Fundamental Theorem for Line Integrals — Question 10

Question 10 – Solution

Strategy. Show that the vector-field integrand is exactly the ordinary derivative of the composite function $f\circ\mathbf r$.

Step 1: Parameter form Write $\mathbf r(t)=\langle x(t),y(t),z(t)\rangle$. By definition, $$\int_C\nabla f\cdot d\mathbf r =\int_a^b\nabla f(\mathbf r(t))\cdot\mathbf r'(t)\,dt.$$ Expanding the dot product gives $$\int_a^b\left(f_xx'(t)+f_yy'(t)+f_zz'(t)\right)dt,$$ where the partial derivatives are evaluated at $\mathbf r(t)$.

Step 2: Apply the chain rule The multivariable chain rule states $$\frac{d}{dt}f(\mathbf r(t)) =f_xx'(t)+f_yy'(t)+f_zz'(t).$$ Thus the line integral is $\int_a^b\frac{d}{dt}f(\mathbf r(t))\,dt$.

Step 3: Apply the one-variable theorem $$\int_a^b\frac{d}{dt}f(\mathbf r(t))\,dt =\boxed{f(\mathbf r(b))-f(\mathbf r(a))}.$$ The smoothness of $\mathbf r$ 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 $a$ and $b$ are interchanged, matching orientation reversal of the line integral.