# Monte carlo integration in spherical coordinates

I was experimenting with creating a code for Montecarlo integration of a function specified in spherical coordinates. As a first straightforward quick examination I determined to write an examination code to get the strong angle under an angle $\theta_m$. For 2 arbitrary number $u$ and also $v$ in $[0,1)$. I create an uniform arbitrary tasting of the round angle making use of $\phi=2\pi u$ and also $\theta =\arccos(1-2v)$.

For N created factors, I have M factors for which $\theta < \theta_m$. My first suggestion was that given that I have an uniform tasting I need to have gotten the proper strong angle $\Omega=2 \pi (1-\cos (\theta_m))$ merely as $4\pi\times \frac{M}{N}$. In fact it resembles I the proper outcome appears just if I make use of:

$$\Omega=\sum_{i=1}^M \frac{4\pi}{N} 2 cos(\theta_i)$$

I can not see the reason that this needs to be proper. The chance circulation function in $\theta$ is $PDF=\frac{1}{2} \sin(\theta)$ so I prefer to anticipate I need to stabilize each factor of the amount by this function yet this does not jobs. What am I doing incorrect and also just how could I warrant the cosinus? Several many thanks!

3
2022-06-07 14:34:49
Source Share
You can substantially streamline this. The angle $\phi$ does not take place anywhere, so you are not in fact making use of $u$ or $\phi$, so we can overlook them. The angle $\theta$ just takes place in the kind $\cos\theta$, so it makes good sense to switch over to $z=\cos\theta$ rather.
Reiterated this way, your inquiry is whether creating (probably evenly dispersed) arbitrary numbers $v$ in $[0,1)$ and also computing $z=1-2v$ will certainly bring about $z>z_m$ in a portion $M/N=2\pi(1-z_m)/(4\pi)=(1-z_m)/2$ of instances. This straight partnership is plainly proper, so your formula is alright and also it appears there have to be something incorrect in your program rather if this is not functioning.