Monte Carlo double integral over surface of $|x|+|y| leq 1$












1












$begingroup$


$iint_{|x|+|y|le1}!x^2,dxdy$



I am supposed to calculate this by using Monte Carlo integration. Can anyone give basic hints or directions? I know the idea behind the Monte Carlo integration method but my brain can't seem to be able to grasp any straw at how to solve this.






monte-carlo







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$endgroup$

















    1












    $begingroup$


    $iint_{|x|+|y|le1}!x^2,dxdy$



    I am supposed to calculate this by using Monte Carlo integration. Can anyone give basic hints or directions? I know the idea behind the Monte Carlo integration method but my brain can't seem to be able to grasp any straw at how to solve this.






    monte-carlo







    share|cite|improve this question











    $endgroup$















      1












      1








      1


      1



      $begingroup$


      $iint_{|x|+|y|le1}!x^2,dxdy$



      I am supposed to calculate this by using Monte Carlo integration. Can anyone give basic hints or directions? I know the idea behind the Monte Carlo integration method but my brain can't seem to be able to grasp any straw at how to solve this.






      monte-carlo







      share|cite|improve this question











      $endgroup$




      $iint_{|x|+|y|le1}!x^2,dxdy$



      I am supposed to calculate this by using Monte Carlo integration. Can anyone give basic hints or directions? I know the idea behind the Monte Carlo integration method but my brain can't seem to be able to grasp any straw at how to solve this.






      monte-carlo




      monte-carlo






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      share|cite|improve this question













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      edited Dec 9 '18 at 20:01









      postmortes

      1,91721117




      1,91721117










      asked Dec 9 '18 at 19:40









      That guy who is bad at mathThat guy who is bad at math

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          $begingroup$

          Enclose your integration in $displaystyle{left[-1,1right)^{,2}}$. The Monte-Carlo integration becomes $left(~overline overline{phantom{AAA}}mbox{means average with an uniform distibution over} left[-1,1right)^{,2}~right)$
          begin{align}
          S_{N} & = sum_{i = 1}^{N}x_{i}^{2}
          left[vphantom{Large A}leftvert x_{i}rightvert +
          leftvert y_{i}rightvert leq 1right]
          \[2mm]
          overline{S_{N}} & = N overline{x^{2}
          left[vphantom{Large A}leftvert xrightvert +
          leftvert yrightvert leq 1right]} =
          Nint_{-1}^{1}int_{-1}^{1}{1 over 4}
          left[vphantom{Large A}leftvert xrightvert +
          leftvert yrightvert leq 1right]x^{2}
          ,mathrm{d}x,mathrm{d}y
          \[5mm]
          & implies int_{-1}^{1}int_{-1}^{1}
          left[vphantom{Large A}leftvert xrightvert +
          leftvert yrightvert leq 1right]x^{2}
          ,mathrm{d}x,mathrm{d}y = 4,{overline{S_{N}} over N}
          approx bbox[10px,#ffd,border:1px groove navy]
          {4,{S_{N} over N}}
          end{align}



          The following code is a $texttt{javascript}$ script which can be run in a terminal with $texttt{node.js}$:




          
"use strict";
          
const ITERATIONS = 10000;
          
let i = 0, theSum = 0, x = null, y = null;
          

          
while (i < ITERATIONS) {
          
      x = 2.0*Math.random() - 1.0;
          
      y = 2.0*Math.random() - 1.0;
          
      if (Math.abs(x) + Math.abs(y) <= 1.0) theSum += x*x;
          
      ++i;
          
}
          

          
console.log(4.0*(theSum/ITERATIONS));


          A typical "run" yields $bbox[10px,#ffd,border:1px groove navy]{displaystyle 0.3302123390009306}$. The exact result is
          $bbox[10px,#ffd,border:1px groove navy]
          {displaystyle{1 over 3}}$          .






          share|cite|improve this answer











          $endgroup$













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            1 Answer
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            0












