Let $M \subseteq \mathbb{R}^n$ be a $n-1$-dimensional manifold, and $N_x M$ the normal vector space of $M$ at a point $x \mathbb{R}^n$, that is, the (1-dimensional) space of vectors that are orthogonal to $\dot{\gamma}(0)$ for each differentiable curve $\gamma$ with $\gamma(0) = x$.
We define $\nu$ as a unit normal field (I'm not sure if that's the common name for it) on $M$ if it's a continuous vector field $\nu: M \to \mathbb{R}^n$ that satisfies:
(i) $\nu(x) \in N_x M$
(ii) $\|\nu(x)\| = 1 \space\space \forall x \in M$
If $M$ has a unit normal field, we define the integral of a vector field $F: M \to \mathbb{R}^n$ over $M$ as
$$\int_{(M, \nu)} F d \overset{\rightarrow}S := \int_M < F(x), \nu(x) > dS$$
where $S$ is the Surface measure, and $<.,.>$ the common scalar product.
(I hope that these are all the common definitions and names for these things, but to be sure, I wanted to point out my definitions first. The question itself starts here.)
Using these definitions, I now want to consider the paraboloid $P$, given by:
$$P = \{(x, y, z) \in \mathbb{R}^3: z = 4 - x^2 - y^2, x^2 + y^2 < 4 \} $$
I first want to find the unit normal field of $P$. Then, I want to integrate the vector field $F(x, y, z) = (x, y, xy + z)^T$ over $(P, \nu)$.
I would love to show some effort on my own, but I frankly don't really know how I can get started.
I know that $x^2 + y^2 = 4$ means that $(x, y)$ are within the open ball $B_4(0)$ of radius $4$ if $(x, y, z) \in P$; therefore I thought about coming up with a transformation so that $P$ is it's graph, something like $h: B_4(0) \to \mathbb{R}, \pmatrix{x \\ y} \mapsto 4 - x^2 - y^2$. Then I would have that $P = \{\pmatrix{x \\ y \\ h(x, y)}: (x, y) \in B_4(0)\}$. Does that help me somehow?
Any help would be appreciated.
Actually, I now got the solution for this excercice, so in case anyone is interested in it, I'm posting it here as an answer.
We first acknowledge that $P$ is the graph of $h(x, y) = 4 - x^2 - y^2, (x, y) \in \Omega = B_2(0)$. This is equivalent to $P$ being the global root-domain of the function $f(x, y, z) = z - h(x, y), (x, y, z) \in \Omega \times \mathbb{R}$.
Now we have that
$\nabla f(x, y, z) = \pmatrix{2x \\ 2y \\1} =: N(x, y)$ is the normal vector at a point $(x, y, z)$, and therefore,
$$\nu(x, y, z) = \frac{1}{\sqrt{1 + 4x^2 +^2}} \pmatrix{2x \\ 2y \\ 1}$$
is a unit normal field of $P$.
Now, to evaluate the integral, we consider the function $F(x, y, z) = \pmatrix{x \\ y \\ x y + z}$ from the question, and insert it into the definition of the integral:
$\int_{(P, \nu)} F d \overset{\rightarrow}S = \int_P < F, \nu > dS = \int_\Omega < F \pmatrix{x \\ y \\ h(x, y)} , N(x, y) > d (x, y) $
$$= \int_\Omega < \pmatrix{x \\ y \\ x y + 4 - x^2 - y^2}, \pmatrix{2x \\ 2y\\ 1} > d(x, y)$$ $$= \int_{B_2(0)} (x^2 + y^2 + 4) d(x, y) + \underbrace{\int_{B_2(0)} x y d(x, y)}_{= 0\text{, because } \Omega \text{ is invariant under } \pmatrix{x \\ y} \mapsto \pmatrix{-x \\ y}} $$
$$\underset{\text{Polar coordinates}}= 2π \int_0^2 (r^2 + 4) r dr = 24π $$