limits of changing the order of a double integral












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I've solved this integral $int_{1}^3 int_{1-y}^{2-y}x^2y,dxdy$ which is 3 1/3. Now I'm trying it with different integral order but can't get the limits right. My try was, $int_{0}^1 int_{1-x}^{2-x}x^2y,dydx$.






integration multiple-integral







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  • $begingroup$
    And the correct one is $int_{-2}^{1}int_{max{1-x,1}}^{min{2-x,3}}x^2y,dy,dx$ (not sure what you do it for).
    $endgroup$
    – metamorphy
    Dec 22 '18 at 6:39


















0












$begingroup$


I've solved this integral $int_{1}^3 int_{1-y}^{2-y}x^2y,dxdy$ which is 3 1/3. Now I'm trying it with different integral order but can't get the limits right. My try was, $int_{0}^1 int_{1-x}^{2-x}x^2y,dydx$.






integration multiple-integral







share|cite|improve this question









$endgroup$












  • $begingroup$
    And the correct one is $int_{-2}^{1}int_{max{1-x,1}}^{min{2-x,3}}x^2y,dy,dx$ (not sure what you do it for).
    $endgroup$
    – metamorphy
    Dec 22 '18 at 6:39
















0












0








0





$begingroup$


I've solved this integral $int_{1}^3 int_{1-y}^{2-y}x^2y,dxdy$ which is 3 1/3. Now I'm trying it with different integral order but can't get the limits right. My try was, $int_{0}^1 int_{1-x}^{2-x}x^2y,dydx$.






integration multiple-integral







share|cite|improve this question









$endgroup$




I've solved this integral $int_{1}^3 int_{1-y}^{2-y}x^2y,dxdy$ which is 3 1/3. Now I'm trying it with different integral order but can't get the limits right. My try was, $int_{0}^1 int_{1-x}^{2-x}x^2y,dydx$.






integration multiple-integral




integration multiple-integral






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asked Dec 22 '18 at 6:29









LLTLLT

32




32












  • $begingroup$
    And the correct one is $int_{-2}^{1}int_{max{1-x,1}}^{min{2-x,3}}x^2y,dy,dx$ (not sure what you do it for).
    $endgroup$
    – metamorphy
    Dec 22 '18 at 6:39




















  • $begingroup$
    And the correct one is $int_{-2}^{1}int_{max{1-x,1}}^{min{2-x,3}}x^2y,dy,dx$ (not sure what you do it for).
    $endgroup$
    – metamorphy
    Dec 22 '18 at 6:39


















$begingroup$
And the correct one is $int_{-2}^{1}int_{max{1-x,1}}^{min{2-x,3}}x^2y,dy,dx$ (not sure what you do it for).
$endgroup$
– metamorphy
Dec 22 '18 at 6:39






$begingroup$
And the correct one is $int_{-2}^{1}int_{max{1-x,1}}^{min{2-x,3}}x^2y,dy,dx$ (not sure what you do it for).
$endgroup$
– metamorphy
Dec 22 '18 at 6:39












3 Answers
3






active

oldest

votes


















2












$begingroup$

A good drawing always helps for these sorts of questions.



enter image description here



As you can see, the region can be broken up into a triangle between $x=-2$ and $-1$, a pair of lines between $x=-1$ and $0$, and another triangle between $x=0$ and $1$.



So hopefully you can see that



$$int_1^3 int_{1-y}^{2-y} x^2y text{d}x text{d}y = int_{-2}^{-1}int_{1-x}^3 x^2y text{d}y text{d}x + int_{-1}^{0} int_{1-x}^{2-x}x^2y text{d}y text{d}x + int_{0}^{1} int_{1}^{2-x}x^2y text{d}y text{d}x$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Cool thx! I drew a graph but forgot the area in the negative area. Now everything’s clear. Cheers.
    $endgroup$
    – LLT
    Dec 22 '18 at 6:57










