tag:blogger.com,1999:blog-2400513859305780710.post1138239860438866289..comments2023-01-28T07:31:02.673-08:00Comments on Mathematical Food For Thought: Step By Step. Topic: Calculus.Jeffrey Wanghttp://www.blogger.com/profile/11114458640271201663noreply@blogger.comBlogger2125tag:blogger.com,1999:blog-2400513859305780710.post-48098654176145673622006-11-14T11:14:15.000-08:002006-11-14T11:14:15.000-08:00The first trick is pretty obvious. Let F = int(b ...The first trick is pretty obvious. Let F = int(b to a) f(x) dx. Then the inside integral is just F * f(y), and the integral of that (with respect to y) is just F * F.<br><br>The second trick represents the volume under a surface z = f(x) f(y) constrained on the positive xy-plane. If the positive xy-plane is viewed in polar coordinates, we want the (outer) integrand to represent the volume of a "radial slice" at a given angle theta. (More explicitly: we want to calculate the volume from theta to theta + dtheta.)<br><br>This "radial slice" is itself an integral, and cutting the "radial slice" perpendicularly produces a rectangle of width L = r theta = dr theta and height g(r, theta), giving our inside integrand. Integrating this expression gives an integrand for the volume of a "radial slice," which we integrate with respect to theta across the positive xy-plane (that is, from 0 to pi/2).t0rajir0unoreply@blogger.comtag:blogger.com,1999:blog-2400513859305780710.post-39643430613482032782006-11-14T12:21:54.000-08:002006-11-14T12:21:54.000-08:00Oh, I forgot to perform the FINAL TRICK. Let f(x)...Oh, I forgot to perform the FINAL TRICK. Let f(x) = e^{-x^2}, right? Okay. So<br><br>(the awesome integral)^2 = (the double integral) = (the radial integral) = int(0 to pi/2) int(0 to infty) r e^{-x^2 - y^2} dr dtheta = int(0 to pi/2) int(0 to infty) r e^{-r^2} dr dtheta = int(0 to pi/2) ( -1/2 e^{-r^2}, r from 0 to infty) dtheta = int(0 to pi/2) (1/2) dtheta = pi/4<br><br>so <br><br>(the awesome integral) = sqrt(pi) / 2t0rajir0unoreply@blogger.com