            $begingroup$

            Enclose your integration in $displaystyle{left[-1,1right)^{,2}}$. The Monte-Carlo integration becomes $left(~overline overline{phantom{AAA}}mbox{means average with an uniform distibution over} left[-1,1right)^{,2}~right)$
            begin{align}
            S_{N} & = sum_{i = 1}^{N}x_{i}^{2}
            left[vphantom{Large A}leftvert x_{i}rightvert +
            leftvert y_{i}rightvert leq 1right]
            \[2mm]
            overline{S_{N}} & = N overline{x^{2}
            left[vphantom{Large A}leftvert xrightvert +
            leftvert yrightvert leq 1right]} =
            Nint_{-1}^{1}int_{-1}^{1}{1 over 4}
            left[vphantom{Large A}leftvert xrightvert +
            leftvert yrightvert leq 1right]x^{2}
            ,mathrm{d}x,mathrm{d}y
            \[5mm]
            & implies int_{-1}^{1}int_{-1}^{1}
            left[vphantom{Large A}leftvert xrightvert +
            leftvert yrightvert leq 1right]x^{2}
            ,mathrm{d}x,mathrm{d}y = 4,{overline{S_{N}} over N}
            approx bbox[10px,#ffd,border:1px groove navy]
            {4,{S_{N} over N}}
            end{align}



            The following code is a $texttt{javascript}$ script which can be run in a terminal with $texttt{node.js}$:




            
"use strict";
            
const ITERATIONS = 10000;
            
let i = 0, theSum = 0, x = null, y = null;
            

            
while (i < ITERATIONS) {
            
      x = 2.0*Math.random() - 1.0;
            
      y = 2.0*Math.random() - 1.0;
            
      if (Math.abs(x) + Math.abs(y) <= 1.0) theSum += x*x;
            
      ++i;
            
}
            

            
console.log(4.0*(theSum/ITERATIONS));


            A typical "run" yields $bbox[10px,#ffd,border:1px groove navy]{displaystyle 0.3302123390009306}$. The exact result is
            $bbox[10px,#ffd,border:1px groove navy]
            {displaystyle{1 over 3}}$            .






            share|cite|improve this answer











            $endgroup$


















              0












              $begingroup$

              Enclose your integration in $displaystyle{left[-1,1right)^{,2}}$. The Monte-Carlo integration becomes $left(~overline overline{phantom{AAA}}mbox{means average with an uniform distibution over} left[-1,1right)^{,2}~right)$
              begin{align}
              S_{N} & = sum_{i = 1}^{N}x_{i}^{2}
              left[vphantom{Large A}leftvert x_{i}rightvert +
              leftvert y_{i}rightvert leq 1right]
              \[2mm]
              overline{S_{N}} & = N overline{x^{2}
              left[vphantom{Large A}leftvert xrightvert +
              leftvert yrightvert leq 1right]} =
              Nint_{-1}^{1}int_{-1}^{1}{1 over 4}
              left[vphantom{Large A}leftvert xrightvert +
              leftvert yrightvert leq 1right]x^{2}
              ,mathrm{d}x,mathrm{d}y
              \[5mm]
              & implies int_{-1}^{1}int_{-1}^{1}
              left[vphantom{Large A}leftvert xrightvert +
              leftvert yrightvert leq 1right]x^{2}
              ,mathrm{d}x,mathrm{d}y = 4,{overline{S_{N}} over N}
              approx bbox[10px,#ffd,border:1px groove navy]
              {4,{S_{N} over N}}
              end{align}



              The following code is a $texttt{javascript}$ script which can be run in a terminal with $texttt{node.js}$:




              
"use strict";
              
const ITERATIONS = 10000;
              
let i = 0, theSum = 0, x = null, y = null;
              

              
while (i < ITERATIONS) {
              
      x = 2.0*Math.random() - 1.0;
              
      y = 2.0*Math.random() - 1.0;
              
      if (Math.abs(x) + Math.abs(y) <= 1.0) theSum += x*x;
              
      ++i;
              
}
              

              
console.log(4.0*(theSum/ITERATIONS));


              A typical "run" yields $bbox[10px,#ffd,border:1px groove navy]{displaystyle 0.3302123390009306}$. The exact result is
              $bbox[10px,#ffd,border:1px groove navy]
              {displaystyle{1 over 3}}$              .