  • $begingroup$
    Sorry, should it be $$int_1^3 int_{1-y}^{2-y} x^2y text{d}x text{d}y = int_{-2}^{-1}int_{1-x}^3 x^2y text{d}y text{d}x + int_{-1}^{0} int_{1-x}^{2-x}x^2y text{d}y text{d}x + int_{0}^{1} int_{1}^{2-x}x^2y text{d}y text{d}x$$. i.e. the upper limit of dy of the first triangle should be 3?
    $endgroup$
    – LLT
    Dec 23 '18 at 6:39












  • $begingroup$
    Indeed it should, that was silly of me
    $endgroup$
    – Cade Reinberger
    Dec 24 '18 at 8:53



















1












$begingroup$

Hint: Try drawing the figure, bounded by
begin{align}
x=2-y\
x=1-y\
y=1\
y=3
end{align}



And then set the limits with y as a function of x first.






share|cite|improve this answer









$endgroup$





















    1












    $begingroup$

    First of all you need to draw a graph of the region of integration which is a parallelogram.



    You notice that you need three integrals to cover the region if you changed the order of integration.



    The x runs from -2 to 1 and you upper and lower limits for y changes on each sub-intervals of length 1.






    share|cite|improve this answer









    $endgroup$













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      3 Answers
      3






      active

      oldest

      votes








      3 Answers
      3






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      2












      $begingroup$

      A good drawing always helps for these sorts of questions.



      enter image description here



      As you can see, the region can be broken up into a triangle between $x=-2$ and $-1$, a pair of lines between $x=-1$ and $0$, and another triangle between $x=0$ and $1$.



      So hopefully you can see that



      $$int_1^3 int_{1-y}^{2-y} x^2y text{d}x text{d}y = int_{-2}^{-1}int_{1-x}^3 x^2y text{d}y text{d}x + int_{-1}^{0} int_{1-x}^{2-x}x^2y text{d}y text{d}x + int_{0}^{1} int_{1}^{2-x}x^2y text{d}y text{d}x$$






      share|cite|improve this answer











      $endgroup$













      • $begingroup$
        Cool thx! I drew a graph but forgot the area in the negative area. Now everything’s clear. Cheers.
        $endgroup$
        – LLT
        Dec 22 '18 at 6:57










      • $begingroup$
        Sorry, should it be $$int_1^3 int_{1-y}^{2-y} x^2y text{d}x text{d}y = int_{-2}^{-1}int_{1-x}^3 x^2y text{d}y text{d}x + int_{-1}^{0} int_{1-x}^{2-x}x^2y text{d}y text{d}x + int_{0}^{1} int_{1}^{2-x}x^2y text{d}y text{d}x$$. i.e. the upper limit of dy of the first triangle should be 3?
        $endgroup$
        – LLT
        Dec 23 '18 at 6:39












      • $begingroup$
        Indeed it should, that was silly of me
        $endgroup$
        – Cade Reinberger
        Dec 24 '18 at 8:53
















      2












      $begingroup$

      A good drawing always helps for these sorts of questions.



      enter image description here



      As you can see, the region can be broken up into a triangle between $x=-2$ and $-1$, a pair of lines between $x=-1$ and $0$, and another triangle between $x=0$ and $1$.



      So hopefully you can see that



      $$int_1^3 int_{1-y}^{2-y} x^2y text{d}x text{d}y = int_{-2}^{-1}int_{1-x}^3 x^2y text{d}y text{d}x + int_{-1}^{0} int_{1-x}^{2-x}x^2y text{d}y text{d}x + int_{0}^{1} int_{1}^{2-x}x^2y text{d}y text{d}x$$






      share|cite|improve this answer











      $endgroup$













      • $begingroup$
        Cool thx! I drew a graph but forgot the area in the negative area. Now everything’s clear. Cheers.
        $endgroup$
        – LLT
        Dec 22 '18 at 6:57










      • $begingroup$
        Sorry, should it be $$int_1^3 int_{1-y}^{2-y} x^2y text{d}x text{d}y = int_{-2}^{-1}int_{1-x}^3 x^2y text{d}y text{d}x + int_{-1}^{0} int_{1-x}^{2-x}x^2y text{d}y text{d}x + int_{0}^{1} int_{1}^{2-x}x^2y text{d}y text{d}x$$. i.e. the upper limit of dy of the first triangle should be 3?
        $endgroup$
        – LLT
        Dec 23 '18 at 6:39












