Which Cross Product for the Desired Orientation of a Hyperboloid ? [Stewart P1103 16.9.8] 
P1103 16.9.$8.$ Evaluate the surface integral $\iint_S \mathbf{F} \cdot d\mathbf{S}$.   $\mathbf{F} = (x^3y,-x^2y^2,-x^2yz)$ and $S$ is the surface of the solid bounded by the hyperboloid $x^2 + y^2 -z^2 = 1$,
  and the planes $z = -2$ and $z = 2$.

Parametrisation for the open middle piece of the paraboloid:
$\mathbf{r}(u,v) = (1\cosh u \cos v, 1 \sinh u \sin v, 1 \sinh u) \, \forall \, -1 \le u \le 1, 0 \le v \le 2\pi$. 
How does one determine: $\partial_{\huge{u}}\mathbf{r} \times \partial_{\huge{v}}\mathbf{r}$ effects the right orientation? If a significant geometric or visual argument is necessary, would you please provide a picture?
Addendum: The question as printed only asks for a computation with Divergence Theorem, but I don't want to use Divergence Thm here; I only want to solve this with piecewise surface integrals.
 A: When a surface is embedded in $\mathbb R^3$, often it is described as a level surface of some scalar field.  Denote such a scalar field as $\phi$.  Any given level surface through a point $p$ has a tangent plane that is orthogonal to the gradient $\nabla \phi$ evaluated at $p$.  You can choose a normal vector $n = \pm \nabla\phi$.
Then, you can test whether this choice of normal vector is inward or outward by taking a divergence of the unit vector:
$$2H = -\nabla \cdot \hat n$$
This is the mean curvature.  If the mean curvature $H$ is positive, then the surface curves toward this choice of normal, and we would regard the normal as being "inward".  If $H$ is negative, then the surface curves away from the normal--the normal is "outward" pointing.  The mean curvature might change sign for some surfaces--this merely means that some surfaces do not have a global notion of inward vs. outward.

For this problem, we can take $\phi(x, y, z) = x^2 + y^2 - z^2$.  The gradient is $\nabla \phi = 2x \hat x + 2 y \hat y - 2z \hat z$.  The normalized gradient is $\hat n = \nabla \phi/\sqrt{4x^2 + 4y^2 + 4z^2} = \nabla \phi/2r$.  Then the mean curvature is
$$2H = -\frac{\nabla^2 \phi}{2r} - (\nabla \phi) \cdot \nabla \frac{1}{2r} = -\frac{1}{r} + \frac{\phi(x,y,z)}{r^3}$$
At $z = \pm 2$, this is definitely negative, as $\phi = 1$ and $r = 3$.
The only thing that remains to do is to compute $\partial_u r \times \partial_v r$ and see whether it is aligned with this choice of $n$ or anti-aligned.  This could be accomplished with a dot product.

Or, you know, you could just draw a picture.
A: The correct parametrization of the "middle piece" $H$ reads
$$\eqalign{{\bf r}(u,v)=(\cosh u\cos v,&\cosh u\sin v,\sinh u)\cr &(-{\rm arsinh}\, 2\leq u\leq{\rm arsinh}\, 2, \ 0\leq v\leq 2\pi)\ .\cr}\tag{1}$$
It follows that
$${\bf r}_u=(\sinh u\cos v,\sinh u\sin v,\cosh u), \quad {\bf r}_v=(-\cosh u\sin v,\cosh u\cos v,0)\ ,$$
so that we obtain
$${\bf r}_u\times {\bf r}_v=(-\cosh^2 u\cos v,\ -\cosh^2 u\sin v,\ \sinh u\cosh u)\ .$$
When $(u,v)=(0,0)$ one gets ${\bf r}_u\times {\bf r}_v=(-1,0,0)$, which shows that the chosen parametrization induces the "inward", i.e.,  "wrong" orientation of $H$. The intended integral $I$ over the outward oriented hyperboloid is therefore given by
$$I=-\int_{\hat H} {\bf F}\cdot ({\bf r}_u\times {\bf r}_v)\ {\rm d}(u,v)\ ,\tag{2}$$
where $\hat H$ denotes the parameter domain described in $(1)$. I leave the plugging in of the various expressions in terms of $u$ and $v$ into $(2)$ to you. When you do the integration with respect to $v$ a lot will simplify due to symmetry. For the integration with respect to $u$ you probably will need that $\cosh({\rm arsinh}\,2)=\sqrt{3}$.
The integrals over the two disks at $z=\pm2$ will vanish due to symmetry. (Note that ${\bf n}=(0,0,\pm1)$ there.)
