Friday, May 17, 2013
Instead of using Green's Theorem, you could evaluate the line integral just as it is, breaking the path C into three separate segments.
Using Stokes' Theorem to establish "surface independence" is the simplest solution to Problem 12, but there are two alternatives. In the first alternative, Stokes' Theorem is used to replace the flux integral with the circulation about the surface's boundary. Since the boundary consists of four line segments, four line integrals are computed and added together. The results are summarized in (A) below.
Alternative (B) involves computing the flux of the curl over S directly, by calculating it for each of the five square faces of S and adding them together. The results are summarized below (but you'll have to do each calculation yourself if you want to check).
Saturday, May 11, 2013
Friday, May 10, 2013
Saturday, May 04, 2013
Saturday, April 13, 2013
Here is the “standard” solution for Problem 2 in rectangular coordinates, with x as the independent variable.
Here are two alternative solutions. In the first, we get a simpler rectangular double integral by letting y be the independent variable. In the second, we find our desired result by using rectangular coordinates for the entire square 0 ≤ x ≤ 2, 0 ≤ y ≤ 2, and then subtracting the integral for the quarter disk, 0 ≤ r ≤ 2, 0 ≤ θ ≤ 2π.
I am not pleased to report that several of you appear to believe a kind of “product rule for antidifferentiation.” That is, if F(x) is the antiderivative of f(x) and G(x) is the antiderivative of g(x), then F(x)G(x) must be the antiderivative of f(x)g(x). There is no such rule! Calculus III students should know better!
You can switch the roles of u and v in the following solution and get an equally valid result (although it changes the sign of J). I gave extra credit for evaluating the transformed integral, although that was contingent on getting the correct new integral in the first place.