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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