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

Posted 6 months ago
618 Views
|
5 Replies
|
4 Total Likes
|
 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
5 Replies
Sort By:
Posted 6 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[2]:= w = ArcTan[2/3]; {xx[Sqrt[2^2 + 3^2]], yy[Sqrt[13]]} Out[3]= {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[13]:= 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[13], {r, 0, Sqrt[2^2 + 3^2]}] Out[13]= 0.139526
Posted 6 months ago
 Thank you very much. This is mostly helpful.
Posted 6 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[13]; then xx[r_] := r^n Cos[w]; yy[r_] := r^n Sin[w]; Indeed In[3]:= w = ArcTan[3/2]; {xx[13^(1/(2 n))], yy[13^(1/(2 n))]} // PowerExpand Out[4]= {2, 3} and In[5]:= 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[6]= (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[7]:= Simplify[integral, n > 0] Out[7]= 0.139526