Surface Integrals of Vector Fields — Question 6

Question 6 – Solution

Strategy. The field has zero normal component on each coordinate face, leaving only the slanted triangular face.

Step 1: Coordinate faces On $x=0$, the outward normal is $-\mathbf i$ and $\mathbf F\cdot(-\mathbf i)=-x=0$. Similarly, the flux densities on $y=0$ and $z=0$ vanish. These three fluxes are all zero.

Step 2: Slanted face Write $z=1-x-y$ over $$D=\{(x,y):x\ge 0,\ y\ge 0,\ x+y\le 1\}.$$ The outward side points away from the origin, and its vector element is $$\boxed{\mathbf n\,dS=\langle 1,1,1\rangle\,dx\,dy}.$$

On this face, $$\mathbf F\cdot\langle 1,1,1\rangle=x+y+z=1.$$

Step 3: Integrate and sum $$\iint_{\partial E}\mathbf F\cdot\mathbf n\,dS =0+\iint_D1\,dA =\boxed{\frac 12}.$$

Verification The field points away from the origin and is tangent to the coordinate faces, so all nonzero flux must leave through the slanted face and must be positive.