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Calculus IIIChapter 6: Surface IntegralsSection 6-5

Stokes' Theorem — Question 4

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Question 4

Let CC be the intersection of the cylinder x2+y2=4x^2+y^2=4 with the plane z=x+yz=x+y. Orient CC consistently with the upward normal to the planar region it bounds. For 𝑭=⟨−y2,x2,0⟩,\mathbf F=\left\langle-\frac y2,\frac x2,0\right\rangle, compute ∮C𝑭⋅d𝒓\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.

Original worksheet page 1: question and worked solution for 6-5-004
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Question 4 – Solution

Strategy. Use the planar patch z=x+yz=x+y over the radius-22 disk; its vertical vector-area component is especially simple.

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

See the diagram in the original worksheet below.

Step 2: Apply Stokes ∮C𝑭⋅d𝒓=∬x2+y2≤4⟨0,0,1⟩⋅⟨−1,−1,1⟩dA=∬x2+y2≤41dA=4π.\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 4π4\pi.

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

Original worksheet page 2: question and worked solution for 6-5-004

Original worksheet layout. Use Enlarge or open the PDF for a closer view.

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