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Mathematics Calculus Wolfram Language Symbolic Computations

[?] Avoid problem with the parametrization of a line integral?

Jaime de la Mota

Posted 7 years ago

POSTED BY: Jaime de la Mota

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Hans Dolhaine

Hans Dolhaine, retired

Posted 7 years ago

But you could change your integral to a real line-integral of 1st kind by deleting the SquareRoot[ 13 ] from your integrand. I take a constant factor out of the integral: In[1]:= xx[t_] := 2 t; yy[t_] := 3 t; varianza = 2; integral = 1/(2 Pivarianza) Integrate[( Exp[(-1.) ((xx[t])^2 + (yy[t])^2)/(2. varianza)]) Sqrt[ D[xx[t], t]^2 + D[yy[t], t]^2], {t, 0, 1}] Out[3]= 0.139526 or In[4]:= xx[t_] := t Cos[ArcTan[3/2]]; yy[t_] := t Sin[ArcTan[3/2]]; varianza = 2; integral = 1/(2 Pivarianza) Integrate[( Exp[(-1.) ((xx[t])^2 + (yy[t])^2)/(2. varianza)]) Sqrt[ D[xx[t], t]^2 + D[yy[t], t]^2], {t, 0, Sqrt[13]}] Out[6]= 0.139526

But you could change your integral to a real line-integral of 1st kind by deleting the SquareRoot[ 13 ] from your integrand. I take a constant factor out of the integral:

In[1]:= xx[t_] := 2 t; yy[t_] := 3 t;
varianza = 2;
integral = 
 1/(2 Pi*varianza) Integrate[( Exp[(-1.) ((xx[t])^2 + (yy[t])^2)/(2. varianza)]) Sqrt[
     D[xx[t], t]^2 + D[yy[t], t]^2], {t, 0, 1}]


Out[3]= 0.139526

In[4]:= xx[t_] := t Cos[ArcTan[3/2]]; yy[t_] := t Sin[ArcTan[3/2]];
varianza = 2;
integral = 
 1/(2 Pi*varianza) Integrate[( Exp[(-1.) ((xx[t])^2 + (yy[t])^2)/(2. varianza)]) Sqrt[
     D[xx[t], t]^2 + D[yy[t], t]^2], {t, 0, Sqrt[13]}]


Out[6]= 0.139526

POSTED BY: Hans Dolhaine

Hans Dolhaine

Hans Dolhaine, retired

Posted 7 years ago

And another remark. If you really want to do a line integral (1st type of line integral of a function f[ x, y ] along a line { x[ t ], y[ t ] } ) this is defined (there are good reasons for this definition) by J = Integral[ f[x[t], y[t] ] Sqrt[D[x[t], t]^2 + D[y[t], t]^2], {t, t1, t2}] giving in your case xx[t_] := 2 t; yy[t_] := 3 t; In[2]:= varianza = 2; integral = Integrate[(Sqrt[2^2 + 3^2] Exp[(-1.) ((xx[t])^2 + (yy[t])^2)/(2. varianza)]/(2 Pi*varianza)) Sqrt[ D[xx[t], t]^2 + D[yy[t], t]^2], {t, 0, 1}] Out[3]= 0.503068

And another remark. If you really want to do a line integral (1st type of line integral of a function f[ x, y ] along a line { x[ t ], y[ t ] } ) this is defined (there are good reasons for this definition) by

J = Integral[ f[x[t], y[t] ]  Sqrt[D[x[t], t]^2 + D[y[t], t]^2],  {t, t1, t2}]

giving in your case

xx[t_] := 2 t; yy[t_] := 3 t;

In[2]:= varianza = 2;
integral = 
 Integrate[(Sqrt[2^2 + 3^2] Exp[(-1.) ((xx[t])^2 + (yy[t])^2)/(2. varianza)]/(2 Pi*varianza)) Sqrt[
   D[xx[t], t]^2 + D[yy[t], t]^2], {t, 0, 1}]

Out[3]= 0.503068

POSTED BY: Hans Dolhaine