Line Integrals of Vector Fields — Question 4

Section 18-4 · Calculus pages 3407–3408

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Question 4

Let CC be the unit circle and let 𝑭(x,y)=⟨−y,x⟩.\mathbf F(x,y)=\langle-y,x\rangle. Compute ∮C𝑭⋅d𝒓\displaystyle\oint_C\mathbf F\cdot d\mathbf r when CC is traversed (a) counterclockwise and (b) clockwise.

Tasks

  1. Parametrize the counterclockwise circle and evaluate the integral.

  2. Use orientation reversal to obtain the clockwise value.

  3. Interpret the signs using the direction of the field.

Question 4 – Solution

Strategy. On the unit circle this field is the counterclockwise unit tangent, making the dot product especially simple.

Step 1: Counterclockwise traversal Take 𝒓(t)=⟨cos⁡t,sin⁡t⟩,0≤t≤2π.\mathbf r(t)=\langle\cos t,\sin t\rangle,\quad 0\le t\le 2\pi. Then 𝑭(𝒓(t))=⟨−sin⁡t,cos⁡t⟩=𝒓′(t).\mathbf F(\mathbf r(t))=\langle-\sin t,\cos t\rangle=\mathbf r'(t).

Diagram shown in the original SVG worksheet.

Consequently, ∮C𝑭⋅d𝒓=∫02π𝒓′(t)⋅𝒓′(t)dt=∫02π1dt=2π.\oint_C\mathbf F\cdot d\mathbf r =\int_0^{2\pi}\mathbf r'(t)\cdot\mathbf r'(t)\,dt =\int_0^{2\pi}1\,dt=\boxed{2\pi}.

Step 2: Reverse the orientation Reversing a vector-field line integral changes its sign, so the clockwise value is −2π.\boxed{-2\pi}.

Verification The field assists counterclockwise motion at every point and opposes clockwise motion at every point, so the respective signs must be positive and negative.