Stokes' Theorem — Question 8

Question 8

Let $C$ be the boundary of the part of the plane $z=2x-y$ above $$0\le x\le 1,\qquad 0\le y\le 2,$$ oriented consistently with the normal having positive $z$-component. Let $$\mathbf F=\frac 12\langle 2z-3y,\ 3x-z,\ y-2x\rangle.$$ Compute $\displaystyle\oint_C\mathbf F\cdot d\mathbf r$.

Tasks

1. Compute the curl of $\mathbf F$.

2. Find the oriented vector surface element for the plane.

3. Apply Stokes’ Theorem and verify the orientation.

Question 8 – Solution

Strategy. The field was chosen to have constant curl, while the plane has a constant vector area element.

Step 1: Curl Direct differentiation gives $$\boxed{\nabla\times\mathbf F=\langle 1,2,3\rangle}.$$

Step 2: Surface element With $z=g(x,y)=2x-y$, $$\boxed{\mathbf n\,dS=\langle-g_x,-g_y,1\rangle\,dA =\langle-2,1,1\rangle\,dA}.$$ Its third component is positive, as required.

Step 3: Apply Stokes The curl flux density is constant: $$\langle 1,2,3\rangle\cdot\langle-2,1,1\rangle=3.$$ The projected rectangle has area $2$, so $$\boxed{\oint_C\mathbf F\cdot d\mathbf r =\iint_R3\,dA=6}.$$

Verification Reversing the traversal of all four edges would correspond to the negative normal and would change the value to $-6$.