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Calculus IIIChapter 5: Line IntegralsSection 5-5

Fundamental Theorem for Line Integrals — Question 7

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Question 7

Let f(x,y)=x3−3xy+y2.f(x,y)=x^3-3xy+y^2. A piecewise smooth path CC travels from A=(0,0)A=(0,0) to (1,0)(1,0), then to (1,1)(1,1), and finally to B=(2,1)B=(2,1).

Tasks

  1. Evaluate ∫C∇f⋅d𝒓\displaystyle\int_C\nabla f\cdot d\mathbf r using the theorem.

  2. Show how the endpoint changes telescope across the three pieces.

  3. Explain why the corners do not invalidate the theorem.

Original worksheet page 1: question and worked solution for 5-5-007
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Question 7 – Solution

Strategy. Apply the theorem to the whole piecewise smooth path, then expose the same cancellation piece by piece.

Step 1: Whole-path calculation f(A)=0,f(B)=23−3(2)(1)+12=8−6+1=3.f(A)=0, \qquad f(B)=2^3-3(2)(1)+1^2=8-6+1=3. Therefore ∫C∇f⋅d𝒓=3.\boxed{\int_C\nabla f\cdot d\mathbf r=3}.

See the diagram in the original worksheet below.

Step 2: Telescope Let P=(1,0)P=(1,0) and Q=(1,1)Q=(1,1). Then ∫C∇f⋅d𝒓=[f(P)−f(A)]+[f(Q)−f(P)]+[f(B)−f(Q)]=f(B)−f(A)=3.\begin{align*} \int_C\nabla f\cdot d\mathbf r &=[f(P)-f(A)]+[f(Q)-f(P)]+[f(B)-f(Q)]\\ &=f(B)-f(A)=3. \end{align*} Every intermediate potential value appears once positively and once negatively.

Step 3: Corners Smoothness is required on each segment, not at the finitely many joining corners. Adding the valid integral on each piece yields the integral on the piecewise smooth path.

Verification Direct values f(P)=1f(P)=1 and f(Q)=−1f(Q)=-1 give contributions 1,−2,41,-2,4, whose sum is 33.

Original worksheet page 2: question and worked solution for 5-5-007

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