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Calculus IIIChapter 6: Surface IntegralsSection 6-5

Stokes' Theorem — Question 5

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Question 5

Let SS be the annulus 1≤x2+y2≤41\le x^2+y^2\le 4 in the plane z=0z=0, oriented upward. Its boundary consists of an outer circle CoC_o and inner circle CiC_i. For 𝑭=⟨−y2,x2,0⟩,\mathbf F=\left\langle-\frac y2,\frac x2,0\right\rangle, evaluate the circulation around the fully oriented boundary ∂S\partial S.

Tasks

  1. Determine the induced direction on each boundary component.

  2. Apply Stokes’ Theorem to the annulus.

  3. Verify the result from the two separate circle integrals.

Original worksheet page 1: question and worked solution for 6-5-005
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Question 5 – Solution

Strategy. For an upward annulus, the outer boundary is counterclockwise and the inner boundary is clockwise.

Step 1: Orient the boundary The right-hand rule gives ∂S=Co−CiCCW.\partial S=C_o-C_i^{\mathrm{CCW}}. Equivalently, traverse the outer circle counterclockwise and the inner circle clockwise.

See the diagram in the original worksheet below.

Step 2: Apply Stokes Since ∇×𝑭=𝒌\nabla\times\mathbf F=\mathbf k, ∮∂S𝑭⋅d𝒓=∬S1dA=π(22−12)=3π.\boxed{\oint_{\partial S}\mathbf F\cdot d\mathbf r =\iint_S1\,dA=\pi(2^2-1^2)=3\pi}.

Step 3: Check components For a counterclockwise circle of radius RR, ∮𝑭⋅d𝒓=πR2.\oint\mathbf F\cdot d\mathbf r=\pi R^2. Thus the outer circle contributes 4π4\pi, while the clockwise inner circle contributes −π-\pi. Their sum is 3π3\pi.

Verification Assigning both circles counterclockwise would give 5π5\pi and violate the induced-boundary convention.

Original worksheet page 2: question and worked solution for 6-5-005

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

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