Find $\mathbf{F}$ such that $\nabla \times \mathbf{F} = (-3xz^2, 0, z^3)$ 
Let $S$ be the surface defined by $z = x^{2} + y^{2}$ for $z \leq 4$, oriented
  with upward-pointing normal.
Use Stokes' theorem to evaluate
  $\iint_{S}\left(\, -3xz^{2}\ ,\ 0\ ,\ z^{3}\,\right)\cdot{\rm d}\mathbf{S}$.
Hint: You may look for a vector field
  $
\mathbf{F}
=
M\left(\,x,y,z\,\right)\mathbf{i} + N\left(\,x,y,z\,\right)\mathbf{j}
$
  such that $\nabla \times \mathbf{F} = (-3xz^2, 0, z^3)$.

The question itself is straightforward except the fact that we need to find out $\mathbf{F}$.
Upon expanding $\nabla \times \mathbf{F}$, we get $$(-\frac{dN}{dz}, \frac{dM}{dz}, \frac{dN}{dx}-\frac{dM}{dy})=(-3xz^2, 0, z^3).$$
Problem arises as I try to find out $M$ and $N$. First, I have $$N = xz^3 + g(x,y)$$for some function $g$.It then gives $$\frac{dM}{dy}=g_x$$
which I cannot proceed. Is it even possible to find out $M$ and $N$ explicitly? Or do I need to solve the above question without finding $M$ and $N$?
 A: $\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
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 \newcommand{\imp}{\Longrightarrow}
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 \newcommand{\norm}[1]{\left\vert\left\vert\, #1\,\right\vert\right\vert}
 \newcommand{\pars}[1]{\left(\, #1 \,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
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\begin{align}&
\left.\begin{array}{rcl}
-\,\partiald{N}{z} & = & -3xz^{2}
\\[1mm]
\partiald{M}{z} & = & 0
\\[1mm]
\partiald{N}{x} - \partiald{M}{y} & = & z^{3}
\end{array}\right\}
\qquad\imp\qquad
\left\{\begin{array}{rcl}
N & = & xz^{3} + \fermi\pars{x,y}
\\[1mm]
M & = & \,{\rm g}\pars{x,y}
\end{array}\right.
\end{align}

Moreover,

$$
z^{3} + \partiald{f}{x} - \partiald{g}{y} = z^{3}
\qquad\imp\qquad\partiald{f}{x} = \partiald{g}{y}\tag{1}
$$

Indeed, we can always add $\ds{\nabla\Psi}$ to any $\ds{\bf F}$ we find because
  $\ds{\nabla\times\pars{{\bf F} + \nabla\Psi} = \nabla\times\bf F}$. For instance, in applying the Stokes Theorem, $\ds{\oint\nabla\Psi\cdot\dd\vec{r}=0}$

The simplest choice, which satisfies $\pars{1}$, is
$\ds{\fermi\pars{x,y} = {\rm g}\pars{x,y} = 0\,,\ \forall\ x,y}$. So,
$$
M=0\,,\qquad N=xz^{3}\qquad\imp\qquad{\bf F}=xz^{3}\,{\bf j}
$$
