I'll show you a different example, from which you should hopefully be able to work out your problem.
I'm going to approximate the area of a circle of radius 1. Firstly, my $x$ and $y$ coordinates will range from -1 to 1.
n=10000;
x=(1-(-1))*rand(n,1)+(-1);
y=(1-(-1))*rand(n,1)+(-1);
Now one of these points $(x,y)$ is inside the circle if the distance from the origin to that point is less than or equal to 1. The distance from the origin to the point is $r=\sqrt{x^2+y^2}$, so calculate that for every point:
r=sqrt(x.^2+y.^2); %// note I used element-wise dot exponentiation
Now see which points lie in the circle:
countInsideCircle=sum(r<=1) %// the number of points less than one unit from the origin
To calculate the area of the circle, first find the proportion of points that lie in the circle:
proportionInsideCircle=countInsideCircle/n
and then multiply by the area of the surrounding square (sides of length 2, so area is 4):
areaOfCircle=4*proportionInsideCircle
I did it and got A=3.1588 which isn't too bad as an approximation to $\pi$.
For your shape, you wont need the radius calculation, and you will just have to modify the countInsideCricle line to include the points you want, and also be careful about the area of your bounding box, it wont be 4.
Hint: s=sum(0<=x && x<=1 && y>=12*cos(x)) should give you the number of points satisfying $0\leq x\leq 1$ and $y\geq 12\cos x$.
The comments are correct that there are various paths and that it
would be helpful if you could give more information.
However, here are some generic investigations that I have found
useful. They might not be the end of your 'validation', but
they may raise issues that point towards the end.
(1) If you have $m = 10^6$ independent iterations in a Monte
Carlo simulation of a mean or proportion, then it is reasonable
to assume that mean or proportion is nearly normal. So find the
relevant SD and use $2SD/\sqrt{m}$ to give a serviceable
estimate of the margin simulation error.
(2) If you have questions about independence (as in Markov Chain
Monte Carlo, for example) then make an ACF plot for lags out
to about 40, and see how long it takes for dependence to decay.
Then 'thin' the results; for instance if ACF for lag 10 and
beyond is essentially 0, then use every 10th observation.
Then adjust m in step (1) accordingly.
(3) If you have questions about the stability of the simulation,
then look at 'traces' of running averages (cumulative sums up to
n divided by n) against n. They may be very rough-looking as the
simulation begins, but they should be quite smooth near the end.
(4) In the case of dependence there is always the risk that
the simulation is 'getting stuck' for periods of time (or forever).
A 'history plot' of the observations against time will help
you to see any serious problems along these lines.
(5) If you are generating several quantities that ought to be
stochastically or functionally independent, make scatter plots
of pairs of such quantities (along with Pearson and Spearman
correlations) to see if there are unexpected relationships.
If there are many such quantities, you might try making 'matrix
plots' of 'pairs' of variables.
(6) Make histograms of individual simulated quantities. Don't
just look at means (or medians) and standard deviations (or
interquartile ranges). A histogram of a random quantity should
have an 'envelope' or 'density estimator' that is nearly the
same as the distribution. If you see unexpected evidence of
strong multiple modes, you should try to figure out why.
(7) Sometimes an easily computed by-product of a simulation is
a quantity that is has a known value or distribution. Compute
them even if they are not part of the reason for doing the
simulation. If they do not have the anticipated values or
approximate the anticipated distributions, that is a sign of
trouble. It is usually better to compare empirical cumulative distribution functions (ECDFs) with known CDFs of distributions
than to compare histograms with known PDFs (no loss of acuity
due to binning with ECDFs).
I doubt that all of these suggestions will be applicable to any
one simulation, but give serious consideration to these
and other numerical and graphical descriptions of the output
of your simulation. You may find unexpected glitches in the
simulation program that are not hard to fix. You may also see
that everything is working wonderfully well, and then you will
feel much better about the usefulness of your simulation.
Most of the words I have put in 'single quotes' have some
chance of fetching useful information in an Internet search if you need further
information. Also, when you know a little more about some of
these methods and about how your simulation works, you might try
posting more specific and technical follow-up questions on
our sister 'statistics' site.
Best Answer
I assume you've got the gist of the idea behind the method from the comments above.
Here's the breakdown of the code:
k = k+1.If you're having trouble implementing this in some specific software (like Python or Matlab), please ask over at Stack Overflow.