Wednesday, August 13, 2008

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:


Morgan Creighton said...

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.

Morgan Creighton said...

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