F is conservative if we can find a scalar funciton f such that grad(f) = F.
This would entail
[tex]\dfrac{\partial f}{\partial x}=10yz[/tex]
[tex]\dfrac{\partial f}{\partial y}=10xz[/tex]
[tex]\dfrac{\partial f}{\partial z}=10xy[/tex]
Integrate both sides of the first equation with respect to x :
[tex]f(x,y,z)=10xyz+g(y,z)[/tex]
Differentiate both sides with respect to y :
[tex]\dfrac{\partial f}{\partial y}=10xz=10xz+\dfrac{\partial g}{\partial y}[/tex]
[tex]\implies\dfrac{\partial g}{\partial y}=0\implies g(y,z)=h(z)[/tex]
Differentiate both sides with respect to z :
[tex]\dfrac{\partial f}{\partial z}=10xy=10xy+\dfrac{\mathrm dh}{\mathrm dz}[/tex]
[tex]\implies\dfrac{\mathrm dh}{\mathrm dz}=0\implies h(z)=C[/tex]
So we have
[tex]f(x,y,z)=10xyz+C[/tex]
that satisfies
[tex]\nabla f(x,y,z)=\mathbf F(x,y,z)[/tex]
and so F is indeed conservative.
