Aproximate Double Integral Result Using The Monte Carlo Method Getting 2 different results












0












$begingroup$


My task is to approximate a result using the Monte Carlo method programatically.
The integral in question is the following:
$iint (2x+y) dxdy$



The integral is delimited by the graph axes (xOy) and by the segment x+y=3.
From the last line I've deduced that the interval for x is [0,3] and for y is [0,3] obtaining some kind of triangle that has the vertices on (0,0), (0,3) and (3,0).



What I've tried so far yielded the following results:



$int^3_0int^3_0 (2x+y) dxdy = int^3_0 (x^2+xy)dy$



Then I've replaced x with the limit ends for it (3,0):



$int^3_0 (x^2+xy)dy = int^3_0 (9+3y) - 0dy$



Then I proceeded to calculate the integral by y.



$int^3_0 (9+3y)dy = 9y+(3y^2)/2$



Then I've replaced the y with the limit terms (3,0):



$9*3+(3*9)/2 - 0 = 27 + 27/2 = 27 + 13,5 = 40,5$



So the final result that I got by manually integrating was 40,5.



Then I tried to aproximate this result using C#, but I'll write in pseudocode:



sum=0

for i->number_of_iterations

    x = random between 0 and 3

    y = random between 0 and 3 - x

    sum = sum + 2 * x + y

end for

sum = sum * (xUpperLimit-xLowerLimit) * (yUpperLimit - yLowerLimit) / number_of_iterations


However, the result that it tends to converge to is 33,75.
What am I doing wrong?
And, is the interval that I've deduced for x and y even the right one?






calculus integration multivariable-calculus







share|cite|improve this question











$endgroup$












  • $begingroup$
    You should compute $$int_0^3 int_0^{3-y} (2x+y) dx dy = int_0^3 ((3-y)^2-(3-y)y) dy$$
    $endgroup$
    – Crostul
    Dec 26 '18 at 11:03
















0












$begingroup$


My task is to approximate a result using the Monte Carlo method programatically.
The integral in question is the following:
$iint (2x+y) dxdy$



The integral is delimited by the graph axes (xOy) and by the segment x+y=3.
From the last line I've deduced that the interval for x is [0,3] and for y is [0,3] obtaining some kind of triangle that has the vertices on (0,0), (0,3) and (3,0).



What I've tried so far yielded the following results:



$int^3_0int^3_0 (2x+y) dxdy = int^3_0 (x^2+xy)dy$



Then I've replaced x with the limit ends for it (3,0):



$int^3_0 (x^2+xy)dy = int^3_0 (9+3y) - 0dy$



Then I proceeded to calculate the integral by y.



$int^3_0 (9+3y)dy = 9y+(3y^2)/2$



Then I've replaced the y with the limit terms (3,0):



$9*3+(3*9)/2 - 0 = 27 + 27/2 = 27 + 13,5 = 40,5$



So the final result that I got by manually integrating was 40,5.



Then I tried to aproximate this result using C#, but I'll write in pseudocode:



sum=0

for i->number_of_iterations

    x = random between 0 and 3

    y = random between 0 and 3 - x

    sum = sum + 2 * x + y

end for

sum = sum * (xUpperLimit-xLowerLimit) * (yUpperLimit - yLowerLimit) / number_of_iterations


However, the result that it tends to converge to is 33,75.
What am I doing wrong?
And, is the interval that I've deduced for x and y even the right one?






calculus integration multivariable-calculus







share|cite|improve this question











$endgroup$












  • $begingroup$
    You should compute $$int_0^3 int_0^{3-y} (2x+y) dx dy = int_0^3 ((3-y)^2-(3-y)y) dy$$
    $endgroup$
    – Crostul
    Dec 26 '18 at 11:03














0












0








0





$begingroup$


My task is to approximate a result using the Monte Carlo method programatically.
The integral in question is the following:
$iint (2x+y) dxdy$



The integral is delimited by the graph axes (xOy) and by the segment x+y=3.
From the last line I've deduced that the interval for x is [0,3] and for y is [0,3] obtaining some kind of triangle that has the vertices on (0,0), (0,3) and (3,0).



What I've tried so far yielded the following results:



$int^3_0int^3_0 (2x+y) dxdy = int^3_0 (x^2+xy)dy$



Then I've replaced x with the limit ends for it (3,0):



$int^3_0 (x^2+xy)dy = int^3_0 (9+3y) - 0dy$



Then I proceeded to calculate the integral by y.



$int^3_0 (9+3y)dy = 9y+(3y^2)/2$



Then I've replaced the y with the limit terms (3,0):



$9*3+(3*9)/2 - 0 = 27 + 27/2 = 27 + 13,5 = 40,5$



So the final result that I got by manually integrating was 40,5.



Then I tried to aproximate this result using C#, but I'll write in pseudocode:



sum=0

for i->number_of_iterations

    x = random between 0 and 3

    y = random between 0 and 3 - x

    sum = sum + 2 * x + y

end for

sum = sum * (xUpperLimit-xLowerLimit) * (yUpperLimit - yLowerLimit) / number_of_iterations


However, the result that it tends to converge to is 33,75.
What am I doing wrong?
And, is the interval that I've deduced for x and y even the right one?






calculus integration multivariable-calculus







share|cite|improve this question











$endgroup$




My task is to approximate a result using the Monte Carlo method programatically.
The integral in question is the following:
$iint (2x+y) dxdy$



The integral is delimited by the graph axes (xOy) and by the segment x+y=3.
From the last line I've deduced that the interval for x is [0,3] and for y is [0,3] obtaining some kind of triangle that has the vertices on (0,0), (0,3) and (3,0).



