Question 7 of 8: Work Done by a Conservative Field Along a Curved Path

Question 7: Work Done by a Conservative Field Along a Curved Path (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. $\mathbf F=(x^2,\,y,\,-z)$; path $\mathbf r(t)=(6t,\,2\cos t,\,2\sin t)$ from $(0,2,0)$ to $(3\pi,0,2)$.

Find. The work $W=\displaystyle\int_C\mathbf F\cdot d\mathbf r$.

Approach. Check whether $\mathbf F$ is conservative — each component depends only on its own variable, so $\mathbf F=\nabla\varphi$ for some scalar potential $\varphi$. If so, $W=\varphi(\text{end})-\varphi(\text{start})$, independent of the actual path.

1. Confirm the parameter range. At $t=0$: $(0,2,0)$✓ (the start point). At $t=\pi/2$: $(3\pi,0,2)$✓ (the end point), so $t$ runs from $0$ to $\pi/2$.
2. Find the potential. Each component of $\mathbf F=(x^2,y,-z)$ depends only on its own variable, so $\mathbf F$ is conservative with potential $$\varphi(x,y,z)=\frac{x^3}3+\frac{y^2}2-\frac{z^2}2\qquad(\text{check: }\nabla\varphi=(x^2,y,-z)=\mathbf F\ \checkmark).$$
3. Evaluate at the endpoints. $$\varphi(0,2,0)=0+2-0=2,\qquad\varphi(3\pi,0,2)=\frac{(3\pi)^3}3+0-2=9\pi^3-2.$$
4. Work by the fundamental theorem for line integrals. $$W=\varphi(3\pi,0,2)-\varphi(0,2,0)=(9\pi^3-2)-2=9\pi^3-4.$$

$$W=\boxed{9\pi^3-4}$$