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Topic: Double integration
Replies: 10   Last Post: Dec 6, 2011 12:42 PM

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Roger Stafford

Posts: 5,929
Registered: 12/7/04
Re: Double integration
Posted: Dec 1, 2011 10:05 PM
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"Marcio Barbalho" <> wrote in message <jb8gd0$52k$>...
> INT (3,5) INT (0,SQRT(25-x^2)) SQRT((16x^2+16y^2-625)/(x^2+y^2-25)) dydx
- - - - - - - - - -
In case you are interested in using numerical methods, in your problem you can change to polar coordinates which will yield an inner integral easily expressed in closed form, thereby avoiding any elliptic integrals. Using this expression you need only perform a single numerical integration with the outer integral having no singularities present.

Changing to polar coordinates: x = r*cos(t) and y = r*sin(t) yields the double integral (using your notation):

INT (0,asec(5/3)) INT (3*sec(t),5) SQRT((625-16*r^2)/(25-r^2))*r dr dt

In the indefinite inner integral here, substituting u = sqrt(25-r^2) produces

INT -SQRT(225+16*u^2) du

which has an easy closed form. This can be used in performing the outer integral numerically. (I don't see any closed form for this outer integration.)

Roger Stafford

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