Search All of the Math Forum:

Views expressed in these public forums are not endorsed by NCTM or The Math Forum.

Notice: We are no longer accepting new posts, but the forums will continue to be readable.

Topic: Rubi 4.5 released
Replies: 21   Last Post: Jun 29, 2014 4:29 AM

 Messages: [ Previous | Next ]
 Nasser Abbasi Posts: 6,677 Registered: 2/7/05
Re: Rubi 4.5 released
Posted: Jun 22, 2014 3:30 PM

On 6/22/2014 7:01 AM, clicliclic@freenet.de wrote:

> It may be easier to use the simplified problem from a later post of this
>
>
> int(int(1/2*(hxz(r*cos(p)-a/2, r*sin(p), z) * vxz(r*cos(p)+a/2,
> r*sin(p), z) + vxz(r*cos(p)-a/2, r*sin(p), z) * hxz(r*cos(p)+a/2,
> r*sin(p), z))*r, r, 0, inf), p, 0, 2*pi)
>
> where:
>
> hxz(x,y,z) := 3*x*z/(x^2+y^2+z^2)^(5/2)
>
> vxz(x,y,z) := x*(sqrt(x^2+y^2+z^2) - z)/((x^2+y^2)*sqrt(x^2+y^2+z^2))
>
> Martin.
>

This is an impossible integrand :) Can't even do the internal one.
Gave up waiting. Tried Rubi 4.5 and Mathematica 9.01.

hxz[x_, y_, z_] := 3*x*z/(x^2 + y^2 + z^2)^(5/2);
vxz[x_, y_, z_] := x*(Sqrt[x^2 + y^2 + z^2] - z)/((x^2 + y^2)*Sqrt[x^2 + y^2 + z^2])
Clear[r, p, a, z];
integrand = 1/2*(hxz[r*Cos[p] - a/2, r*Sin[p], z]*vxz[r*Cos[p] + a/2, r*Sin[p], z] +
vxz[r*Cos[p] - a/2, r*Sin[p], z]*hxz[r*Cos[p] + a/2, r*Sin[p], z])*r;

integrand = FullSimplify[integrand]

(48*r*z*(a - 2*r*Cos[p])*(a + 2*r*Cos[p])*(-(1/((2*z)/(a^2 + 4*(r^2 + z^2)
- 4*a*r*Cos[p])^2 + 1/(a^2 + 4*(r^2 + z^2) - 4*a*r*Cos[p])^(3/2)))
- 1/((2*z)/(a^2 + 4*(r^2 + z^2) + 4*a*r*Cos[p])^2 +
1/(a^2 + 4*(r^2 + z^2) + 4*a*r*Cos[p])^(3/2))))/
((a^2 + 4*(r^2 + z^2) - 4*a*r*Cos[p])^(5/2)*
(a^2 + 4*(r^2 + z^2) + 4*a*r*Cos[p])^(5/2))

Int[integrand, r]
.... no answer after 1/2 hr. Aborted.

Please give a simpler one :)

--Nasser