Line Integrals of Vector Fields — Question 5

Question 5 – Solution

Strategy. Substitute all three coordinates before taking the dot product with the three-dimensional velocity.

Step 1: Field and velocity $$\mathbf F(\mathbf r(t))=\langle t^3,t,t^2\rangle,\qquad \mathbf r'(t)=\langle 1,2t,3t^2\rangle.$$

Step 2: Dot and integrate $$\begin{align*} \mathbf F(\mathbf r(t))\cdot\mathbf r'(t) &=t^3+2t^2+3t^4,\\ \int_C\mathbf F\cdot d\mathbf r &=\left[\frac{t^4}{4}+\frac{2t^3}{3}+\frac{3t^5}{5}\right]_0^1\\ &=\frac 14+\frac 23+\frac 35 =\boxed{\frac{91}{60}}. \end{align*}$$

Step 3: Differential-form check Along the curve, $$z\,dx=t^3dt,\quad x\,dy=t(2t\,dt)=2t^2dt,\quad y\,dz=t^2(3t^2\,dt)=3t^4dt.$$ These are exactly the three terms above.

Verification Every term is nonnegative on $[0,1]$, and the common-denominator sum $(15+40+36)/60=91/60$ confirms the arithmetic.