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Integration In Matlab

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Pranjal Pathak
Pranjal Pathak 2012-3-23
Thanks to andrei bobrov for your answer! Yes R1=R2. Your coding seems to work, but the value(answer) is a huge one to know the exact value. Can we command in MATLAB to give the calculated result only upto 4 decimal place.
syms r R1=sqrt(3)*(2*r.^2-1) R2=sqrt(3)*(2*r.^2-1) b = .7; S=2*pi*imag(int(exp(1i*b*R1)*r,r,0,1)*int(R2*exp(-1i*b*R1)*r,r,0,1)) out = vpa(S);
  2 个评论
Jan
Jan 2012-3-23
Please post comments to answers in the corresponding thread.
Alexander
Alexander 2012-3-23
Also, if something works for you, could you please accept the answer? So others know that the matter is resolved.

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Alexander
Alexander 2012-3-23
Just use double:
syms r
R1=sqrt(3)*(2*r.^2-1)
R2=sqrt(3)*(2*r.^2-1)
b = .7;
S=2*pi*imag(int(exp(1i*b*R1)*r,r,0,1)*int(R2*exp(-1i*b*R1)*r,r,0,1))
out = vpa(S);
double(out)
ans =
-0.7308
Or if you don't need vpa you can do out = double(S) directly.

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