Surface Integrals — Question 10

Question 10 – Solution

Strategy. Recognize $\mathbf s(p,q)=\mathbf r(1-p,q)$, then compare cross products by magnitude rather than sign.

Step 1: Image and orientation The reflection $(u,v)=(1-p,q)$ maps the unit square onto itself, so the images agree. Moreover, $$\mathbf r_u\times\mathbf r_v=\langle-1,-2,1\rangle, \qquad \mathbf s_p\times\mathbf s_q=\langle 1,2,-1\rangle.$$ The normals are negatives, so the orientations are opposite; both magnitudes are $\sqrt 6$.

Step 2: First parametrization $$\iint_S(x+y)\,dS =\int_0^1\int_0^1(u+v)\sqrt 6\,du\,dv =\boxed{\sqrt 6}.$$

Step 3: Second parametrization $$\iint_S(x+y)\,dS =\int_0^1\int_0^1(1-p+q)\sqrt 6\,dp\,dq =\boxed{\sqrt 6}.$$

Verification A scalar surface integral uses $\lVert\mathbf r_u\times\mathbf r_v\rVert$, so reversing the cross product does not change $dS$. Here the reflection has Jacobian determinant $-1$, whose absolute value is $1$, confirming the equality under change of parameters.