Surface Integrals of Vector Fields — Question 9

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 $\mathbf s(p,q)=\mathbf r(q,p)$, the images are the same. However, $$\mathbf r_u\times\mathbf r_v=\langle-1,-2,1\rangle, \qquad \mathbf s_p\times\mathbf s_q=\langle 1,2,-1\rangle.$$

Step 2: Compute signed fluxes Since $\mathbf F=\langle 1,0,0\rangle$ and both parameter domains have area $1$, $$\iint\mathbf F\cdot(\mathbf r_u\times\mathbf r_v)\,du\,dv =\boxed{-1},$$ whereas $$\iint\mathbf F\cdot(\mathbf s_p\times\mathbf s_q)\,dp\,dq =\boxed{1}.$$

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

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.