Surface Integrals of Vector Fields — Question 1

Question 1

Let $S$ be the part of the plane $z=1+x+y$ above the unit square $0\le x,y\le 1$, oriented upward. Let $$\mathbf F(x,y,z)=\langle x,y,z\rangle.$$

Tasks

1. Find the upward-oriented vector surface element.

2. Compute the flux $\displaystyle\iint_S\mathbf F\cdot\mathbf n\,dS$.

3. State how the result changes under downward orientation.

Question 1 – Solution

Strategy. Parametrize the graph by $(x,y)$ and retain the cross product itself, not merely its magnitude.

Step 1: Oriented element Use $$\mathbf r(x,y)=\langle x,y,1+x+y\rangle.$$ Then $$\mathbf r_x=\langle 1,0,1\rangle,\qquad \mathbf r_y=\langle 0,1,1\rangle,$$ so the upward vector area element is $$\boxed{\mathbf n\,dS=(\mathbf r_x\times\mathbf r_y)\,dx\,dy =\langle-1,-1,1\rangle\,dx\,dy}.$$

Step 2: Dot product On the surface, $z=1+x+y$, and hence $$\mathbf F\cdot(\mathbf r_x\times\mathbf r_y) =-x-y+z=1.$$ Therefore $$\boxed{\iint_S\mathbf F\cdot\mathbf n\,dS =\int_0^1\int_0^1 1\,dx\,dy=1}.$$

Step 3: Reverse orientation A downward normal is the negative of the chosen one, so the downward flux is $\boxed{-1}$.

Verification The cross product has positive third component, confirming the upward orientation; changing only the orientation must negate a flux integral.