Stokes' Theorem โ€” Question 1

Section 19-5 ยท Calculus pages 3561โ€“3562

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Question 1

Let CC be the unit circle x2+y2=1x^2+y^2=1 in the plane z=0z=0, oriented counterclockwise when viewed from above. For ๐‘ญ(x,y,z)=โŸจโˆ’y,x,zโŸฉ,\mathbf F(x,y,z)=\langle-y,x,z\rangle, evaluate โˆฎC๐‘ญโ‹…d๐’“\displaystyle\oint_C\mathbf F\cdot d\mathbf r using Stokesโ€™ Theorem.

Tasks

  1. Choose an oriented spanning surface and compute โˆ‡ร—๐‘ญ\nabla\times\mathbf F.

  2. Apply Stokesโ€™ Theorem.

  3. Verify the answer by direct parametrization of CC.

Question 1 โ€“ Solution

Strategy. Span the circle by its flat disk; the prescribed counterclockwise direction selects the upward normal.

Step 1: Curl and orientation โˆ‡ร—๐‘ญ=โˆฃ๐’Š๐’‹๐’Œโˆ‚xโˆ‚yโˆ‚zโˆ’yxzโˆฃ=โŸจ0,0,2โŸฉ.\nabla\times\mathbf F =\begin{vmatrix}\mathbf i&\mathbf j&\mathbf k\\ \partial_x&\partial_y&\partial_z\\-y&x&z\end{vmatrix} =\langle 0,0,2\rangle. Let SS be the unit disk with ๐’=๐’Œ\mathbf n=\mathbf k.

Diagram shown in the original SVG worksheet.

Step 2: Apply Stokes โˆฎC๐‘ญโ‹…d๐’“=โˆฌS(โˆ‡ร—๐‘ญ)โ‹…๐’dS=โˆฌS2dA=2ฯ€.\boxed{\oint_C\mathbf F\cdot d\mathbf r =\iint_S(\nabla\times\mathbf F)\cdot\mathbf n\,dS =\iint_S2\,dA=2\pi}.

Step 3: Direct check With ๐’“(t)=โŸจcos⁡t,sin⁡t,0โŸฉ\mathbf r(t)=\langle\cos t,\sin t,0\rangle, ๐‘ญ(๐’“(t))=โŸจโˆ’sin⁡t,cos⁡t,0โŸฉ=๐’“โ€ฒ(t).\mathbf F(\mathbf r(t))=\langle-\sin t,\cos t,0\rangle =\mathbf r'(t). Therefore โˆซ02ฯ€๐‘ญโ‹…๐’“โ€ฒdt=โˆซ02ฯ€1dt=2ฯ€\int_0^{2\pi}\mathbf F\cdot\mathbf r'\,dt=\int_0^{2\pi}1\,dt=2\pi.

Verification The boundary is counterclockwise from the tip of ๐’Œ\mathbf k, so the orientation pairing is correct.