Surface Integrals of Vector Fields β€” Question 9

Section 19-4 Β· Calculus pages 3557–3558

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Question 9

A plane patch over the unit square has two parametrizations 𝒓(u,v)=⟨u,v,u+2v⟩,𝒔(p,q)=⟨q,p,q+2p⟩.\mathbf r(u,v)=\langle u,v,u+2v\rangle, \qquad \mathbf s(p,q)=\langle q,p,q+2p\rangle. For 𝑭=⟨1,0,0⟩\mathbf F=\langle 1,0,0\rangle, compare the flux values produced by the ordered cross products of the two parametrizations.

Tasks

  1. Show that 𝒔\mathbf s swaps the parameters of 𝒓\mathbf r.

  2. Compute both signed flux integrals.

  3. Explain how to make the two computations represent the same geometric orientation.

Question 9 – Solution

Strategy. Swapping parameters reverses the cross product, so a vector surface integral detects a sign change that a scalar surface integral would not.

Step 1: Compare cross products Because 𝒔(p,q)=𝒓(q,p)\mathbf s(p,q)=\mathbf r(q,p), the images are the same. However, 𝒓u×𝒓v=βŸ¨βˆ’1,βˆ’2,1⟩,𝒔p×𝒔q=⟨1,2,βˆ’1⟩.\mathbf r_u\times\mathbf r_v=\langle-1,-2,1\rangle, \qquad \mathbf s_p\times\mathbf s_q=\langle 1,2,-1\rangle.

Diagram shown in the original SVG worksheet.

Step 2: Compute signed fluxes Since 𝑭=⟨1,0,0⟩\mathbf F=\langle 1,0,0\rangle and both parameter domains have area 11, βˆ¬π‘­β‹…(𝒓u×𝒓v)dudv=βˆ’1,\iint\mathbf F\cdot(\mathbf r_u\times\mathbf r_v)\,du\,dv =\boxed{-1}, whereas βˆ¬π‘­β‹…(𝒔p×𝒔q)dpdq=1.\iint\mathbf F\cdot(\mathbf s_p\times\mathbf s_q)\,dp\,dq =\boxed{1}.

Step 3: Match orientation If the orientation from 𝒓u×𝒓v\mathbf r_u\times\mathbf r_v is prescribed geometrically, then the 𝒔\mathbf s calculation must use βˆ’(𝒔p×𝒔q)-(\mathbf s_p\times\mathbf s_q). It then also gives βˆ’1-1.

Verification The two raw answers are negatives, exactly as required because the ordered normals are negatives. Flux is invariant under reparametrization only after the geometric orientation is held fixed.