              share|cite|improve this answer











              $endgroup$
















                0












                0








                0





                $begingroup$

                Enclose your integration in $displaystyle{left[-1,1right)^{,2}}$. The Monte-Carlo integration becomes $left(~overline overline{phantom{AAA}}mbox{means average with an uniform distibution over} left[-1,1right)^{,2}~right)$
                begin{align}
                S_{N} & = sum_{i = 1}^{N}x_{i}^{2}
                left[vphantom{Large A}leftvert x_{i}rightvert +
                leftvert y_{i}rightvert leq 1right]
                \[2mm]
                overline{S_{N}} & = N overline{x^{2}
                left[vphantom{Large A}leftvert xrightvert +
                leftvert yrightvert leq 1right]} =
                Nint_{-1}^{1}int_{-1}^{1}{1 over 4}
                left[vphantom{Large A}leftvert xrightvert +
                leftvert yrightvert leq 1right]x^{2}
                ,mathrm{d}x,mathrm{d}y
                \[5mm]
                & implies int_{-1}^{1}int_{-1}^{1}
                left[vphantom{Large A}leftvert xrightvert +
                leftvert yrightvert leq 1right]x^{2}
                ,mathrm{d}x,mathrm{d}y = 4,{overline{S_{N}} over N}
                approx bbox[10px,#ffd,border:1px groove navy]
                {4,{S_{N} over N}}
                end{align}



                The following code is a $texttt{javascript}$ script which can be run in a terminal with $texttt{node.js}$:




                
"use strict";
                
const ITERATIONS = 10000;
                
let i = 0, theSum = 0, x = null, y = null;
                

                
while (i < ITERATIONS) {
                
      x = 2.0*Math.random() - 1.0;
                
      y = 2.0*Math.random() - 1.0;
                
      if (Math.abs(x) + Math.abs(y) <= 1.0) theSum += x*x;
                
      ++i;
                
}
                

                
console.log(4.0*(theSum/ITERATIONS));


                A typical "run" yields $bbox[10px,#ffd,border:1px groove navy]{displaystyle 0.3302123390009306}$. The exact result is
                $bbox[10px,#ffd,border:1px groove navy]
                {displaystyle{1 over 3}}$                .






                share|cite|improve this answer











                $endgroup$



                Enclose your integration in $displaystyle{left[-1,1right)^{,2}}$. The Monte-Carlo integration becomes $left(~overline overline{phantom{AAA}}mbox{means average with an uniform distibution over} left[-1,1right)^{,2}~right)$
                begin{align}
                S_{N} & = sum_{i = 1}^{N}x_{i}^{2}
                left[vphantom{Large A}leftvert x_{i}rightvert +
                leftvert y_{i}rightvert leq 1right]
                \[2mm]
                overline{S_{N}} & = N overline{x^{2}
                left[vphantom{Large A}leftvert xrightvert +
                leftvert yrightvert leq 1right]} =
                Nint_{-1}^{1}int_{-1}^{1}{1 over 4}
                left[vphantom{Large A}leftvert xrightvert +
                leftvert yrightvert leq 1right]x^{2}
                ,mathrm{d}x,mathrm{d}y
                \[5mm]
                & implies int_{-1}^{1}int_{-1}^{1}
                left[vphantom{Large A}leftvert xrightvert +
                leftvert yrightvert leq 1right]x^{2}
                ,mathrm{d}x,mathrm{d}y = 4,{overline{S_{N}} over N}
                approx bbox[10px,#ffd,border:1px groove navy]
                {4,{S_{N} over N}}
                end{align}



                The following code is a $texttt{javascript}$ script which can be run in a terminal with $texttt{node.js}$:




                
"use strict";
                
const ITERATIONS = 10000;
                
let i = 0, theSum = 0, x = null, y = null;
                

                
while (i < ITERATIONS) {
                
      x = 2.0*Math.random() - 1.0;
                
      y = 2.0*Math.random() - 1.0;
                
      if (Math.abs(x) + Math.abs(y) <= 1.0) theSum += x*x;
                
      ++i;
                
}
                

                
console.log(4.0*(theSum/ITERATIONS));


                A typical "run" yields $bbox[10px,#ffd,border:1px groove navy]{displaystyle 0.3302123390009306}$. The exact result is
                $bbox[10px,#ffd,border:1px groove navy]
                {displaystyle{1 over 3}}$                .







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited Dec 10 '18 at 18:36

























                answered Dec 9 '18 at 21:56









                Felix MarinFelix Marin

                67.8k7107142




                67.8k7107142






























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