      • $begingroup$
        Indeed it should, that was silly of me
        $endgroup$
        – Cade Reinberger
        Dec 24 '18 at 8:53














      2












      2








      2





      $begingroup$

      A good drawing always helps for these sorts of questions.



      enter image description here



      As you can see, the region can be broken up into a triangle between $x=-2$ and $-1$, a pair of lines between $x=-1$ and $0$, and another triangle between $x=0$ and $1$.



      So hopefully you can see that



      $$int_1^3 int_{1-y}^{2-y} x^2y text{d}x text{d}y = int_{-2}^{-1}int_{1-x}^3 x^2y text{d}y text{d}x + int_{-1}^{0} int_{1-x}^{2-x}x^2y text{d}y text{d}x + int_{0}^{1} int_{1}^{2-x}x^2y text{d}y text{d}x$$






      share|cite|improve this answer











      $endgroup$



      A good drawing always helps for these sorts of questions.



      enter image description here



      As you can see, the region can be broken up into a triangle between $x=-2$ and $-1$, a pair of lines between $x=-1$ and $0$, and another triangle between $x=0$ and $1$.



      So hopefully you can see that



      $$int_1^3 int_{1-y}^{2-y} x^2y text{d}x text{d}y = int_{-2}^{-1}int_{1-x}^3 x^2y text{d}y text{d}x + int_{-1}^{0} int_{1-x}^{2-x}x^2y text{d}y text{d}x + int_{0}^{1} int_{1}^{2-x}x^2y text{d}y text{d}x$$







      share|cite|improve this answer














      share|cite|improve this answer



      share|cite|improve this answer








      edited Dec 24 '18 at 8:53

























      answered Dec 22 '18 at 6:41









      Cade ReinbergerCade Reinberger

      38317




      38317












      • $begingroup$
        Cool thx! I drew a graph but forgot the area in the negative area. Now everything’s clear. Cheers.
        $endgroup$
        – LLT
        Dec 22 '18 at 6:57










      • $begingroup$
        Sorry, should it be $$int_1^3 int_{1-y}^{2-y} x^2y text{d}x text{d}y = int_{-2}^{-1}int_{1-x}^3 x^2y text{d}y text{d}x + int_{-1}^{0} int_{1-x}^{2-x}x^2y text{d}y text{d}x + int_{0}^{1} int_{1}^{2-x}x^2y text{d}y text{d}x$$. i.e. the upper limit of dy of the first triangle should be 3?
        $endgroup$
        – LLT
        Dec 23 '18 at 6:39












      • $begingroup$
        Indeed it should, that was silly of me
        $endgroup$
        – Cade Reinberger
        Dec 24 '18 at 8:53


















      • $begingroup$
        Cool thx! I drew a graph but forgot the area in the negative area. Now everything’s clear. Cheers.
        $endgroup$
        – LLT
        Dec 22 '18 at 6:57










      • $begingroup$
        Sorry, should it be $$int_1^3 int_{1-y}^{2-y} x^2y text{d}x text{d}y = int_{-2}^{-1}int_{1-x}^3 x^2y text{d}y text{d}x + int_{-1}^{0} int_{1-x}^{2-x}x^2y text{d}y text{d}x + int_{0}^{1} int_{1}^{2-x}x^2y text{d}y text{d}x$$. i.e. the upper limit of dy of the first triangle should be 3?
        $endgroup$
        – LLT
        Dec 23 '18 at 6:39












      • $begingroup$
        Indeed it should, that was silly of me
        $endgroup$
        – Cade Reinberger
        Dec 24 '18 at 8:53
















      $begingroup$
      Cool thx! I drew a graph but forgot the area in the negative area. Now everything’s clear. Cheers.
      $endgroup$
      – LLT
      Dec 22 '18 at 6:57




