Stokes' Theorem: Question 6-5.9 & Solution

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

Determine the induced direction of $C$ as viewed from above.
Compute the curl flux through the five faces.
Verify the result using the missing bottom square.

Original worksheet page 1: question and worked solution for 6-5-009

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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.

See the diagram in the original worksheet below.

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 $C$ .

Original worksheet page 2: question and worked solution for 6-5-009

Stokes' Theorem — Question 9

Question 9

Question 9 – Solution