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Another Stokes' theorem question?
Verify Stokes' theorem for the helicoid Ψ(r,θ)=(rcosθ,rsinθ,θ) where (r,θ) lies in the rectangle [0,1]×[0,π/2], and F is the vector field F=(2z,6x,3y).
First, compute the surface integral: ∫∫(∇×F)⋅dS=a∫b c∫d f(r,θ)drdθ. I need to find f, a, b, c, and d.
Then, I have to compute the line integral on that part of the boundary from (1,0,0) to (0,1,π/2).
∫C F⋅ds=a∫b g(θ)dθ. Now, I need to find a,b, and g.
1 Answer
- kbLv 79 years agoFavorite Answer
1) Since curl F = <3, 2, 6>, and Ψ_r x Ψ_θ = <sin θ, -cos θ, r>,
∫∫s (∇×F)⋅dS
= ∫∫ <3, 2, 6> ⋅ <sin θ, -cos θ, r> dA
= ∫(r = 0 to 1) ∫(θ = 0 to π/2) (3 sin θ - 2 cos θ + 6r) dθ dr
= ∫(r = 0 to 1) (-3 cos θ - 2 sin θ + 6rθ) {for θ = 0 to π/2} dr
= ∫(r = 0 to 1) (3 - 2 + 3πr) dr
= 1 + 3π/2.
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2) There are 4 parts to the boundary of the helicoid.
The vertices are (0, 0, 0), (0, 0, π/2), (0, 1, π/2), (1, 0, 0).
(i) C₁: (0, 0, 0) to (0, 0, π/2)
Via r = 0 ==> Ψ(0, θ) = <0, 0, θ>,
∫c₁ F ⋅ dr = ∫(θ = 0 to π/2) <2θ, 0, 0> ⋅ <0, 0, 1> dθ = 0.
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(ii) C₂: (0, 0, π/2) to (0, 1, π/2)
Via θ = π/2 ==> Ψ(r, π/2) = <0, r, π/2>,
∫c₂ F ⋅ dr = ∫(r = 0 to 1) <π, 0, 3r> ⋅ <0, 1, 0> dθ = 0.
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(iii) C₃: (0, 1, π/2) to (0, 1, 0)
Via r = 1 ==> Ψ(1, θ) = <cos θ, sin θ, θ>,
∫c₃ F ⋅ dr = ∫(θ = 0 to π/2) <2θ, 6 cos θ, 3 sin θ> ⋅ <-sin θ, cos θ, 1> dθ
..............= ∫(θ = 0 to π/2) [-2θ sin θ + 6 cos^2(θ) + 3 sin θ] dθ
..............= ∫(θ = 0 to π/2) [-2θ sin θ + 3(1 + cos(2θ)) + 3 sin θ] dθ
..............= [(2θ cos θ - 2 sin θ) + 3(θ + sin(2θ)/2) - 3 cos θ] {for θ = 0 to π/2}
..............= 1 + 3π/2.
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(iv) C₄: (0, 1, 0) to (0, 0, 0)
Via θ = 0 ==> Ψ(r, 0) = <r, 0, 0>,
∫c₄ F ⋅ dr = ∫(r = 0 to 1) <0, r, 0> ⋅ <1, 0, 0> dθ = 0.
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Adding the results from (i) - (iv) together yields
∫c F ⋅ dr = 1 + 3π/2.
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I hope this helps!
