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Integration In Matlab

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Pranjal Pathak
Pranjal Pathak 2012년 3월 23일
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);
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Jan
Jan 2012년 3월 23일
Please post comments to answers in the corresponding thread.
Alexander
Alexander 2012년 3월 23일
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 2012년 3월 23일
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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