Re: integral question
- To: mathgroup at smc.vnet.net
- Subject: [mg49863] Re: integral question
- From: Paul Abbott <paul at physics.uwa.edu.au>
- Date: Wed, 4 Aug 2004 10:46:51 -0400 (EDT)
- Organization: The University of Western Australia
- References: <cen83i$jr$1@smc.vnet.net>
- Sender: owner-wri-mathgroup at wolfram.com
In article <cen83i$jr$1 at smc.vnet.net>, "quest04" <na at na.na> wrote: > 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. You are computing a line integral. See, e.g., http://ltcconline.net/greenl/courses/202/vectorIntegration/lineIntegrals. htm In general, if you parameterize x and y as a function of t then the line integral of f[x,y] from {x[a],y[a]} to {x[b],y[b]} is Integrate[f[x[t], y[t]] Sqrt[x'[t]^2 + y'[t]^2], {t, a, b}] Your function is r[x_,y_] = x^2+y^2; and the (straight line) parameterization is x[t_] = t; y[t_] = t; where a = 0 and b = 1. Then the line integral is Integrate[r[x[t], y[t]] Sqrt[x'[t]^2 + y'[t]^2], {t, 0, 1}] Cheers, Paul -- Paul Abbott Phone: +61 8 9380 2734 School of Physics, M013 Fax: +61 8 9380 1014 The University of Western Australia (CRICOS Provider No 00126G) 35 Stirling Highway Crawley WA 6009 mailto:paul at physics.uwa.edu.au AUSTRALIA http://physics.uwa.edu.au/~paul