Finding a surface $S$ with property that the vector field $F$ has no work done on any path

Question

$F(x,y,z) = (2x+y+z)i + (2y+z+x)j + (2z+x+y)k$.
I previously proved that $F$ is conservative and so this means any closed loop on any piecewise differentiable surface evaluates the work done to be zero. (My interpretation, not the solution).

The question asks:
"Find a surface $S$ with the property that no work is done by $F$ when the particle moves along any path that begins at a point $(x_0,y_0,z_0)$ on $S$ and also ends at a point $(x_1,y_1,z_1)$ on $S$."

It seems to me that it wants a surface $S$ such that it's still zero when the two points are distinct? How would I do this or am I interpreting it wrong?

Answer 1

The force is conservative, so it's representable as the gradient of a potential field.

Looking at $F(x,y,z)=\nabla\phi(x,y,z)$, it's quite easy to guess (or calculate) $\phi$, then the equipotential surfaces are exactly the surfaces you want.

$\phi=\frac{1}{2}((x+y)^2+(y+z)^2+(z+x)^2)$ And the equipotential surfaces are a family of ellipsoids.