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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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