Copasetic Flow

This one kinda speaks for itself: Cool 12second Math Tricks: Odd Areas on 12seconds.tv

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 evalua...

This installment in the ‘It’s Obvious. Not!’ series relates to the second edition of the book “div grad curl and all that” by H.M. Schey, published by W. W. Norton. Near the end of the example I referenced here, the author of “div grad curl and all that” states that the following integral is ‘trivial’ and results in an answer of 1/6 pi, (specifically, this falls on page 26 of the second edition). As far as I can tell, the solution is more tedious than it is trivial. I’m hoping there really is a trivial solution. If you know it, please add it to the comments below. I’m posting two versions of the ‘tedious’ solution here. The integral in question: The author suggests switching to polar coordinates before solving the integral using the following substitutions: The substitution that’s not mentioned is: So, now to solve the ‘trivial’ integral, first use the substitutions mentioned above: Factoring out the -r squared term in square root: Using the trigonometry...