Fall 2010 UBC

Student Question

Q. I encountered a problem when I was trying to understand one of the examples in the divergence theorem chapter. In 17.9,the second part of the question of example 2, the example calculates the region E that lies between the closed surfaces S1 and S2. They assigned n = -n1 and n= n2 on S1 and S2 respectively. I was wondering why one of them is negative and the other positive? If using this example as a guide, does it mean that for every region that is bounded inside another region, then the n will always be negative?

A fine question.  The student is asking about the text just after Example 2 in Section 17.9.

To use the Divergence Theorem, we have to identify an inside volume and its boundary skin.  This inside volume is like the flesh of an avocado, not counting the pit.  (Or the white of an egg, not counting the yolk.)  The boundary of this volume is not just the outer skin of the avocado, but also the skin of the pit.  To orient the surface (which consists of these two pieces) outward, we have to point away from the flesh in both cases (outward means away from the volume).  On the skin of the avocado, that’s an outward normal, \mathbf{n}_2 in the book.  On the skin of the pit, that’s a normal pointing away from the flesh — and toward the pit.  That’s -\mathbf{n}_1 in the book.

I found Chapter 17 Review, #38 to be a very good review question.  I recommend it.

I’m getting a lot of “is this on the exam?” and “will you give us this formula”?  I’m specifying some answers to these things here:

0) Formulas.  I’ll put the following curvature formula on the exam if you will need to use it, because it is, in my estimation, too time-consuming or just plain annoying to re-derive yourself:

\kappa = \frac{| \mathbf{r}'(t) \times \mathbf{r}''(t)|}{|\mathbf{r}'(t)|^3}

The other curvature formulas are respectively a) the definition (change in unit tangent with respect to arclength) and b) derived in one step using chain rule (see equation 9 in Chapter 14.3).

1) Yes, I could ask you to compute the centre of mass or moment of inertia, but I would remind you of the formulas if I do this.  Don’t pointlessly memorize the formulas, but it can be instructive to ponder why they are what they are.

2) Yes, I could ask you about tangential and normal components of acceleration, and paths of projectiles.  You should know the formulas, but these formulas can be re-derived if necessary, so the best is to study that derivation, as it is excellent review of the concepts of Chapter 14.  In doing so, you’ll probably memorise the formulas by accident, as they’re very simple.

3) Yes, I could ask you about Kepler’s laws of motion, but we didn’t do much more with them than learn their statements.  So just understand the statements.

4) Yes, The Law of Gravitation is a fine example for vector fields.  Be familiar with the example.  You should probably know the form of the field (how does it depend on \mathbf{r} or r), but don’t memorise the constants!

I will add to this list if there are specific topics you’re wondering about.  But the general answer is “yes, if it’s part of the course, it could be on the exam!”

The solution packet made by the UBC Math Club has some errors in it.  I will post the errors here (I will update this post if there are more reported to me).  After the exam, I will report these errors to the math club so that they can improve next year’s packet.

Some errors are quite serious, and some are excellent learning opportunities.  Please report any new errors you find to me.

April 2007, Problem 7 (d)

This is a TERRIBLE mistake.  It highlights something I’ve been trying to explain all semester!  (And it looks like they tried to correct it from a previous also wrong answer!)  The answer is that \mathbf{F} is NOT conservative, because if it were, then all line integrals on closed loops would be zero, and we found a counterexample in part (c).  The fact that \nabla \times \mathbf{F} = 0 would usually tell us that \mathbf{F} is zero, but that test doesn’t work here, because the domain of \mathbf{F} is not simply connected.

December 2009, Problem 5

The solutions give an incorrect domain to the parametrisation.  They give 0 \le x \le 3 and 0 \le y \le 3, which is a piece of the surface lying over the 1st quadrant of the xy-plane.  This surface is a parabolic bowl, facing downward, and shifted up, and we are asked to use the part of the bowl above the xy plane.  This piece looks like an upside-down bowl.  However, the solutions give a domain that would give just the noon-to-3-o’clock portion of the bowl.  If you have any doubts, please graph the two parametrised surfaces.  (In particular, if we dipped the bowl down into a sink of water, it would hold air.  But the piece of surface over the domain the solutions suggest wouldn’t hold any air!)  This error is an excellent learning opportunity — please do investigate what I’ve described in this paragraph.

