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Re: Definite integral

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  • Subject: [mg116213] Re: Definite integral
  • From: Leonid Shifrin <lshifr at>
  • Date: Sun, 6 Feb 2011 05:35:05 -0500 (EST)


Judging by your equations, you have a cubic curve rather than a parabola.
Could this
be a mistake? For a cubic curve, you get a fourth order general polynomial
under the
square root for the computation of the length, so there is no surprise that
it is hard to
compute. The indefinite integral does actually return an answer, but it is
huge and hardly
useful for anything.


On Sat, Feb 5, 2011 at 1:45 PM, <carlos at> wrote:

Have problems getting Mathematica to compute this
> definite integral (length of a parabola, for a graduate homework).
> Only completes in reasonable time if theta=0; else freezes.
> Any advice? Thanks.
> ClearAll[theta,Ls,m,g,=CF=B5,d]; xhat=x/Ls; theta=Pi/6;
> H=m*g*Ls^2/(8*d); epsilon=Simplify[m*g*Ls*Sin[theta]/H];
> dzhat=xhat*(1-xhat)/2*(1+(epsilon/6)*(1-2*xhat));
> zhat=xhat*Tan[theta]+dzhat;
> z=zhat*m*g*Sec[theta]*Ls^2/H; z=Simplify[z];
> Print["z=",z]; dzdx=Simplify[D[z,x]];
> Print["dzdx=",dzdx]; ds=Sqrt[1+dzdx^2];
> L=Integrate[ds,{x,0,Ls},Assumptions-
> >x>=0&&x<=Ls,Ls>0&&m>0&&g>0&&d>0&&theta>=0];
> L=Simplify[L,Ls>0&&L>0]; Print["L=",L//TraditionalForm];

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