Fundamental Theorem for Line Integrals — Question 1

Section 18-5 · Calculus pages 3421–3422

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Question 1

Let f(x,y,z)=x2y+ez,f(x,y,z)=x^2y+e^z, and let CC be any smooth curve from A=(1,0,0)A=(1,0,0) to B=(2,1,ln⁡2)B=(2,1,\ln 2). Evaluate ∫C∇f⋅d𝒓.\int_C \nabla f\cdot d\mathbf r.

Tasks

  1. State the Fundamental Theorem for Line Integrals.

  2. Evaluate the integral without parametrizing CC.

  3. Explain why the unspecified shape of CC is irrelevant.

Question 1 – Solution

Strategy. Since the field is explicitly ∇f\nabla f, apply the endpoint formula ∫C∇f⋅d𝒓=f(B)−f(A)\int_C\nabla f\cdot d\mathbf r=f(B)-f(A).

Step 1: Evaluate the endpoint values f(B)=22(1)+eln⁡2=4+2=6,f(A)=12(0)+e0=1.f(B)=2^2(1)+e^{\ln 2}=4+2=6, \qquad f(A)=1^2(0)+e^0=1.

Diagram shown in the original SVG worksheet.

Step 2: Apply the theorem ∫C∇f⋅d𝒓=f(B)−f(A)=6−1=5.\int_C\nabla f\cdot d\mathbf r=f(B)-f(A)=6-1=\boxed{5}.

Step 3: Explain path independence Along any smooth parametrization 𝒓(t)\mathbf r(t), ∇f(𝒓(t))⋅𝒓′(t)\nabla f(\mathbf r(t))\cdot\mathbf r'(t) is the derivative of f(𝒓(t))f(\mathbf r(t)). Its integral therefore records only the net change in ff.

Verification The logarithmic coordinate was chosen so that eln⁡2=2e^{\ln 2}=2; using ln⁡2\ln 2 itself in f(B)f(B) would be an endpoint-substitution error.