Example:
Q:
f(x,y) = 3x^2 + 5y + 1
Integrate over 0 \le x \le 2, 1 \le y \le 2
A:
We can see this is a half pipe diving to the right.
\int_1^2 \int_0^2 (3x^2 + 5y + 1) dx dy = \int_1^2 (x^3 + 5xy + x) |_0^2 dy \\ = \int_1^2 10y + 10 dy \\ = (5y^2 + 10y)|_1^2 \\ = 25
The only special thing we did was solve x and treat y as a constant
A: Now we’ll do it again but with Monte Carlo Integration
We’ll generate twenty random numbers (ten numbers for x, and ten numbers for y).
Remember — To get numbers in range, we’ll have to use multiplication/division/addition/subtraction.
x_1 = 1.74254 \\ x_2 = 1.1909 \\ x_3 = 0.73206 \\ x_4 = 0.74202 \\ x_5 = 0.77804 \\ x_6 = 1.9432 \\ x_7 = 1.15826 \\ x_8 = 1.93952 \\ x_9 = 1.6486 \\ x_{10} = 1.85054 \\
y_1 = 1.24556 \\ y_2 = 1.48978 \\ y_3 = 1.615108 \\ y_4 = 1.23733 \\ y_5 = 1.6437 \\ y_6 = 1.26651 \\ y_7 = 1.8392 \\ y_8 = 1.81442 \\ y_9 = \\ y_{10} = \\ TODO finish copying random numbers
??? = 16.33713695
??? = 12.6834 TODO
??? = 10. TODO
??? = 8.838 TODO
??? = 11.0345 TODO
??? =
??? = 14.200 TODO
??? = 19.135263434912 TODO
??? = 18.22574588
??? = 20.5268948748
–
NOw let’s compute
\frac{\sum_i f(x_i, y_i)}{n} = \frac{150.3261015198}{10} = 15.03261015198
Therefore, the volume that we are looking for is the above number times 2 times 1. (TODO why? something to do with the area of the base of the shape formed by the limits?)
30.06522
“Four lines of code, basically, but very effective”
Example: Derivation of ???
x^2 + y^2 + (z-1)^2 = 1
TODO Graph with Plantuml (read doc, they have many wacky things)
This is a sphere centered at (0,0,1) with a radius of 1.
We can’t use integral because this isn’t a function, and it’s a closed surface.
Here’s what we’re gonna do:
The box will be 2x2x2.
We’re going to use a Monte Carlo technique where we:
\text{Constraints} \begin{cases} -1 \le x_i \le 1 \\ -1 \le y_i \le 1 \\ 0 \le z_i \le 2 \\ \end{cases}
^ just a formal way of saying, “the points must be in the box”
x_i^2 + y_i^2 + (z_i-1)^2 \le 1
^ just a formal way of saying, “these points are in the sphere”
Let the count be k, V_c be the volume of the cube, and V_s be the volume of the sphere.
Then, the volume of V_s can be approximated
V_s \sim \frac{k}{n} V_c
Example: Application
We’ll do twenty points.
TODO I’ll do this myself later dawg, I do not want to copy these damn points down.
To get the points in range, multiply by 2 and subtract 1.
In class we got like \frac{11}{20}(8)=4.4 > ^ this result from class (to be overwritten by my work) is pretty lucky, because the range could’ve been ??? to ???. > - You want to do way more points.
The true volume is 4.18879
Suppose the intersection of this conical shape cutting into the sphere.
\begin{aligned} x^2 + y^2 + (z-1)^2 &= 1 \\ \sqrt{x^2 + y^2} &= z^2 \end{aligned}
We want to find the area of the shape formed by the intersection.
TODO draw this stumpy icecream cone looking thing
Q: Find the volume
A:
The box will be the same size as before, so the constraints for the random points are the same.
The differnce is the constraints to check if points fall in the ice cream cone will be
x^2 + y^2 + (z-1)^2 \le 1 \\ \sqrt{x^2 + y^2} - z^2 \le 0
Note we moves the z^2 term left in the bottom function for niceness ??? TODO
Aside — The real volume is just V_c + \frac{V_s}{2}.
