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MathGroup Archive 2004

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Re: vector integral

  • To: mathgroup at smc.vnet.net
  • Subject: [mg49865] Re: vector integral
  • From: "Dr. Wolfgang Hintze" <weh at snafu.de>
  • Date: Wed, 4 Aug 2004 10:47:01 -0400 (EDT)
  • References: <cen6tg$ai$1@smc.vnet.net>
  • Sender: owner-wri-mathgroup at wolfram.com

If I understand your problem correctly I would restate it as follows.
The integral you're looking for is a vector because dr is a vector. 
Hence it is defined more completely as (denoting vectors by a_ etc.)

I_ = { Integrate[ r^2 , x], Integrate [r^2 , y] }

You have now to define the path in the x-y plane along which the 
integral is to be taken, and then express the relation between y and x 
along that path, so that you have y=y[x] and dy = D[y,x] dx and there 
remains an integral over x for both components of I_. Alternatively, and 
more genrally you can express both x and y as function od a parameter, 
say t, and transform the integrals into ones over t.

Here are some simple examples (direct method first)

1) from {0,0} to {a,0}

Here we have dy = 0. Furthermore we have y=0 hence r^2 = x^2. Thus

I_ = { Integrate[x^2,{x,0,a}], 0 } = { (1/3) a^3, 0 }

2) Your example path is the line from {0,0} to {1,1}.
Hence dy=dx, y=x which gives r^2 = x^2 + y^2 = 2 x^2. Thus

I_ = { Integrate[2 x^2,{x,0,1}], Integrate[2 x^2,{x,0,1}] }
= { (2/3) , 2/3 }

3) Part of the unit circle starting at {1,0} going to angle phi 
(parameter t)

x = Cos[t], y = Sin[t]
dx = - Sin[t] dt, dy = Cos[t] dt
r^2 = 1

I_ = { Integrate[ -Sin[t],{t,0,phi}], Integrate[ Cos[t], {t,0,phi}]
= { Cos[phi] - 1, Sin[phi] }

Hope this helps.

Regards,
Wolfgang



quest04 wrote:

> 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] along vector defined by 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!
> 
> 
> 


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