Stokes' Theorem — Question 9

Question 9

Let $S$ consist of the top and four side faces of the unit cube $0\le x,y,z\le 1$, but not the bottom face. Give $S$ the outward orientation. Its boundary $C$ is the rim of the missing bottom square. For $$\mathbf F=\left\langle-\frac y2,\frac x2,0\right\rangle,$$ compute $\displaystyle\oint_C\mathbf F\cdot d\mathbf r$ using Stokes’ Theorem on the composite surface.

Tasks

1. Determine the induced direction of $C$ as viewed from above.

2. Compute the curl flux through the five faces.

3. Verify the result using the missing bottom square.

Question 9 – Solution

Strategy. The curl is vertical, so the four vertical faces contribute zero; only the top face remains.

Step 1: Boundary direction The outward orientation on the four sides induces counterclockwise traversal of the bottom rim when viewed from above.

Step 2: Flux through $S$ Since $$\nabla\times\mathbf F=\mathbf k,$$ its dot product with every horizontal side-face normal is zero. On the top, the outward normal is $\mathbf k$, so $$\boxed{\oint_C\mathbf F\cdot d\mathbf r =\iint_S(\nabla\times\mathbf F)\cdot\mathbf n\,dS =0+\iint_{[0,1]^2}1\,dA=1}.$$

Step 3: Replace the surface The upward-oriented bottom square has the same counterclockwise boundary $C$. Applying Stokes to that single square also gives $\iint 1\,dA=1$.

Verification Using the outward normal $-\mathbf k$ on the missing bottom face would induce the opposite boundary direction; it is not the compatible replacement for the stated $C$.