Line Integrals of Vector Fields — Question 2

Question 2 – Solution

Strategy. Dot the constant force with the velocity, then use the displacement as an independent check.

Step 1: Velocity and dot product $$\mathbf r'(t)=\langle-\sin t,\cos t,1\rangle,$$ so $$\mathbf F\cdot\mathbf r'(t)=-2\sin t-\cos t+3.$$

Step 2: Work $$\begin{align*} W&=\int_0^{2\pi}(-2\sin t-\cos t+3)\,dt\\ &=0+0+6\pi=\boxed{6\pi}. \end{align*}$$ The horizontal coordinates complete a full turn, so their signed contributions cancel. The vertical rise is $2\pi$, producing work $3(2\pi)$.

Verification The net displacement is $\mathbf r(2\pi)-\mathbf r(0)=\langle 0,0,2\pi\rangle$. For a constant force, its dot product with this displacement is $\langle 2,-1,3\rangle\cdot\langle 0,0,2\pi\rangle=6\pi$.