What I've tried so far yielded the following results:



$int^3_0int^3_0 (2x+y) dxdy = int^3_0 (x^2+xy)dy$



Then I've replaced x with the limit ends for it (3,0):



$int^3_0 (x^2+xy)dy = int^3_0 (9+3y) - 0dy$



Then I proceeded to calculate the integral by y.



$int^3_0 (9+3y)dy = 9y+(3y^2)/2$



Then I've replaced the y with the limit terms (3,0):



$9*3+(3*9)/2 - 0 = 27 + 27/2 = 27 + 13,5 = 40,5$



So the final result that I got by manually integrating was 40,5.



Then I tried to aproximate this result using C#, but I'll write in pseudocode:



sum=0

for i->number_of_iterations

    x = random between 0 and 3

    y = random between 0 and 3 - x

    sum = sum + 2 * x + y

end for

sum = sum * (xUpperLimit-xLowerLimit) * (yUpperLimit - yLowerLimit) / number_of_iterations


However, the result that it tends to converge to is 33,75.
What am I doing wrong?
And, is the interval that I've deduced for x and y even the right one?






calculus integration multivariable-calculus




calculus integration multivariable-calculus






share|cite|improve this question















share|cite|improve this question













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edited Dec 26 '18 at 11:25









Key Flex

8,58761233




8,58761233










asked Dec 26 '18 at 10:56









Bărhălescu Vlad FlorianBărhălescu Vlad Florian

32




32












  • $begingroup$
    You should compute $$int_0^3 int_0^{3-y} (2x+y) dx dy = int_0^3 ((3-y)^2-(3-y)y) dy$$
    $endgroup$
    – Crostul
    Dec 26 '18 at 11:03


















  • $begingroup$
    You should compute $$int_0^3 int_0^{3-y} (2x+y) dx dy = int_0^3 ((3-y)^2-(3-y)y) dy$$
    $endgroup$
    – Crostul
    Dec 26 '18 at 11:03
















$begingroup$
You should compute $$int_0^3 int_0^{3-y} (2x+y) dx dy = int_0^3 ((3-y)^2-(3-y)y) dy$$
$endgroup$
– Crostul
Dec 26 '18 at 11:03




$begingroup$
You should compute $$int_0^3 int_0^{3-y} (2x+y) dx dy = int_0^3 ((3-y)^2-(3-y)y) dy$$
$endgroup$
– Crostul
Dec 26 '18 at 11:03










1 Answer
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You are choosing $x$ from a uniform distribution on $[0,3]$, but this is wrong: over the triangle of integration, $x$ is more likely to be small than large. This is the source of your error.



You have to choose a point $(x,y)$ from a uniform distribution over the triangle of integration. The simplest way is to choose $x$ and $y$ separately from uniform distibutions over $[0,3]$, and reject those $(x,y)$ pairs with $x+y>3$. Or to avoid waste, instead of rejecting such pairs you can use $(3-x,3-y)$ instead.






share|cite|improve this answer









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    1 Answer
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    $begingroup$

    You are choosing $x$ from a uniform distribution on $[0,3]$, but this is wrong: over the triangle of integration, $x$ is more likely to be small than large. This is the source of your error.



    You have to choose a point $(x,y)$ from a uniform distribution over the triangle of integration. The simplest way is to choose $x$ and $y$ separately from uniform distibutions over $[0,3]$, and reject those $(x,y)$ pairs with $x+y>3$. Or to avoid waste, instead of rejecting such pairs you can use $(3-x,3-y)$ instead.






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      You are choosing $x$ from a uniform distribution on $[0,3]$, but this is wrong: over the triangle of integration, $x$ is more likely to be small than large. This is the source of your error.



      You have to choose a point $(x,y)$ from a uniform distribution over the triangle of integration. The simplest way is to choose $x$ and $y$ separately from uniform distibutions over $[0,3]$, and reject those $(x,y)$ pairs with $x+y>3$. Or to avoid waste, instead of rejecting such pairs you can use $(3-x,3-y)$ instead.






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        You are choosing $x$ from a uniform distribution on $[0,3]$, but this is wrong: over the triangle of integration, $x$ is more likely to be small than large. This is the source of your error.



        You have to choose a point $(x,y)$ from a uniform distribution over the triangle of integration. The simplest way is to choose $x$ and $y$ separately from uniform distibutions over $[0,3]$, and reject those $(x,y)$ pairs with $x+y>3$. Or to avoid waste, instead of rejecting such pairs you can use $(3-x,3-y)$ instead.






        share|cite|improve this answer









        $endgroup$



        You are choosing $x$ from a uniform distribution on $[0,3]$, but this is wrong: over the triangle of integration, $x$ is more likely to be small than large. This is the source of your error.



        You have to choose a point $(x,y)$ from a uniform distribution over the triangle of integration. The simplest way is to choose $x$ and $y$ separately from uniform distibutions over $[0,3]$, and reject those $(x,y)$ pairs with $x+y>3$. Or to avoid waste, instead of rejecting such pairs you can use $(3-x,3-y)$ instead.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 26 '18 at 11:20









        TonyKTonyK

        43.1k356135




        43.1k356135






























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