Example Problem
Use a parameterisation to find the flux of $F = (3xy, 0, -z)$ outward (normal away from the z-axis) through the cone $z = 9\sqrt{x^2 + y^2}$, $0 \le z \le 9$.
My Work
We will use cylindrical coordinates to parameterise the cone.
The formula for the flux through a surface is $\iint_S F(S(\rho, \theta)) \cdot \overrightarrow{N} dt$ where $S(\rho, \theta)$ is the parameterised surface and $\overrightarrow{N} = \dfrac{\partial{S}}{\partial{\rho}} \times \dfrac{\partial{S}}{\partial{\theta}}$.
The parameterised surface is $S(\rho, \theta) = (\rho \cos(\theta), \rho \sin(\theta), 9\rho) \forall \rho \in [0, 1], \theta \in [0, 2\pi]$.
$\dfrac{\partial{S}}{\partial{\rho}} = (\cos(\theta), \sin(\theta), 9)$
$\dfrac{\partial{S}}{\partial{\theta}} = (-\rho \sin(\theta), \rho \cos(\theta), 0)$
$\overrightarrow{N} = \dfrac{\partial{S}}{\partial{\rho}} \times \dfrac{\partial{S}}{\partial{\theta}} = (-9\rho \cos(\theta), -9\rho \sin(\theta), \rho)$
$F(S(\rho, \theta)) = (3\rho^2 \cos(\theta)\sin(\theta), 0, -9\rho)$
Flux $= \int^{2\pi}_{\theta = 0} \int^{1}_{0} (3\rho^2 \cos(\theta)\sin(\theta), 0, -9\rho) \cdot (-9\rho \cos(\theta), -9\rho \sin(\theta), \rho) d\rho d\theta$
$= -6\pi$
Question and Explanation
I mostly understand flux problems, but one component that I have no idea how to do is to find the outward-pointing normal vector (as opposed to the inward-pointing normal vector). At the moment, I just take the cross product of the partial derivatives of the parameterised surface and hope I get the correct normal vector, which is obviously a serious deficiency in my knowledge.
How does one ensure that they get the outward-pointing normal vector, $\overrightarrow{N}$? I have no idea how to do this generally, across all (closed) surface flux problems. I would prefer an analytical method for finding this rather than graphing the surfaces every time, since I really don't think graphing everything is a robust/generalisable strategy — especially in higher dimensions.
The above problem is an example of how I solve these problems. As you can see, I just calculated $\overrightarrow{N}$ with no way of knowing whether it was the outward-pointing normal vector. In this specific example, I have a feeling that it is actually the incorrect (inward-pointing) normal vector. This is an example of the deficiency in my skills.
I would greatly appreciate it if people could please take the time to explain a methodology for ensuring that I find the outward-pointing normal vector when solving these types of problems. I've been doing a lot of research on this, but I cannot find a clear explanation of a methodology. I would very much appreciate help from MSE members.
Best Answer
It is a fact of life that $S^0$ consists of two points. This then implies that the normal line $n$ through a point $p$ of a hypersurface $S\subset{\mathbb R}^n$ has a natural origin, namely $p$, but two unit vectors spanning $n$, which differ by a factor $-1$. In a concrete geometrical or physical situation involving a two-dimensional surface $S\subset{\mathbb R}^3$ one has a particular "positive" normal ${\bf n}$ in mind, which one then describes as "outward", or "upward", etc., depending on the circumstances.
Assume that you are given a surface $S$ in geometrical or colloquial terms, say "a sphere with center $M$ and radius $r$, oriented outwards", and you then consult a catalogue of surface representations in order to obtain a parametric representation of $S$, then there is only a $50\%$ chance that the two parameters $u_1$, $u_2$ used generate via ${\bf f}_{.1}\times{\bf f}_{.2}$ the desired "orientation" of ${\bf n}$. For instance, in conection with spherical coordinates $(\phi,\theta)$ it plays a rôle which of $\phi$ and $\theta$ denotes the geographical latitude, and whether $\theta=0$ corresponds to the equator, or to the north pole.
To sum it up: There is no general rule. Given the desired "orientation" as well as a parametrization of the surface $S$ you have to determine "geometrically", i.e. by checking the resulting ${\bf f}_{.1}\times{\bf f}_{.2}$ at a convenient point of the parameter domain whether you have picked the "right orientation".