Re: Integration probelm
- To: mathgroup at smc.vnet.net
- Subject: [mg122050] Re: Integration probelm
- From: Heike Gramberg <heike.gramberg at gmail.com>
- Date: Tue, 11 Oct 2011 04:22:24 -0400 (EDT)
- Delivered-to: l-mathgroup@mail-archive0.wolfram.com
- References: <201110100827.EAA15353@smc.vnet.net>
All three of these integrals can be calculated after making the
substitution y=-r Sin[t]. For example for the first integral
(note that since -Sqrt[3]/2 r<y<0, the bounds for t become 0<t<Pi/3)
exp = Simplify[(1/(2 Sqrt[3]) r^2 (-Sqrt[r^2 + 2 y (y - Sqrt[3]
Sqrt[r^2 - y^2])])) /. {y -> -r Sin[t]}, r > 0 && 0 < t < Pi/3];
(* multiply exp with dy/dt == -r Cos[t] and integrate. *)
Integrate[-r Cos[t] exp, {t, 0, Pi/3}]
which gives
1/36 (9 + Sqrt[3] \[Pi]) r^4
The other two can be done in exactly the same way.
Heike.
On 10 Oct 2011, at 10:27, Jing wrote:
> Based on what I have done, I find that the Mathmatic can not integrate
the following expression:
>
> 1. Integrate[
> 1/(2 Sqrt[3])
> r^2 (-Sqrt[r^2 + 2 y (y - Sqrt[3] Sqrt[r^2 - y^2])]), {y, -r*
> Sqrt[3]/2, 0},
> Assumptions -> {y < 0 && r > 0 && y^2 < r^2 && r^2 < 4 y^2/3}]
>
> 2. Integrate[
> (y + Sqrt[3] Sqrt[r^2 - y^2]) ArcSec[(2 r)/(
> y + Sqrt[3] Sqrt[r^2 - y^2])], {y, -r*Sqrt[3]/2, 0},
> Assumptions -> {y < 0 && r > 0 && y^2 < r^2 && r^2 < 4 y^2/3}]
>
> 3.Integrate[
> 1/72 (8 Sqrt[3] r^3 -
> Sqrt[3] r^2 Sqrt[r^2 + 2 y (y - Sqrt[3] Sqrt[r^2 - y^2])] -
> 2 Sqrt[3] y^2 Sqrt[r^2 + 2 y (y - Sqrt[3] Sqrt[r^2 - y^2])] +
> 6 y Sqrt[(r^2 - y^2) (r^2 +
> 2 y (y - Sqrt[3] Sqrt[r^2 - y^2]))]), {y, -r*Sqrt[3]/2, 0},
> Assumptions -> {y < 0 && r > 0 && y^2 < r^2 && r^2 < 4 y^2/3}]
>
> When I try to integrate the above equations, the Mathmatic doesn't
work.
>
> I hope someone can help me to figure it out.
>
> Cheers
- References:
- Integration probelm
- From: Jing <jing.guo89@yahoo.com>
- Integration probelm