How can I deduce a lower hemisphere's boundary's orientation? Source: Stewart. Calculus: Early Transcendentals (6 edn 2007). p. 1098. §16.8, Exercise #19.
Despite reading this, I don't understand how to deduce the red sentence beneath.

$\text{19}.$ If $S$ is a sphere and $\mathbf{F}$ satisfies the hypotheses of Stokes' Theorem, show that $\iint_S \text{curl} \, \mathbf{F} \cdot d\mathbf{S} = 0 $.
Author's Solution:
Assume $S$ is centered at the origin with radius $a$. Let $U$ and $L$ be the upper and lower hemispheres, respectively, of $S.$ Then by Stokes' Theorem: $\iint_S (\nabla \times \mathbf{F}) \cdot d\mathbf{S} = \iint_U (\nabla \times \mathbf{F}) \cdot d\mathbf{U} + \iint_L (\nabla \times \mathbf{F}) \cdot d\mathbf{L} = \oint_{\partial U} \! \mathbf{F}\cdot d\mathbf{r} + \oint_{\partial L} \! \mathbf{F}\cdot d\mathbf{r}  $
But ${\partial U}$ is $x^2 + y^2 = a^2$ oriented counterclockwise  and $\color{red}{{\partial L} \text{ is the same circle oriented clockwise}}$. (Rest of solution omitted)

Criterion A for Orientation of Boundary Curve (from Stewart p. 1093):

If you walk in the positive direction around $\partial S$ and your head points in the direction of $\mathbf{\hat{n}}$, then the surface will always be on your left.

Criterion B (from Thomas' Calculus, 2009 12 edn, p 963):

If the thumb of a right hand points along $\mathbf{\hat{n}}$, then the fingers curl in the direction of $\partial S$.

I know that the orientation of $S$ determines the positive orientation of $\partial S$. Also, the outward unit normals determine any closed surface's positive orientation . Here, the outward unit normals have $z$-component $< 0$ which yields the person in red.

$\color{purple}{\text{By Criterion A, the red person's head is pointing downwards (in the $-z$ direction)}}.$
$\color{purple}{\text{The purple direction is compulsory for the surface to be on that person's left.}}$
By Criterion B, my thumb must point in the $-z$ direction. So my fingers curl in the orange direction.
Why's there a discrepancy between Criteria A and B? Where's the mistake?

$\Large{\text{Supplementary to Prof Shifrin's Answer:}}$
Here's my interpretation: The red person must travel in the purple direction, for the surface to be on her left. Then the purple direction is counterclockwise when viewed from above the $xy$-plane, BUT when viewed from below (as the red person is doing), the purple direction is CLOCKwise.
I had to draw a clock in my updated picture to spot this. Without this clock, I still don't understand why the purple direction can be both clockwise and counterclockwise, depending on from where I look at ahe $xy$-plane. Is there a more intuitive or natural explanation, without drawing clocks?

$\Large{\text{2nd Supplementary to Prof Shifrin's Answer:}}$
I still don't understand how to spot that reflections across the $xy$-plane reserve orientation (but preserve direction), without looking at a sheet of paper from above and below.
$\Large{1.}$ Why did you choose the basis {$j, -i, k$}, instead of the standard basis for $\mathbb{R}^3$?
$\Large{2.}$ Doesn't $-i := (-1,0,0)$ imply that the tip of the vector, lying on the $x$-axis, point towards $-\infty.$ How does it "still point inward"?
 A: In order to understand orientation on this intuitive level work with the following paradigm: Assume that the $(x,y)$-plane $z=0$  of $(x,y,z)$-space is oriented "upwards", in other words: that $n=(0,0,1)$ is the "positive" normal, and consider the unit disk $D$ in the $(x,y)$-plane with its boundary $\partial D$. Then the positive orientation of $\partial D$ is for all of us the counterclockwise orientation, as seen from the tip of $n$, or from high up on the $z$-axis.
This is in accordance with your Criterion A: A man walking along $\partial D$ with his head in direction $n$, i.e., upright on the $(x,y)$-plane,  has $D$ to his left.
It is also in accordance with your Criterion B: Position your right hand at $(1,0)\in\partial D$, the thumb in direction $n$. Then the fingers will curl around $\partial D$ in the counterclockwise direction, as seen from above.
Now to your lower hemisphere: A (red) man walking along $\partial L$ upright on the sphere, i.e., with the length of his body in the plane $z=0$, will have $L$ to his left when he walks clockwise, as seen by a spectator high up on the $z$-axis. The same spectator will perceive the other man walking along $\partial U$ having $U$ to his left as walking in the counterclockwise direction.
By the way: It is not necessary to cut up the sphere. It is enough to note that $\partial S^2$ for a full sphere $S^2$ is $0$.
A: You're wrong about criterion A. (Fantastic artwork, BTW.) For the lower hemisphere (think disk with downward-pointing normal) to be on your left, you must walk clockwise.
A: With many thanks to Profs Schrin and Blatter's response, I had an epiphany on how to comprehend/intuit, without looking at a sheet of paper from above and below, that when viewed from below, the purple direction is clockwise. The key picture is:

To do so, I followed Prof Blatter's advice just to look at the unit disk. The revealing moment came after drawing the clock on the unit disk. I saw that when viewed from above (the $xy$-plane), $\color{yellowgreen}?$ = 3 so purple = counterclockwise. On the other hand, when viewed from below, so when I look at the $xy$-plane from the tip of $(0, 0, -1)$ or from the point of view of the red person, $\color{yellowgreen}?$ = 9 so purple = clockwise.
The short answer is that because, for example,  $ \color{yellowgreen}{? =} \left\{\begin{array}{lr}
      \color{yellowgreen}3&\color{yellowgreen}{,\text{when viewed from above = tip of } (0,0,1)} \\
      \color{yellowgreen}9&\color{yellowgreen}{,\text{when viewed from below = tip of } (0,0,-1)}
    \end{array}
  \right. $
therefore when viewed from the other side of the $xy$-plane, a clock changes its order : 1 $\rightarrow$ 11, 2 $\rightarrow$ 10, 3 $\rightarrow$ 9, 4 $\rightarrow$ 8, 5 $\rightarrow$ 7 while 12 and 6 o'clocks remain the same. This explains why reflections across the $xy$-plane reverse orientation.
A: My Picture for Prof Blatter's Second Last Paragraph:

A: My Picture for Prof Shifrin's Answer on June 29th 2013: Green is for $\pm\mathbf{\hat{j}}$, red for $-\mathbf{\hat{i}}$, purple for $\pm\mathbf{\hat{k}}$. All hands are right hands herein. 

A: My Picture for Prof Shifrin's Answer on June 30th 2013. As before, green is for $\pm\mathbf{\hat{j}}$, red for $-\mathbf{\hat{i}}$, purple for $\pm\mathbf{\hat{k}}$.

