Question

Find the work done by the force field f(x, y, z) = y + z, x + z, x + y on a particle that moves along the line segment from (1, 0, 0) to (5, 3, 2).

Answer 1

Note that the vector field is irrotational, since

$\nabla\times\mathbf f(x,y,z)=\mathbf 0$

which means $\mathbf f$ is conservative. This means there is some scalar potential function $f(x,y,z)$ such that $\nabla f(x,y,z)=\mathbf f(x,y,z)$. If we can find such a function, then we only need to evaluate the difference of $f(5,3,2)$ and $f(1,0,0)$ (because the gradient theorem holds in this case).

We're looking for a function $f(x,y,z)$ that satisfies

$\dfrac{\partial f}{\partial x}=y+z$
$\dfrac{\partial f}{\partial y}=x+z$
$\dfrac{\partial f}{\partial z}=x+y$

Integrate the first equation with respect to $x$ to get

$f(x,y,z)=xy+xz+g(y,z)$

Differentiate with respect to $y$, then we must have

$\dfrac{\partial f}{\partial y}=x+z=x+\dfrac{\partial g}{\partial y}$
$\implies\dfrac{\partial g}{\partial y}=z$
$\implies g(y,z)=yz+h(z)$
$\implies f(x,y,z)=xy+xz+yz+h(z)$

Differentiate with respect to $z$, and we have to have

$\dfrac{\partial f}{\partial z}=x+y=x+y+\dfrac{\mathrm dh}{\mathrm dz}$
$\implies\dfrac{\mathrm dh}{\mathrm dz}=0$
$\implies h(z)=C$
$\implies f(x,y,z)=xy+xz+yz+C$

Now, by the gradient theorem, we have

$\text{work}=\displaystyle\int_{\mathcal C}\mathbf f\cdot\mathrm d\mathbf r=f(5,3,2)-f(1,0,0)=31$

where $\mathcal C$ is the line segment.