December 2009, Problem 3 (c)

The curl is computed incorrectly (the first component) but it does not seem to affect the final answer.

April 2009, Problem 9, True-False #4

The answer is TRUE.  (The solutions claim it is false).  Both expressions are well-defined.  The right-hand side is zero because the curl of the gradient is always zero.  The left hand side is zero because whatever \nabla f is, it is parallel to itself, and parallel vectors have zero cross-product.

April 2009, Problem 4

An algebra error in the integral.  Your classmate explains it well:

“The solution applies Green’s theorem, which I can follow, but the integration from the 3rd to 4th lines of the computation doesn’t seem to be right. When I did it myself before looking at the solutions, I got 54 as my answer.. and I think the difference in in the integration. Simplifying the expression in the integral gives -8x^3 +24x^2 then they antidifferentiated it to be -2x^4+12x^3.  I think it should be -2x^4+8x^3, though.. which changes the final answer to 54.”

April 2007, #4

The integral is set up completely wrong.  And they should do it directly anyway; it’s much better to use the Divergence Theorem.  This is the same question as Stewart, Section 17.9 #11, and the answer is 32 \pi / 3.

April 2007, #5

There’s an algebra error in the integral.

April 2007, #8

There’s a missing r in the double integral.

April 2009, #8

There’s an algebra error in the integral.  The answer should be 36 \pi.

April 2005, #4

There’s an error in the algebra of u’s and v’s somewhere.

April 2009, #4

The algebra seems highly suspect (integral calculation).

December 2009, #3 (c) (ii)

Typo: it should read ax^2 + bxy + cy^2.

December 2009, #6 (c)

The solution makes vague reference to the “winding number” but it seems to claim the integral is 4\pi for any field with a hole in its domain.  That’s not true: the integral must depend on the vector field.  Not all vector fields with a hole at the origin have integrals 4\pi for curves circling the origin twice.  (Imagine multiplying the field by 7, for example; the value of the integral would change.)  So the solution isn’t a solution, it’s just a guess.

MORE IN THE COMMENTS BELOW!

Office Hours Post

Come back to this post to see when/where office hours will be (I’ll update it with info as it becomes available):
MONDAY (Dec 13):  6-8 pm MATX 1118 (Math Annex, just south of Math)

TUESDAY (Dec 14): 1:30 – 3 pm MATX 1102 (Math Annex, just south of Math)

WEDNESDAY (Dec 15): 5 pm – 7:30 pm MATH 204.

THURSDAY (Dec 16): YOUR EXAM!!

A very good review question for the final exam:

If S_1 is the upper hemisphere (z \ge 0) of a unit sphere oriented upward, and S_2 is the unit disk in the xy-plane oriented upward, is \iint_{S_1} \mathbf{F} \cdot d\mathbf{S} = \iint_{S_2} \mathbf{F} \cdot d\mathbf{S}?

What if I claimed that \iint_{S_1} \mathbf{F} \cdot d\mathbf{S} > \iint_{S_2} \mathbf{F} \cdot d\mathbf{S}.  Would you think I was lying?  If not, what could you conclude about the divergence of \mathbf{F}?

Answer in the comments if you like; I’ll post a further discussion tomorrow.

[EDIT:  Haven’t had any comments or questions about this one.  It is designed to remind you that you can’t always change one surface into another with the same boundary, but it requires that divergence is zero.]

An example computing flux

I’ve made a pencast of a flux example that might help you review some concepts from the last week of class. A pencast is a sort of animated lecture note.  You can watch it as a video, and see me write and hear me speak as I go through the example, or you can click anywhere on the notes to hear what I say as I write that penstroke.  It’s about 9 minutes long.  You’ll get the best experience if you click “fullscreen” so you can see everything in good detail.

Your feedback and comments would be helpful.  (I can see that a few of the penstrokes got recorded incorrectly by the pen, but I hope they don’t interfere with understanding.  Your feedback is appreciated!)

