[tex]\vec F(x,y,z)=z\,\vec\imath+2yz\,\vec\jmath+(x+y^2)\,\vec k[/tex]
a. The divergence of [tex]\vec F[/tex] is
[tex]\nabla\cdot\vec F=\dfrac{\partial(z)}{\partial x}+\dfrac{\partial(2yz)}{\partial y}+\dfrac{\partial(x+y^2)}{\partial z}[/tex]
[tex]\boxed{\nabla\cdot\vec F=2z}[/tex]
b. The curl of [tex]\vec F[/tex] is
[tex]\nabla\times\vec F=\left(\dfrac{\partial(x+y^2)}{\partial y}-\dfrac{\partial(2yz)}{\partial z}\right)\,\vec\imath+\left(\dfrac{\partial(z)}{\partial z}-\dfrac{\partial(x+y^2)}{\partial x}\right)\,\vec\jmath+\left(\dfrac{\partial(2yz)}{\partial x}-\dfrac{\partial(z)}{\partial y}\right)\,\vec k[/tex]
[tex]\boxed{\nabla\times\vec F=\vec0}[/tex]
c. [tex]\vec F[/tex] is conserviatve if there is a scalar function [tex]f[/tex] for which [tex]\vec F=\nabla f[/tex]. This means we would need
[tex]\dfrac{\partial f}{\partial x}=z[/tex]
[tex]\dfrac{\partial f}{\partial y}=2yz[/tex]
[tex]\dfrac{\partial f}{\partial z}=x+y^2[/tex]
From these conditions we get
[tex]f(x,y,z)=xz+g(y,z)[/tex]
[tex]\dfrac{\partial f}{\partial y}=\dfrac{\partial g}{\partial y}=2yz\implies g(y,z)=y^2z+h(z)[/tex]
[tex]f(x,y,z)=xz+y^2z+h(z)[/tex]
[tex]\dfrac{\partial f}{\partial z}=x+y^2+\dfrac{\mathrm dh}{\mathrm dz}=x+y^2\implies\dfrac{\mathrm dh}{\mathrm dz}=0\implies h(z)=C[/tex]
So we do find a potential function [tex]f[/tex],
[tex]\boxed{f(x,y,z)=xz+y^2z+C}[/tex]
and [tex]\vec F[/tex] is indeed conservative.
d. Since [tex]\vec F[/tex] is conservative, and [tex]C[/tex] is closed circle, the integral of [tex]\vec F[/tex] along [tex]C[/tex] is 0.
e. Since [tex]\vec F[/tex] is conservative, its integral along [tex]C[/tex] depends only on the endpoints. In particular,
[tex]\displaystyle\int_C\nabla f\cdot(\mathrm dx,\mathrm dy,\mathrm dz)=f(1,1,1)-f(0,0,0)=\boxed{2}[/tex]
