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>