Here’s a direct link, in case the embedded video isn’t working for you (for me, I couldn’t click “fullscreen” on the embedded version):  M317F2010UBC – An example computing flux
brought to you by Livescribe

Vodpod videos no longer available.

About “Detecting Holes”

I made mention in class that

1) We always have curl \nabla f = 0.  In other words, if \mathbf{F} is conservative, then it has zero curl.

2) When you have a field \mathbf{F} which is defined on a simply connected domain, then if curl \mathbf{F} = 0, it must be that \mathbf{F} = \nabla f.  That is, on simply connected domains, all fields with zero curl come from the gradient of a function (i.e. are conservative).

A few notes to ward off common confusions:

A) If \mathbf{F} is defined on a domain with a hole in it, and curl \mathbf{F} = 0, then we don’t know anything about whether it is conservative.  The gravitational field is an example that has a hole in it, but it’s still conservative.

B) If \mathbf{F} is defined on a domain with a hole in it, and curl \mathbf{F} = 0, then we could consider just part of that domain that doesn’t have a hole in it, and on that restricted domain, \mathbf{F} must be conservative.  So being conservative depends on the domain!  A field can be conservative on one part of its domain without being conservative on all of it.  The pdf exam conceptual review packet “Important Example” number one works like this.

Finally [this paragraph is extra-curricular material not necessary for the exam, but for general interest], in class I said that we can detect holes in a domain by using the fact that the implication “zero curl implies conservative” only works on simply connected domains.  More precisely, what I mean is that if the domain has a hole in it, then there will be some field that has zero curl but is not conservative.  If there is no hole, then there will be no such field.  (Of course, it’s not obvious how to check whether some field with a given property exists among all infinitely many possible fields, so I don’t mean this as a practical detection tool, at least not without doing more theoretical work.)

Here are a selection of Students’ Study Tips.  Feel free to post comments and more tips in the comments section.

On saving time integrating

“When taking a double integral of a function f(x,y) dx dy.  If you can write f(x,y) as g(x)h(y), then you can take the integrals separately.  That is, f(x,y) dx dy = g(x) dx h(y) dy.  It can save a lot of time!”  – Huge

[EDIT:  This is a great suggestion, but you have to be careful about the limits.  For example, consider \int_0^1 \int_x^{x^2} xy \; dydx.  According to this suggestion, you can turn it into \int_0^1 \int_x^{x^2} y \; dy \; x \;dx.  But you can’t change it into \int_0^1 x \; dx \int_x^{x^2} y \; dy because the latter won’t simplify to a number! ]

On surface integral formulas

Anonymous was pleased to clear up the following confusion about surface integrals.  The surface integral formula

\iint_S \mathbf{F} \cdot \mathbf{n} dS

becomes

\iint_D \mathbf{F} \cdot (\mathbf{r}_u \times \mathbf{r}_v) dA

when you choose your parametrisation.  At first, this seems wrong, because the first formula appears to have a unit normal, while the second does not.  But in fact, this is because dS becomes | \mathbf{r}_u \times \mathbf{r}_v |dA, so we observe some cancellation:

\iint_S \mathbf{F} \cdot \mathbf{n} dS = \iint_D \mathbf{F} \cdot \frac{\mathbf{r}_u \times \mathbf{r}_v}{| \mathbf{r}_u \times \mathbf{r}_v | } | \mathbf{r}_u \times \mathbf{r}_v | dA = \iint_D \mathbf{F} \cdot (\mathbf{r}_u \times \mathbf{r}_v) dA

On the meaning of “closed”

When I say “closed” in this class, I mean “has no boundary,” and this occurs in two situations:

A “closed loop” is a curve with the same endpoint as starting point.

A “closed surface” is a surface which cuts out a finite piece of 3D space, i.e. has a finite inside distinct from its outside.  This also has no boundary or “edge”.

You may also come across the term “closed” in mathematics to mean “complement of an open set.”  (You have probably heard of a “closed interval” in this sense.)  In this course we don’t talk about open sets much, and I don’t think we’ve talked about closed sets at all.  So when I say closed, I am speaking of a curve or a surface, and I mean that it has no boundary.

