Re: change of variable in multiple integrations...
- From: "Proginoskes" <proginoskes@xxxxxxxxxxxxx>
- Date: 3 Aug 2005 22:52:04 -0700
kiki wrote:
> "Proginoskes" <proginoskes@xxxxxxxxxxxxx> wrote in message
> news:1123058829.408325.6100@xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx
> >
> > kiki wrote:
> >> Hi all,
> >>
> >> I understand the change of variables in bi-varite integration:
> >>
> >> For example, if you have a<x1<x2<b, and you want to do integration:
> >>
> >> Integrate(Integrate(f(x1, x2), w.r.t x2 from x1 to b), w.r.t x1 from a to
> >> b)
> >>
> >> If I change order of integration, it becomes
> >>
> >> Integrate(Integrate(f(x1, x2), w.r.t x1 from a to x2), w.r.t x2 from a to
> >> b)
> >>
> >> This I understand well.
> >>
> >> How about more variables: for example, a<x1<x2<x3<x4<b,
> >>
> >> Original ordering: (first of all, is this a correct one)?
> >>
> >> Integrate(Integrate(Integrate(Integrate(f(x1, x2, x3, x4), w.r.t x4 from
> >> x3
> >> to b), w.r.t x3 from x2 to b, w.r.t x2 from x1 to b, w.r.t x1 from a to
> >> b)
> >>
> >> Then change the order of x2 and x4: (is the following correct?)
> >>
> >> Integrate(Integrate(Integrate(Integrate(f(x1, x2, x3, x4), w.r.t x2 from
> >> x1
> >> to x3), w.r.t x3 from x1 to x4, w.r.t x4 from x1 to b, w.r.t x1 from a to
> >> b)
> >
> > Yes, but I think your answer will be off by a sign. When you change
> > variables, you need to multiply f(...) by the determinant of a certain
> > matrix. This is why when you do integration with polar coordinates,
> > you have the extra r factor. In short:
> >
> > Change from (x,y) to (r,T) [T for theta]:
> > x = r cos T
> > y = r sin T
> >
> > dx/dr = cos T, dx/dT = -r sin T
> > dy/dr = sin T, dy/dT = r cos T
> >
> > | cos T -r sin T |
> > | sin T r cos T |
> >
> > = r cos^T - (-r) sin^T = r (cos^T + sin^T) = r.
> >
> > Thus: int (f(x,y) dy dx) = int(f(r cos T, r sin T) r dr dT).
> >
> >> As you can see, these kind things can be very confusing when number of
> >> variables become larger... is there any general rules governing such
> >> techniques?
> >
> > Yes. Look up "Jacobian" in a 3-semester calculus textbook, or
> > http://mathworld.wolfram.com/Jacobian.html .
> >
> > --- Christopher Heckman
> >
> > P.S. This seems correct; if anyone sees any obvious mistakes, please
> > let me know, because I'll be teaching this in a month or so and haven't
> > touched Calc III since my undergraduate days ...
> >
> > P.P.S. No, it was my senior year in high school, Fall 1988.
> >
>
> Hi Chris,
>
> I understood the Jacobian and change of polar coordinates etc. But I still
> did not see where did I miss a "sign"?
>
> Could you please elaborate directly on this problem...
You're swapping variables x2 and x4. This is like using the following
transformation:
y1 = x1
y2 = x4
y3 = x3
y4 = x2
Now you need to calculate partial derivatives. The Jacobian becomes
| 1 0 0 0 |
| 0 0 0 1 |
| 0 0 1 0 |
| 0 1 0 0 |
The determinant of this matrix is -1, so the new integral is of
f(x1,x4,x3,x2) times -1, not just f(x1,x4,x3,x2).
(Multiplying by -1 changes the sign.)
--- Christopher Heckman
.
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- change of variable in multiple integrations...
- From: kiki
- Re: change of variable in multiple integrations...
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- Re: change of variable in multiple integrations...
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