Stokes' Theorem — Question 4

Question 4

Let $C$ be the intersection of the cylinder $x^2+y^2=4$ with the plane $z=x+y$. Orient $C$ consistently with the upward normal to the planar region it bounds. For $$\mathbf F=\left\langle-\frac y2,\frac x2,0\right\rangle,$$ compute $\displaystyle\oint_C\mathbf F\cdot d\mathbf r$.

Tasks

1. Identify a convenient spanning surface and its vector element.

2. Compute the curl flux.

3. Explain why the tilt of the curve does not alter the result.

Question 4 – Solution

Strategy. Use the planar patch $z=x+y$ over the radius-$2$ disk; its vertical vector-area component is especially simple.

Step 1: Curl and vector area We have $$\nabla\times\mathbf F=\langle 0,0,1\rangle.$$ For $z=x+y$, the upward vector element is $$\mathbf n\,dS=\langle-1,-1,1\rangle\,dA.$$

Step 2: Apply Stokes $$\begin{align*} \oint_C\mathbf F\cdot d\mathbf r &=\iint_{x^2+y^2\le 4} \langle 0,0,1\rangle\cdot\langle-1,-1,1\rangle\,dA\\ &=\iint_{x^2+y^2\le 4}1\,dA =\boxed{4\pi}. \end{align*}$$

Step 3: Interpret The curl is vertical, so only the vertical component of the oriented area vector matters. That component equals the area element of the horizontal projection, whose area is $4\pi$.

Verification Reversing the curve would force the downward normal and change the answer to $-4\pi$.