      $begingroup$
      Cool thx! I drew a graph but forgot the area in the negative area. Now everything’s clear. Cheers.
      $endgroup$
      – LLT
      Dec 22 '18 at 6:57












      $begingroup$
      Sorry, should it be $$int_1^3 int_{1-y}^{2-y} x^2y text{d}x text{d}y = int_{-2}^{-1}int_{1-x}^3 x^2y text{d}y text{d}x + int_{-1}^{0} int_{1-x}^{2-x}x^2y text{d}y text{d}x + int_{0}^{1} int_{1}^{2-x}x^2y text{d}y text{d}x$$. i.e. the upper limit of dy of the first triangle should be 3?
      $endgroup$
      – LLT
      Dec 23 '18 at 6:39






      $begingroup$
      Sorry, should it be $$int_1^3 int_{1-y}^{2-y} x^2y text{d}x text{d}y = int_{-2}^{-1}int_{1-x}^3 x^2y text{d}y text{d}x + int_{-1}^{0} int_{1-x}^{2-x}x^2y text{d}y text{d}x + int_{0}^{1} int_{1}^{2-x}x^2y text{d}y text{d}x$$. i.e. the upper limit of dy of the first triangle should be 3?
      $endgroup$
      – LLT
      Dec 23 '18 at 6:39














      $begingroup$
      Indeed it should, that was silly of me
      $endgroup$
      – Cade Reinberger
      Dec 24 '18 at 8:53




      $begingroup$
      Indeed it should, that was silly of me
      $endgroup$
      – Cade Reinberger
      Dec 24 '18 at 8:53











      1












      $begingroup$

      Hint: Try drawing the figure, bounded by
      begin{align}
      x=2-y\
      x=1-y\
      y=1\
      y=3
      end{align}



      And then set the limits with y as a function of x first.






      share|cite|improve this answer









      $endgroup$


















        1












        $begingroup$

        Hint: Try drawing the figure, bounded by
        begin{align}
        x=2-y\
        x=1-y\
        y=1\
        y=3
        end{align}



        And then set the limits with y as a function of x first.






        share|cite|improve this answer









        $endgroup$
















          1












          1








          1





          $begingroup$

          Hint: Try drawing the figure, bounded by
          begin{align}
          x=2-y\
          x=1-y\
          y=1\
          y=3
          end{align}



          And then set the limits with y as a function of x first.






          share|cite|improve this answer









          $endgroup$



          Hint: Try drawing the figure, bounded by
          begin{align}
          x=2-y\
          x=1-y\
          y=1\
          y=3
          end{align}



          And then set the limits with y as a function of x first.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 22 '18 at 6:39









          mm-crjmm-crj

          425213




          425213























              1












              $begingroup$

              First of all you need to draw a graph of the region of integration which is a parallelogram.



              You notice that you need three integrals to cover the region if you changed the order of integration.



              The x runs from -2 to 1 and you upper and lower limits for y changes on each sub-intervals of length 1.






              share|cite|improve this answer









              $endgroup$


















                1












                $begingroup$

                First of all you need to draw a graph of the region of integration which is a parallelogram.



                You notice that you need three integrals to cover the region if you changed the order of integration.



                The x runs from -2 to 1 and you upper and lower limits for y changes on each sub-intervals of length 1.






                share|cite|improve this answer









                $endgroup$
















                  1












                  1








                  1





                  $begingroup$

                  First of all you need to draw a graph of the region of integration which is a parallelogram.



                  You notice that you need three integrals to cover the region if you changed the order of integration.



                  The x runs from -2 to 1 and you upper and lower limits for y changes on each sub-intervals of length 1.






                  share|cite|improve this answer









                  $endgroup$



                  First of all you need to draw a graph of the region of integration which is a parallelogram.



                  You notice that you need three integrals to cover the region if you changed the order of integration.



                  The x runs from -2 to 1 and you upper and lower limits for y changes on each sub-intervals of length 1.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 22 '18 at 6:49









                  Mohammad Riazi-KermaniMohammad Riazi-Kermani

                  41.6k42061




                  41.6k42061






























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