The gradient theorem applies here, because we can find a scalar function f for which ∇ f (or the gradient of f ) is equal to the underlying vector field:
[tex]\nabla f(x,y,z)=\langle2xy,x^2-z^2,-2yz\rangle[/tex]
We have
[tex]\dfrac{\partial f}{\partial x}=2xy\implies f(x,y,z)=x^2y+g(y,z)[/tex]
[tex]\dfrac{\partial f}{\partial y}=x^2-z^2=x^2+\dfrac{\partial g}{\partial y}\implies\dfrac{\partial g}{\partial y}=-z^2\implies g(y,z)=-yz^2+h(z)[/tex]
[tex]\dfrac{\partial f}{\partial z}=-2yz=-2yz+\dfrac{\mathrm dh}{\mathrm dz}\implies\dfrac{\mathrm dh}{\mathrm dz}=0\implies h(z)=C[/tex]
where C is an arbitrary constant.
So we found
[tex]f(x,y,z)=x^2y-yz^2+C[/tex]
and by the gradient theorem,
[tex]\displaystyle\int_{(0,0,0)}^{(1,2,3)}\nabla f\cdot\langle\mathrm dx,\mathrm dy,\mathrm dz\rangle=f(1,2,3)-f(0,0,0)=\boxed{-16}[/tex]
