### Stokes Theorem: Keeping the circle flat and integrating theta

This installment of “It’s Obvious. Not!” looks at:

Book: “div grad curl and all that”

Edition: second

Author: H. M. Schey

Publisher W. W. Norton.

Page: 95

As is often the case in this excellent book, the author illustrates theory with a concrete example. The example demonstrates that Stokes Theorem works for a vector field described by:

where Stokes Theorem is evaluated using the path of the circumference of a unit circle in the x-y plane and using the surface enclosed by the unit circle on the x-y plane. I ran into a few minor points of confusion working through the example, and I’ve added my intermediate steps below.

When evaluating the line integral, the book immediately moves from:

immediately to:

My confusion:

What happened to the dx and the dz terms?

Remembering that the path lies entirely in the x-y plane, z is always equal to 0. So, the dx term above drops out. Also, because z is a constant value in the x-y plane, x dz always evaluates to 0.

The next step in the example evaluates the integral

The author points out that the integral is parameterized over theta, (hence the introduction of the derivative with respect to theta above and reminds the readers that:

My confusion:

Where did the cosine squared theta term come from and how does the final integral evaluate to pi?

Substituting for x in the second term of the integral evaluation above gives

OK, now we have one cosine theta into the mix, but where does the second one come from? Keeping in mind that:

and the derivative of the sine of theta is the cosine of theta, we have:

My difficulty arose in remembering to treat the y in dy as an independent quantity. In other words, I forgot that I could do the following:

so now, we have cosine squared theta as above and we only need to evaluate the integral. Using the integral tables at Wikipedia, we get:

Really interesting post, Thanks! One of my quibbles with Schey is that he provides an explicit formula for unit tangent vector t^ in Cartesian coordinates, but doesn't explore it in other coordinate systems, even in the problems. (Except for an incidental reference at the tail end of the book.)

You write that "because z is a constant value in the x-y plane, xdz always evaluates to 0". That's a neat way to think about it, which I haven't considered. The way I looked at that was that the unit vector k^ is always perpendicular to the unit tangent vector t^. So their dot product must vanish.

By the way, you don't need a table to evaluate the integral of a square of a sine or cosine. If you remember the well-worn identity that sin(a+b) = sin(a)cos(b) + cos(a)sin(b), then you can show that 2cos^2(theta) = 1 + cos(2theta), which is easier to integrate.

Again, thanks for the interesting post.
Whoops, I meant cos(a+b) = cos(a)cos(b) - sin(a)sin(b), and the identity that cos^2 + sin^2 = 1.

### Cool Math Tricks: Deriving the Divergence, (Del or Nabla) into New (Cylindrical) Coordinate Systems

The following is a pretty lengthy procedure, but converting the divergence, (nabla, del) operator between coordinate systems comes up pretty often. While there are tables for converting between common coordinate systems, there seem to be fewer explanations of the procedure for deriving the conversion, so here goes!

What do we actually want?

To convert the Cartesian nabla

to the nabla for another coordinate system, say… cylindrical coordinates.

What we’ll need:

1. The Cartesian Nabla:

2. A set of equations relating the Cartesian coordinates to cylindrical coordinates:

3. A set of equations relating the Cartesian basis vectors to the basis vectors of the new coordinate system:

How to do it:

Use the chain rule for differentiation to convert the derivatives with respect to the Cartesian variables to derivatives with respect to the cylindrical variables.

The chain rule can be used to convert a differential operator in terms of one variable into a series of differential operators in terms of othe…

### Lost Phone

We were incredibly lucky to have both been in university settings when our kids were born.  When No. 1 arrived, we were both still grad students.  Not long after No. 2 arrived, (about 10 days to be exact), mom-person defended her dissertation and gained the appellation prependage Dr.

While there are lots of perks attendant to grad school, not the least of them phenomenal health insurance, that’s not the one that’s come to mind for me just now.  The one I’m most grateful for at the moment with respect to our kids was the opportunities for sheer independence.  Most days, we’d meet for lunch on the quad of whatever university we were hanging out at at the time, (physics research requires a bit of travel), to eat lunch.  During those lunches, the kids could crawl, toddle, or jog off into the distance.  There were no roads, and therefore no cars.  And, I realize now with a certain wistful bliss I had no knowledge of at the time, there were also very few people at hand that new what a baby…

### Lab Book 2014_07_10 More NaI Characterization

Summary: Much more plunking around with the NaI detector and sources today.  A Pb shield was built to eliminate cosmic ray muons as well as potassium 40 radiation from the concreted building.  The spectra are much cleaner, but still don't have the count rates or distinctive peaks that are expected.
New to the experiment?  Scroll to the bottom to see background and get caught up.
Lab Book Threshold for the QVT is currently set at -1.49 volts.  Remember to divide this by 100 to get the actual threshold voltage. A new spectrum recording the lines of all three sources, Cs 137, Co 60, and Sr 90, was started at approximately 10:55. Took data for about an hour.
Started the Cs 137 only spectrum at about 11:55 AM

Here’s the no-source background from yesterday
In comparison, here’s the 3 source spectrum from this morning.

The three source spectrum shows peak structure not exhibited by the background alone. I forgot to take scope pictures of the Cs137 run. I do however, have the printout, and…