}={sin2ϕcosθi+sin2ϕsinθj+cosϕsinϕ(cos2θ+sin2θ)k}

rϕ×rθ={sin2ϕcosθi+sin2ϕsinθj+cosϕsinϕk} {cos2θ+sin2θ=1}

Find F[r(ϕ,θ)]

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