TITLE: Which Cross Product for the Desired Orientation of a Sphere ? [Stewart P1091 16.7.23]
QUESTION [6 upvotes]: P1086: For a closed surface, the positive orientation is the one for which the normal vectors point outward from the surface, and inward-pointing normals give the negative orientation.
P1087: If $S$ is a smooth orientable surface given in parametric form by a vector function $\mathbf{r}(u,v)$, then it is automatically supplied with the orientation of the unit normal  $\mathbf{n} = \cfrac{\partial_u\mathbf{r} \times \partial_v\mathbf{r}}{\vert \partial_u\mathbf{r} \times \partial_v \mathbf{r} \vert} $...
P1093: The orientation of a surface S induces the positive orientation of the boundary curve C shown in the figure. This means that if one walks in the positive direction around the curve with one's head pointing in the direction of $\mathbf{n}$, then the surface is always on one's left.

How does one determine whether $\partial_{\huge{u}}\mathbf{r} \times \partial_{\huge{v}}\mathbf{r} \quad \text{ or } \quad \partial_{\huge{v}}\mathbf{r} \times \partial_{\huge{u}}\mathbf{r} \quad  $ (negatives of each other) matches the desired orientation?
Since a surface may be hard to sketch (especially under exam conditions), I was hoping for an argument that isn't geometric or visual.  But if geometry and visualisation are the easiest,  would you please provide pictures for your explanations?


P1091 16.7.$23 \text{ generalised.}$ $\mathbf{F} = (x,-z,y)$ and $S$ is the part of $x^2 + y^2 + z^2 = p$ in the first octant and oriented towards $(0,0,0)$. Evaluate the surface integral $\iint_S \mathbf{F} \cdot d\mathbf{S}$. For closed surfaces, use the positive (outward) orientation.

Solution: Since $S$ is a sphere, parameterize with $r(\theta, \phi) = (p\sin \theta \cos \theta, p \sin \theta \sin \theta, p \cos \phi)$.
Then $\mathbf{F[r(\theta, \phi)]} \cdot \color{red}{(\partial_{\theta} r \times \ \partial_{\phi} r )} = p^3 \sin^3 \theta \cos^2 \theta \qquad (♦)$
Then $\iint_S \mathbf{F} \cdot d\mathbf{S} = \iint_D \mathbf{F} \cdot (\partial_{\theta} r \times \ \partial_{\phi} r ) \, dA = p^3\int^{2\pi}_0 \cos^2 \theta \, d\theta \int^{\pi/2}_9 \sin^3 \phi \, d\phi = ... = p^3 \quad \pi \quad 1/3.$
The answer is given as $ -p^\color{red}{2} \quad \pi \quad 1/3 $.
How would've one determined that the cross product in (♦) coloured in red is wrong,
and that it should've been $\color{green}{ \partial_{\large{\phi}} r \times \partial_{\large{\theta}} r  }$ ?
Predicated on user Dan's Answer:

REPLY [2 votes]: A surface $S$ as such is a "two-dimensional smooth set of points" embedded in three-space. At each point $p\in S$ we have a tangent plane $T_p(\sim{\mathbb R}^2)$ with origin at $p$, and this tangent plane has a well defined orthogonal complement $N:=T_p^\perp$, a line through $p$ with origin at 
$p$. On this line one can measure lengths, but a-priori it does not have a  positive, let alone: correct, sense of direction.
For certain purposes, in particular when it comes to the computation of flows, one would like to single out one of the two possible senses as positive, and this in a manner depending continuously on $p$. This can be done in various ways, either by words like "outward", "inward", "upward", "to the right" referring to the implied $(x,y,z)$-coordinate system, or by saying that for a certain parametrization $(u,v)\mapsto{\bf r}(u,v)$ the orientation induced by ${\bf r}_u\times{\bf r}_v$ (in this order!) is positive. Such an explicit directive has to be given by the person that hands you the surface and tells you to do something with it; it does not come out of thin air.
When you are given a geometric description of $S$ (e.g., "the piece of an ellipsoidal surface with axes $\ldots$ bounded by $\ldots$" ), together with the intended orientation, and you look up a parametric representation of $S$ in a catalogue then you have to verify using geometric vizualization (even in an exam situation) whether ${\bf r}_u\times{\bf r}_v$ induces the intended orientation or not.