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integration problem

  • To: mathgroup at smc.vnet.net
  • Subject: [mg122236] integration problem
  • From: Jing <jing.guo89 at yahoo.com>
  • Date: Sat, 22 Oct 2011 06:05:16 -0400 (EDT)
  • Delivered-to: l-mathgroup@mail-archive0.wolfram.com

Hi,

I want to integration two eqations:

1/4 (-Sqrt[3] x + y) Sqrt[r^2 - 1/4 (-Sqrt[3] x + y)^2];

and -r^2 ArcCos[(-Sqrt[3]x+y)/(2r)]/2

For both of them, x is integrated from -Sqrt[r^2-y^2] to -y/Sqrt[3]+s; y is from (-Sqrt[3]s-Sqrt[12r^2-9s^2])/4 to (-Sqrt[3]s+Sqrt[12r^2-9s^2])/4. s, r are two constant and s>0 and Sqrt[3]s/2<r<s.

For example, for the first equation:
m18 = Integrate[
  1/4 (-Sqrt[3] x + y) Sqrt[
   r^2 - 1/4 (-Sqrt[3] x + y)^2], {x, -Sqrt[r^2 - y^2], -y/Sqrt[3] + 
    s }, Assumptions -> {y < 0 && r > 0 && r^2 > 4 y^2/3 && s > 0 && 
     Sqrt[3] s/2 < r < s}]

Result: 1/72 (-3 Sqrt[3] s^2 Sqrt[4 r^2 - 3 s^2 + 4 Sqrt[3] s y - 4 y^2] + 
   12 s y Sqrt[4 r^2 - 3 s^2 + 4 Sqrt[3] s y - 4 y^2] - 
   4 Sqrt[3] y^2 Sqrt[4 r^2 - 3 s^2 + 4 Sqrt[3] s y - 4 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]))] + 
   Sqrt[3] r^2 (4 Sqrt[4 r^2 - 3 s^2 + 4 Sqrt[3] s y - 4 y^2] - Sqrt[
      r^2 + 2 y (y - Sqrt[3] Sqrt[r^2 - y^2])]))

I used to ask a question quite same, someone suggest to me that I can use y=-r*sin(t), this variable alteration and then do the integration. I folowed this time, 

exp1 = Simplify[m18 /. {y -> -r Sin[j]}, r > 0 && 0 < j < Pi/3]
result:
1/72 (2 r^2 Sqrt[
    6 r^2 - 9 s^2 + 6 r^2 Cos[2 j] - 12 Sqrt[3] r s Sin[j]] - 
   3 s^2 Sqrt[
    6 r^2 - 9 s^2 + 6 r^2 Cos[2 j] - 12 Sqrt[3] r s Sin[j]] - 
   12 r s Sin[j] Sqrt[
    2 r^2 - 3 s^2 + 2 r^2 Cos[2 j] - 4 Sqrt[3] r s Sin[j]] - 
   3 r^3 Sin[2 j] Sqrt[2 - Cos[2 j] + Sqrt[3] Sin[2 j]] - 
   2 r^3 Sqrt[6 - 3 Cos[2 j] + 3 Sqrt[3] Sin[2 j]] + 
   Sqrt[3] r^2 Cos[
     2 j] (2 Sqrt[
       2 r^2 - 3 s^2 + 2 r^2 Cos[2 j] - 4 Sqrt[3] r s Sin[j]] + 
      r Sqrt[2 - Cos[2 j] + Sqrt[3] Sin[2 j]]))

Then I hope to do the integration and subsitute the two limits for y( if I directly put the two bounds of y into the integration, it will take over half hour to run and finally, it will show "no memroy"]:
m19 = Integrate[-r Cos[j] exp1, j]
It gives me really long euqtion, about hundreds lines long.
   
For the second equation, 
the result is really long, it is about hundreds line long and there is I inside the equation.

I don't know why "I" this symbol will appear in the equation.

Can someone help me to solve this integration problems.

Thanks.

Jing



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