Question

Hey guys! I'm doing an exercise using Monte Carlo method. It's about calculating the center of mass of a two-dimensional figure . I've solved it but problem it's about calculating the error for the x and y coordinates of the center of mass.

Answer 1

\[x_g=\frac{ \int\limits xdm }{ \int\limits dm }\]
\[y_g=\frac{ \int\limits ydm }{ \int\limits dm }\]
where \[dm=\rho dxdy\]

Answer 2

How do I find the errors for \[x_g\] and \[y_g\]? I've tried error propagation but it wasn't correct.

Answer 3

where is the monte carlo method here

Answer 4

\[I \approx V \langle f\rangle \pm V \sqrt{\frac{ \langle f^2\rangle - \langle f\rangle^2 }{ N }}\]

where \[\langle f\rangle=\frac{1}{N} \sum_{i=1}^N f({\mathbf{x}}_i)\] and \[\langle f^2\rangle=\frac{1}{N} \sum_{i=1}^N f^2({\mathbf{x}}_i)\] (V is the multidimensional volume and N the number of points used in the method)

Answer 5

So we can do an approximation of the integrals using this method, but how about the errors of the center of mass coordinates for my problem?

Answer 6

error estimation can be done by considering the difference between your successive iterations.
Since it is based on random sampling, the results would vary with each run!

Answer 7

So what can I do?