Stokes' Theorem — Question 5

Section 19-5 · Calculus pages 3569–3570

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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.

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.

Diagram shown in the original SVG worksheet.

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.