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Calculus IIIChapter 6: Surface IntegralsSection 6-5

Stokes' Theorem — Question 8

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Question 8

Let CC be the boundary of the part of the plane z=2x−yz=2x-y above 0≤x≤1,0≤y≤2,0\le x\le 1,\qquad 0\le y\le 2, oriented consistently with the normal having positive zz-component. Let 𝑭=12⟨2z−3y,3x−z,y−2x⟩.\mathbf F=\frac 12\langle 2z-3y,\ 3x-z,\ y-2x\rangle. Compute ∮C𝑭⋅d𝒓\displaystyle\oint_C\mathbf F\cdot d\mathbf r.

Tasks

  1. Compute the curl of 𝑭\mathbf F.

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

  3. Apply Stokes’ Theorem and verify the orientation.

Original worksheet page 1: question and worked solution for 6-5-008
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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 ∇×𝑭=⟨1,2,3⟩.\boxed{\nabla\times\mathbf F=\langle 1,2,3\rangle}.

Step 2: Surface element With z=g(x,y)=2x−yz=g(x,y)=2x-y, 𝒏dS=⟨−gx,−gy,1⟩dA=⟨−2,1,1⟩dA.\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.

See the diagram in the original worksheet below.

Step 3: Apply Stokes The curl flux density is constant: ⟨1,2,3⟩⋅⟨−2,1,1⟩=3.\langle 1,2,3\rangle\cdot\langle-2,1,1\rangle=3. The projected rectangle has area 22, so ∮C𝑭⋅d𝒓=∬R3dA=6.\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 −6-6.

Original worksheet page 2: question and worked solution for 6-5-008

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

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