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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.

enter image description here

$\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"?

Pedro
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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 $0$.

Christian Blatter
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  • Thank you for your answer. You wrote: "Now to your lower hemisphere... as seen by a spectator $\color{red}{\text{high up.}}$" Since the red person's point of view is from the tip of the vector $(0, 0, -1)$, shouldn't this end with "...as seen by a spectator $\color{limegreen}{\text{from below}}$"? –  Jun 20 '13 at 07:09
  • @LaPrevoyance: See my edit. – Christian Blatter Jun 20 '13 at 07:57
  • Thank you for your edit. In my Second Answer to this question (the bottom of this Question thread), I've drawn a picture for your second last paragraph. Does it match your meaning 100%? –  Jun 29 '13 at 07:47
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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.

Ted Shifrin
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  • The terminology is confusing, admittedly. When we say the red person faces and walks clockwise, we mean that from our frame of reference, looking from above, not from his own frame of reference. This isn't surprising: If I look down on the $xy$-plane I must rotate counterclockwise to get from the $x$-axis to the $y$-axis; but if I look up (i.e., from underneath) I must rotate clockwise. To convince yourself, draw the axes on a piece of paper and look from both sides. ... By the way, j'adore votre nom :P – Ted Shifrin May 19 '13 at 17:34
  • Professeur, merci beaucoup pour votre réponse et votre reconnaissance de mon nom. After drawing the axes on a piece of paper and looking at it from below, I saw that I must rotate clockwise to get from the $x$-axis to $y$-axis. However, I'm wondering if I could've understood this withOUT looking at a sheet of paper from below? Is there a more abstract or clever way to perceive this change from counterclockwise to clockwise? How can I understand this while looking at the sheet of paper from above? –  May 23 '13 at 17:50
  • M LaPrevoyance, the key thing is that reflection across the $xy$-plane is orientation-reversing. This reflection turns your up-standing (:)) red person upside down but does not change the direction he is pointing along the curve. Say he is at the point $(1,0,0)$, so $\vec j$ is the counterclockwise-pointing tangent to the circle. Reflection sends the basis $\vec j, -\vec i,\vec k$ to the basis $\vec j, -\vec i, -\vec k$. For $-\vec k$ to be the normal vector to our downward-pointing disk, since $-\vec i$ still points inward, we must change the sign and make $-\vec j$ our tangent vector. – Ted Shifrin May 23 '13 at 19:02
  • Professeur, I thank you deeply for your multiple answers. I apologise for this delay, but even after a few weeks, I'm still stuck. Could you please look at my 3 second supplementary questions in my original post? –  Jun 18 '13 at 10:18
  • I'm at the point $(1,0,0)$. The vector $\vec j$ points tangent to the circle (the direction I'm walking), the vector $-\vec i$ points inward into the disk (think of flattening out the hemisphere), and the vector $\vec k$ points in the direction of my head. Note that everything is happening as I walk around the circle and I'm at the point $(1,0,0)$. – Ted Shifrin Jun 18 '13 at 12:54
  • You keep forgetting that you're standing at the point $(1,0,0)$. Put your tails of your vectors at the point. Then $(1,0,0)$ is the outward-pointing normal and its negative is the inward-pointing normal. – Ted Shifrin Jun 20 '13 at 11:13
  • Thank you very much again. I now understand that $-i$ points inward into the disk. We want $-k$ to be the normal vector to the downward-pointing disk. However, why do we need $-j$ as our NEW tangent vector? Also, for the upper hemisphere, why pick {$j, -i, k$} as the basis, and not the standard {$i, j, k$}? –  Jun 29 '13 at 07:56
  • See my response two above. First vector — tangent vector; second — pointing into the surface; third — outward normal to the surface. This must make a right-handed system. If the second two vectors are $-\vec i$ and $-\vec k$, the first must be $-\vec j$ if we are to have a right-handed system. (J'arrive en France à la fin de juillet :)) – Ted Shifrin Jun 29 '13 at 12:48
  • Thanks so much. I've drawn a picture in a separate Answer below for your latest response. Is it 100% right? Unfortunately, I still can't fathom: For the upper hemisphere, why pick {$j, -i, k$} as the basis, and not the standard {$i, j, k$}? –  Jun 30 '13 at 03:27
  • Think about how you apply the right-hand rule: You point your fingers along the curve [first vector], curl them into the surface [second vector], and then your thumb points in the direction of the outward normal [third vector]. – Ted Shifrin Jun 30 '13 at 03:38
  • After rumination, I understand that the RH rule applied for Criterion B would result in my thumb signifying $\mathbf{\hat{k}} = \text{ third vector }$. However, I can't grasp what would lead me to "point fingers along the curve" to obtain $\mathbf{\hat{j}}$ [first vector] and to "curl them into the surface" to obtain $-\mathbf{\hat{i}}$ [second vector]? –  Jul 08 '13 at 07:07
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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:

enter image description here

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.

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My Picture for Prof Blatter's Second Last Paragraph:

enter image description here

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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.

enter image description here

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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 $\pm\mathbf{\hat{k}}$.

enter image description here