Fundamental Theorem for Line Integrals โ€” Question 10

Section 18-5 ยท Calculus pages 3439โ€“3440

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Question 10

Let ff have continuous first partial derivatives on an open region containing a smooth curve ๐’“(t)\mathbf r(t), aโ‰คtโ‰คba\le t\le b. Prove โˆซCโˆ‡fโ‹…d๐’“=f(๐’“(b))โˆ’f(๐’“(a)).\int_C\nabla f\cdot d\mathbf r=f(\mathbf r(b))-f(\mathbf r(a)).

Tasks

  1. Rewrite the line integral using the parametrization.

  2. Apply the multivariable chain rule.

  3. Finish with the one-variable Fundamental Theorem of Calculus and identify the hypotheses used.

Question 10 โ€“ Solution

Strategy. Show that the vector-field integrand is exactly the ordinary derivative of the composite function fโˆ˜๐’“f\circ\mathbf r.

Step 1: Parameter form Write ๐’“(t)=โŸจx(t),y(t),z(t)โŸฉ\mathbf r(t)=\langle x(t),y(t),z(t)\rangle. By definition, โˆซCโˆ‡fโ‹…d๐’“=โˆซabโˆ‡f(๐’“(t))โ‹…๐’“โ€ฒ(t)dt.\int_C\nabla f\cdot d\mathbf r =\int_a^b\nabla f(\mathbf r(t))\cdot\mathbf r'(t)\,dt. Expanding the dot product gives โˆซab(fxxโ€ฒ(t)+fyyโ€ฒ(t)+fzzโ€ฒ(t))dt,\int_a^b\left(f_xx'(t)+f_yy'(t)+f_zz'(t)\right)dt, where the partial derivatives are evaluated at ๐’“(t)\mathbf r(t).

Step 2: Apply the chain rule The multivariable chain rule states ddtf(๐’“(t))=fxxโ€ฒ(t)+fyyโ€ฒ(t)+fzzโ€ฒ(t).\frac{d}{dt}f(\mathbf r(t)) =f_xx'(t)+f_yy'(t)+f_zz'(t). Thus the line integral is โˆซabddtf(๐’“(t))dt\int_a^b\frac{d}{dt}f(\mathbf r(t))\,dt.

Step 3: Apply the one-variable theorem โˆซabddtf(๐’“(t))dt=f(๐’“(b))โˆ’f(๐’“(a)).\int_a^b\frac{d}{dt}f(\mathbf r(t))\,dt =\boxed{f(\mathbf r(b))-f(\mathbf r(a))}. The smoothness of ๐’“\mathbf r and continuity of the first partial derivatives make the chain-rule derivative continuous and integrable. The same proof applies piece by piece to a piecewise smooth curve, with intermediate endpoint values canceling.

Verification The right side reverses sign when aa and bb are interchanged, matching orientation reversal of the line integral.