       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/2 r<y<0, the bounds for t become 0<t<Pi/3)

exp = Simplify[(1/(2 Sqrt) r^2 (-Sqrt[r^2 + 2 y (y - Sqrt
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 \[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)
>  r^2 (-Sqrt[r^2 + 2 y (y - Sqrt Sqrt[r^2 - y^2])]), {y, -r*
>   Sqrt/2, 0},
> Assumptions -> {y < 0 && r > 0 && y^2 < r^2 && r^2 < 4 y^2/3}]
>
> 2. Integrate[
> (y + Sqrt Sqrt[r^2 - y^2]) ArcSec[(2 r)/(
>  y + Sqrt Sqrt[r^2 - y^2])], {y, -r*Sqrt/2, 0},
> Assumptions -> {y < 0 && r > 0 && y^2 < r^2 && r^2 < 4 y^2/3}]
>
> 3.Integrate[
> 1/72 (8 Sqrt r^3 -
>    Sqrt r^2 Sqrt[r^2 + 2 y (y - Sqrt Sqrt[r^2 - y^2])] -
>    2 Sqrt y^2 Sqrt[r^2 + 2 y (y - Sqrt Sqrt[r^2 - y^2])] +
>    6 y Sqrt[(r^2 - y^2) (r^2 +
>        2 y (y - Sqrt Sqrt[r^2 - y^2]))]), {y, -r*Sqrt/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

```

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