V_c = \frac{1}{3}\pi r^2 h = \frac{1}{3} \pi \\ V_s = \frac{4}{3}\pi r^3 = \frac{2}{3} \pi \\~\\ V_{ic} = \pi
Here are the twenty random points we’ll use:
Modify the following table of random numbers to be in the respective 3 ranges:
[-1,1] [-1,1] [0,2]
0.95526 0.97036 0.09884
0.63532 0.87558 0.54776
0.72402 0.82532 0.88717
0.30007 0.24039 0.40403
0.64942 0.98282 0.99464
0.49327 0.33884 0.45765
0.03533 0.27944 0.43261
0.52069 0.40464 0.67015
0.45893 0.02859 0.66381
0.80286 0.49054 0.08719
0.69975 0.05308 0.29838
0.34732 0.54225 0.24882
0.42259 0.63095 0.40953
0.11053 0.41241 0.77606
0.29834 0.56380 0.30713
0.04171 0.76315 0.78369
0.14949 0.54690 0.79315
0.82239 0.86686 0.12750
0.51108 0.58095 0.04933
0.20729 0.58790 0.02552In-Range:
0.91052 0.94072 0.19768
0.27064 0.75116 1.09552
0.44804 0.65064 1.77434
-0.39986 -0.51922 0.80806
0.29884 0.96564 1.98928
-0.01346 -0.32232 0.91530
-0.92934 -0.44112 0.86522
0.04138 -0.19072 1.34030
-0.08214 -0.94282 1.32762
0.60572 -0.01892 0.17438
0.39950 -0.89384 0.59676
-0.30536 0.08450 0.49764
-0.15482 0.26190 0.81906
-0.77894 -0.17518 1.55212
-0.40332 0.12760 0.61426
-0.91658 0.52630 1.56738
-0.70102 0.09380 1.58630
0.64478 0.73372 0.25500
0.02216 0.16190 0.09866
-0.58542 0.17580 0.05104x^2 + y^2 + (z-1)^2 \le 1 \\ \sqrt{x^2 + y^2} - z^2 \le 0
f(x,y,z) := x^2 + y^2 + (z-1)^2
g(x,y,z) := sqrt(x^2 + y^2) - z^2
h(x,y,z) := if(f(x,y,z) <= 1, [ if(g(x,y,z) <= 0, [ true ], [false]) ], [false])
h( 0.91052 , 0.94072 , 0.19768 ) = 0
h( 0.27064 , 0.75116 , 1.09552 ) = 1
h( 0.44804 , 0.65064 , 1.77434 ) = 0
h(-0.39986 ,-0.51922 , 0.80806 ) = 0
h( 0.29884 , 0.96564 , 1.98928 ) = 0
h(-0.01346 ,-0.32232 , 0.91530 ) = 1
h(-0.92934 ,-0.44112 , 0.86522 ) = 0
h( 0.04138 ,-0.19072 , 1.34030 ) = 1
h(-0.08214 ,-0.94282 , 1.32762 ) = 0
h( 0.60572 ,-0.01892 , 0.17438 ) = 0
h( 0.39950 ,-0.89384 , 0.59676 ) = 0
h(-0.30536 , 0.08450 , 0.49764 ) = 0
h(-0.15482 , 0.26190 , 0.81906 ) = 1
h(-0.77894 ,-0.17518 , 1.55212 ) = 1
h(-0.40332 , 0.12760 , 0.61426 ) = 0
h(-0.91658 , 0.52630 , 1.56738 ) = 0
h(-0.70102 , 0.09380 , 1.58630 ) = 1
h( 0.64478 , 0.73372 , 0.25500 ) = 0
h( 0.02216 , 0.16190 , 0.09866 ) = 0
h(-0.58542 , 0.17580 , 0.05104 ) = 0TODO I screwed up somewhere, k is supposed to be 9 dawg. ughughughughuhgu
Thus,
\begin{aligned} V_s &\sim \frac{k}{n} V_c \\ &\approx \frac{6}{20} (8) \\ &\approx \frac{6}{20} (8) \\ &\approx 2.4 \end{aligned}
Ugh what the hell did i do wrong.