Find the flux of $\mathbb{v} = (x^2-2xz, -2xy, z^2-x)$ downwards through the paraboloid $z = 1 - x^2 - y^2$ 
Consider the vector field $\mathbb{v} = \operatorname{curl}\mathbb{u}$, where
  $\mathbb{u} = (xy, xz^2, x^2y)$.
  Find the flux of $\mathbb{v}$
  downwards (negative z-component) through the paraboloid $z = 1 - x^2 - y^2$, where $x \geq 0$
  and $z \geq 0$.

So, $\mathbb{v} = (x^2-2xz, -2xy, z^2-x)$. Tried then apply the divergence theorem, but found out that it probably wouldn't work as the surface is open.
Kind of lost on where to go from here. I think the next step would be to just compute
$$\iint_{S} \mathbb{v}\cdot\mathbb{n}\;dS,$$
but I have to parametrize the surface first? No idea how to do that, or if I'm going in the right direction.
 A: Hints: You may parametrize the paraboloid with the coordinates
$$\begin{cases}
x(\rho,\theta) = \rho\cos\theta, \\ y(\rho,\theta) = \rho\sin\theta, \\ z(\rho,\theta) = 1 - x(\rho,\theta)^2 - y(\rho,\theta)^2 = 1-\rho^2
\end{cases}$$
where $\rho$ and $\theta$ range in a suitable domain (look at the way your paraboloid was defined), which is going to determine the limits of your integrals later on.
Then the “directed surface element” $d\mathbf a = \hat{\mathbf n}\ da$ is given by
$$\pm \left(\frac{\partial x}{\partial \rho}, \frac{\partial y}{\partial \rho}, \frac{\partial z}{\partial \rho}\right) \times \left(\frac{\partial x}{\partial \theta}, \frac{\partial y}{\partial \theta} , \frac{\partial z}{\partial \theta} \right) \ d\rho\ d\theta, $$
where the sign depends on the choice of normal direction.
Much simpler solution: the full paraboloid $z=1-x^2-y^2$, together with the planes $x=0$ and $z=0$, bound the compact set
$$K = \left\{ (x,y,z)\in \mathbb R^3\ |\ x\in [0,1], y\in \left[-\sqrt{1-x^2},\sqrt{1-x^2}\right],\ z\in [0,1-x^2-y^2]\right\} ,$$
which is half of a “capped paraboloid” resting on the $xy$-plane.
In other words, the boundary $\partial K$ is comprised of a vertical wall lying on the $yz$-plane, a horizontal wall lying on the $xy$-plane, and the portion $S$ of the paraboloid through which we are supposed to calculate the flux.

The total flux through $\partial K$, by the divergence theorem, is equal to the integral of the divergence over $K$ (which is easy: what is the divergence of a curl?), so the flux through $S$ is the same as the flux through the two walls, modulo sign...
A: $\mathbb{v} = (x^2-2xz, -xy, z^2-x)$.
Then, when $z=0\implies(x^2, -2xy, -x)$.
\begin{align}
\iint_S(x^2, -2xy, -x)\cdot(0, 0, -1)\;dS &= \iint_Sx\;dS \\
&= \int_0^1\int_{-\pi/2}^{\pi/2} (r\cos{\theta})r\;d\theta dr \\
&= \int_0^1 2r^2 \;d\theta dr \\
&= 2/3.
\end{align}
The integral for the other wall also becomes zero like the divergence of the $\operatorname{curl}\mathbb{u}$. This leaves us with the answer $2/3$.
Check out @giobrach's answer to see some more details to the solution.
