Work Done by Force Field (Part 3)

We can parametrize the path of the particle: $\begin{cases} \vec{r}(t) = 3t\hat{i}+4t\hat{j}+5t\hat{k}\\ 0 \leq t \leq 1 \end{cases}$ The work done is then: $\begin{aligned} W &= \int_0^1 \vec{F} \cdot d \vec{r}\\ &= \int_0^1 \left((4t)(5t)\hat{i}+(3t)(5t)\hat{j}+(3t)(4t)\hat{k}\right) \cdot \left(3\hat{i}+4\hat{j}+5\hat{k}\right) dt\\ &= \int_0^1 \left(20t^2\hat{i}+15t^2\hat{j}+12t^2\hat{k}\right) \cdot \left(3\hat{i}+4\hat{j}+5\hat{k}\right) dt\\ &= \int_0^1 \left(60t^2+60t^2+60t^2\right) dt\\ &= \int_0^1 180t^2 dt\\ &= 60t^3 \Big|_0^1\\ &= \boxed{60} \end{aligned}$

See @Alan Enrique Ontiveros Salazar for a demonstration of how to parametrize and integrate to get the solution. Another way is to recognize that $\vec{F}$ is a curl-free (conservative) vector field. It can thus be thought of as the gradient of a potential function. The potential function in this case is $U(x,y,z) = xyz$ .

To find the work done, evaluate the potential function at both end points and take the difference:

$\Delta U = U(3,4,5) - U(0,0,0) = 3 \cdot 4 \cdot 5 - 0 \cdot 0 \cdot 0 = 60$