Basically 3-D Darts throwing program

[KevinNukkE]

Homework Statement

Two spheres with radius 1 cm and centers (-.25,0,0) and (.25,0,0) are contained in a box bounded by planes x=±2, y=±2, z=±2. Use Monte Carlo to compute the volume fraction bounded by the spheres in the box. Sample a point (x,y,z) randomly within the box. If the point is inside either of the spheres, tally 1 for that point. The volume fraction is just the tally divided by the number of trials

Homework Equations

I am using Matlab and its horrible, terrible random number generator.

The problem seems to me like a 3-D version of the darts game. My questions are:
A. Do I need to repeat each plane variable 4 times to get the full circle?
B. Should I multiply my random numbers for x, y, and z by 2 (for both positive and negative) to have them "bound" inside the box?

The Attempt at a Solution

I did this for the 2-D Darts game and came up with

function darts
Nc=0;
n=100000000;
tic;
for i = 1:n
x = rand(1,n);
y = rand(1,n);
if x.^2+y.^2<1
Nc = Nc+1;
end
end
dartstime = n/toc;
pies = 4 * Nc;

Using the same principle method I have

%%Throwing darts positive
x = rand;
y = rand;
z = rand;
%x and y plane
if x^2+y^2<0.5
Nc = Nc+1;
end
%Repeat 4 times for each of the three? then for the negative circle?
%x and z plane
if x^2+z^2<0.5
Nc = Nc+1;
end
%y and z plane
if y^2+z^2<0.5
Nc = Nc+1;
end
%%Throwing Darts negative
x1 = -rand;
y1 = -rand;
z1 = -rand;
if x1^2+y1^2<0.5
Nc = Nc+1;
end
if x1^2+z1^2<0.5
Nc = Nc+1;
end
if y1^2+z1^2<0.5
Nc = Nc+1;
end
disp(Nc);

 

[KataKoniK]

disp(x);
disp(y);
disp(z);
Dartstime = n/toc;
volume = 4 * Nc;
volume = volume * 2;
disp(volume);

To answer your questions:

A. No, you do not need to repeat each plane variable 4 times. The spheres are 3-dimensional objects, so you only need to check if the point is inside the sphere at any given coordinate (x,y,z).

B. Yes, you should multiply your random numbers for x, y, and z by 2 to have them "bound" inside the box. This ensures that the points are within the bounds of the box (x=±2, y=±2, z=±2).

Your code looks correct, but there are a few things you can improve upon. Instead of using a for loop to generate the random numbers, you can use the "rand" function to generate all the random numbers at once. Also, instead of using multiple if statements, you can use a single if statement with logical operators to check if the point is inside the sphere. Here is an example of how your code can be improved:

% Define the radius of the spheres and the bounds of the box
r = 1; % radius of the spheres
xmin = -2; % minimum x coordinate of the box
xmax = 2; % maximum x coordinate of the box
ymin = -2; % minimum y coordinate of the box
ymax = 2; % maximum y coordinate of the box
zmin = -2; % minimum z coordinate of the box
zmax = 2; % maximum z coordinate of the box

% Generate n random points within the box
n = 100000000; % number of random points
x = (xmax-xmin).*rand(n,1) + xmin; % generate n random x coordinates
y = (ymax-ymin).*rand(n,1) + ymin; % generate n random y coordinates
z = (zmax-zmin).*rand(n,1) + zmin; % generate n random z coordinates

% Check if the points are inside the spheres
Nc = sum((x.^2 + y.^2 + z.^2) <= r^2); % count the number of points inside the spheres

% Calculate the volume fraction
volume_fraction = Nc/n; % divide the tally by the number of trials

% Multiply by 2 to account