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MathGroup Archive 2004

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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)         
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Crawley WA 6009                      mailto:paul at physics.uwa.edu.au 
AUSTRALIA                            http://physics.uwa.edu.au/~paul


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