— Kate (channelling Pippi Longstocking)

On visualising graphs

“Coming from a Math background and not a physics background, I always had trouble visualizing graphs in 3-D. I usually just refused, put it off, or tried to graph it online. Sometimes I would even plug in number after number until I somewhat could visualize what it looked like. It seems extremely trivial now, but I learned to cover up one of the variables and graph only the other 2, and then repeat this process for the other two combinations. So, take F=<P,Q,R>, I would cover up the P value and know that I am looking down the x-axis onto the yz plane (this is what I never understood before; the fact that I was looking down the axis that I was covering up). I would graph this and then repeat by covering up the Q value, which corresponds to looking down the y-axis onto the xz plane, and the R value which corresponds to looking down the z-axis which looks onto the xy plane. Once I had a graph for each plane (xy, xz, and yz), it became much easier to put it all together and see the graph in 3-D. It’s so easy now but has honestly saved me so much time and frustration! It has also helped me get a better understanding of calculus in general as I am more capable of visualizing what happens to surfaces. ” – Anonymous

On remembering the curl formula

“I always had trouble with how to remember which P-partial went with which R-partial, etc., when we’re trying to find 3-D vector field is conservative by equating certain partial derivatives. So, I finally realized there was a trick to remember this:

I first start out by writing, PQR in a column on the top of my exam, since they’re in alphabetical order. In the next column, shift PQR down by one spot, to obtain QRP
P = Q
Q = R
R = P

Then, the variable you’re taking the partial with respect to corresponds with where the corresponding variable in the column to the right is, looking at the vector equation.
For example, thinking of <P,Q,R> as <x,y,z> :
Py  = Qx (since Q is in the y-spot and P is in the x-spot)
Qz = Ry (since R is in the z-spot and Q is in the y-spot)
Rx = Pz (since P is in the x-spot and R is in the z-spot)

Then, to check, you should have the capital letters (field components) in alphabetical order on the left side, and the ‘with respect to’ variables in alphabetical order on the right side. :)” — Anonymous

A useful website (from Anonymous)

http://tutorial.math.lamar.edu/Classes/CalcIII/CalcIII.aspx

Curl and divergence:  which is which?

“At first, I keep messing up Curl and Divergence. I couldn’t remember which uses cross product and which uses dot product. After a while, I remembered it by thinking that Curl starts with a ‘C’ like ‘C’ross product and Divergence starts with a ‘D’ like ‘D’ot product.It’s that simple. =)” — Anonymous

Orientation of surfaces with several parts

“If you are integrating on a surface that consist of a least two different pieces, then the orientation of each integration has to agree with the others. Pay special attention if the question is asking for outward orientation instead of upward. For example, a hemisphere with its bottom disc will have normal vectors that are pointing in different directions for the two pieces. For the part of the hemisphere, its normal is pointing more upwards, but for the bottom disc, it’s pointing downward. ” — Jimmy Wales’s Personal Appeal

On prerequisites and review

“To do well in this course, I would HIGHLY recommend doing a thorough review of 3D surface/volume parameterizations. While it has been a year since I took MATH 253, I felt as though I remembered enough of it to be able to get by. I soon found out, however, that much of the material that is covered in MATH 317 draws heavily on what was taught in Multivariable Calculus, and it would have been wise for me to ensure that my understanding of the material from that class was solid. Much of my trouble with the material in this course isn’t a result of the new concepts being difficult to grasp, but rather because I am not as familiar with material that was learned in previous courses.”  — Anonymous

On choosing the orientation

“When computing for the normal sometimes it is confusing to set the order of the vector multiplication.  Just cross product the two vectors in whatever order and then later on visualize the actual situation or graph and set the sign accordingly without thinking about it too much during the calculation.”   — Anonymous

FRIDAY GAMESHOW!

On Friday, we’ll talk a bit more about the handout I gave on Wednesday — please bring it again.  Then we’ll have a Math 317 Review Gameshow!  You can choose to be a non-participating observer only, but if you want to play, make yourselves into teams of 3-4 people (I’ll help you find teammates in class if you don’t already have some in mind).  Start thinking now of your awesome mathematical team name!  And bring your lecture notes and text (it’s open book).

Expect snacks and prizes!

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