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

Posted 3 months ago
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 Hello everyone. I have parametrized a straight line to go from point A (0,0) to point B (2,3) with two different aproaches. The first, is to make x=2t and y=3t and 0 Answer
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Posted 3 months ago
 1) your integral is not really a line-integral. 2) your ArcTan seems to be wrong:define, as you do xx[r_] := r Cos[w]; yy[r_] := r Sin[w]; then with In:= w = ArcTan[2/3]; {xx[Sqrt[2^2 + 3^2]], yy[Sqrt]} Out= {3, 2} and that is not your B.Then if you change your parameter from t to r your Integral should read (rule of Substitution) Integrate[f[x[t], y[t]], t] == Integrate[f[x[t[r]], y[t[r]]] D[t, r], r] Taking into account that your t-Domain ( 0, 1 ) is mapped unto the r -Domain ( 0, Sqrt[ 13 ] ) one can say t = r / Sqrt[ 13 ] Hence D[ t, r ] = 1 / Sqrt[ 13 ] With this and the correct ArcTan you get In:= varianza = 2; w = ArcTan[3/2]; Integrate[(Sqrt[2^2 + 3^2] Exp[(-1.) ((xx[r])^2 + (yy[r])^2)/(2. varianza)]/(2 Pi*varianza)) 1/Sqrt, {r, 0, Sqrt[2^2 + 3^2]}] Out= 0.139526 Answer
Posted 3 months ago
 Thank you very much. This is mostly helpful. Answer
Posted 3 months ago
 Should you want to proceed from {0,0} to {2,3} in a nonlinear way (nonlinear means, the straight line remains straight, but is divided in non-uniform intervals), e.g. t = r^n/Sqrt; then xx[r_] := r^n Cos[w]; yy[r_] := r^n Sin[w]; Indeed In:= w = ArcTan[3/2]; {xx[13^(1/(2 n))], yy[13^(1/(2 n))]} // PowerExpand Out= {2, 3} and In:= varianza = 2; integral = Integrate[(Sqrt[2^2 + 3^2] Exp[(-1.) ((xx[r])^2 + (yy[r])^2)/(2. varianza)]/(2 Pi*varianza)) D[t, r], {r, 0,13^(1/(2 n))}] Out= (1/(4 \[Pi]))n If[n > 0, 1.75333/n, Integrate[E^(-0.25 r^(2 n)) r^(-1 + n), {r, 0, 13^(1/(2 n))}, Assumptions -> n <= 0]] Looks complicated, but In:= Simplify[integral, n > 0] Out= 0.139526 Answer
Posted 3 months 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:= 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= 0.503068 Answer
Posted 3 months 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:= 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= 0.139526 or In:= 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}] Out= 0.139526 Answer
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