RE: integral question

*To*: mathgroup at smc.vnet.net*Subject*: [mg49853] RE: [mg49834] integral question*From*: "David Park" <djmp at earthlink.net>*Date*: Wed, 4 Aug 2004 10:46:31 -0400 (EDT)*Sender*: owner-wri-mathgroup at wolfram.com

Doing this in some detail, we have for the original integral in terms of r... eqn0 = r^2 == x^2 + y^2 Sqrt /@ % Simplify[%, r > 0] eqn1 = Reverse[%] giving r^2 == x^2 + y^2 Sqrt[r^2] == Sqrt[x^2 + y^2] Sqrt[x^2 + y^2] == r r == Sqrt[x^2 + y^2] Now finding the r limits along your line. eqn1 /. Thread[{x, y} -> {0, 0}] eqn1 /. Thread[{x, y} -> {1, 1}] giving r == 0 r == Sqrt[2] Integrate[r^2, {r, 0, Sqrt[2]}] (2*Sqrt[2])/3 Converting into an integral with x and dx. We first solve for dr (== Dt[r]) eqn0 % /. y -> x Dt[%] drrule = Solve[%, Dt[r]][[1,1]] giving r^2 == x^2 + y^2 r^2 == 2*x^2 2 r Dt[r] == 4 x Dt[x] Dt[r] -> (2*x*Dt[x])/r Then calculating the new integral in terms of x and dx... r^2Dt[r] % /. drrule % /. r -> Sqrt[x^2 + y^2] % /. y -> x Simplify[%, x > 0] giving r^2*Dt[r] 2*r*x*Dt[x] 2*x*Sqrt[x^2 + y^2]*Dt[x] 2*Sqrt[2]*x*Sqrt[x^2]*Dt[x] 2*Sqrt[2]*x^2*Dt[x] We can then do the integral over x. Integrate[2*Sqrt[2]*x^2, {x, 0, 1}] (2*Sqrt[2])/3 David Park djmp at earthlink.net http://home.earthlink.net/~djmp/ From: quest04 [mailto:na at na.na] To: mathgroup at smc.vnet.net Hello, I have a question and was wondering if someone could help me with it. This is a general vector calculus problem, not specific to Mathematica. I have a simple integral as follows: Given r^2= x^2+y^2, solve Integral[r^2, dr] from point1 (0,0) to point2 (1,1), which would be evaluated from r=0 to r=sqrt[2] and gives answer = 2*sqrt[2]/3. The above is pretty simple, however, I am not sure how to formulate the problem when I convert the 'dr' back to cartesian coordinates as follows: Integrate [x^2+y^2, d????] and the limits??? WHat should 'dr' be in terms of dx? if my integrand is directly x^2+y^2. It cannot be a double integral, since the original problem was a single integral, but then how do you solve this?? Can anyone help? thanks!