Line Integrals of Vector Fields β€” Question 3

Section 18-4 Β· Calculus pages 3405–3406

Show solutionHide solution

Question 3

Let CC travel from A=(0,0)A=(0,0) to B=(2,0)B=(2,0) and then from BB to C=(2,1)C=(2,1). For 𝑭(x,y)=βŸ¨βˆ’y,x⟩,\mathbf F(x,y)=\langle-y,x\rangle, evaluate ∫C𝑭⋅d𝒓\displaystyle\int_C\mathbf F\cdot d\mathbf r.

Tasks

  1. Parametrize each segment with the stated orientation.

  2. Compute the contribution from each segment.

  3. Explain geometrically why the first contribution vanishes.

Question 3 – Solution

Strategy. Split the piecewise path and add the two oriented line integrals.

Step 1: Horizontal segment Use 𝒓1(t)=⟨t,0⟩\mathbf r_1(t)=\langle t,0\rangle, 0≀t≀20\le t\le 2. Then 𝑭(𝒓1(t))=⟨0,t⟩,𝒓1β€²(t)=⟨1,0⟩,𝑭⋅𝒓1β€²=0.\mathbf F(\mathbf r_1(t))=\langle 0,t\rangle,\qquad \mathbf r_1'(t)=\langle 1,0\rangle,\qquad \mathbf F\cdot\mathbf r_1'=0.

Diagram shown in the original SVG worksheet.

The field is vertical along this segment while the motion is horizontal, so the vectors are perpendicular.

Step 2: Vertical segment Use 𝒓2(s)=⟨2,s⟩\mathbf r_2(s)=\langle 2,s\rangle, 0≀s≀10\le s\le 1. Thus 𝑭(𝒓2(s))=βŸ¨βˆ’s,2⟩,𝒓2β€²(s)=⟨0,1⟩,\mathbf F(\mathbf r_2(s))=\langle-s,2\rangle,\qquad \mathbf r_2'(s)=\langle 0,1\rangle, and ∫C2𝑭⋅d𝒓=∫012ds=2.\int_{C_2}\mathbf F\cdot d\mathbf r=\int_0^1 2\,ds=2.

Step 3: Add Hence ∫C𝑭⋅d𝒓=0+2=2.\boxed{\int_C\mathbf F\cdot d\mathbf r=0+2=2}.

Verification Only the upward component on the vertical segment contributes, and that component is the constant 22 over a distance of 11.