Double Integral over region defined by inequality












0












$begingroup$


$$iint_{G}!x^2,mathrm{d}xmathrm{d}y$$



where $G := left{(x,y)inmathbb{R}^{2},;,|x|+|y| le 1right}$



How does one go about finding the boundaries of these types of integrals? I did fail at searching for examples like this as I don't even know their name(if they do have a specific one).



Oh and also Happy New Year in advance!






integration multivariable-calculus multiple-integral







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  • 1




    $begingroup$
    Making a graph of this is helpful
    $endgroup$
    – Milan Stojanovic
    Dec 30 '18 at 21:11










  • $begingroup$
    If you made a graph you would see that this is symetrical , so you can just evalute integral when $x+y < 1$ and multiply by four
    $endgroup$
    – Milan Stojanovic
    Dec 30 '18 at 21:16
















0












$begingroup$


$$iint_{G}!x^2,mathrm{d}xmathrm{d}y$$



where $G := left{(x,y)inmathbb{R}^{2},;,|x|+|y| le 1right}$



How does one go about finding the boundaries of these types of integrals? I did fail at searching for examples like this as I don't even know their name(if they do have a specific one).



Oh and also Happy New Year in advance!






integration multivariable-calculus multiple-integral







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Making a graph of this is helpful
    $endgroup$
    – Milan Stojanovic
    Dec 30 '18 at 21:11










  • $begingroup$
    If you made a graph you would see that this is symetrical , so you can just evalute integral when $x+y < 1$ and multiply by four
    $endgroup$
    – Milan Stojanovic
    Dec 30 '18 at 21:16














0












0








0


1



$begingroup$


$$iint_{G}!x^2,mathrm{d}xmathrm{d}y$$



where $G := left{(x,y)inmathbb{R}^{2},;,|x|+|y| le 1right}$



How does one go about finding the boundaries of these types of integrals? I did fail at searching for examples like this as I don't even know their name(if they do have a specific one).



Oh and also Happy New Year in advance!






integration multivariable-calculus multiple-integral







share|cite|improve this question











$endgroup$




$$iint_{G}!x^2,mathrm{d}xmathrm{d}y$$



where $G := left{(x,y)inmathbb{R}^{2},;,|x|+|y| le 1right}$



How does one go about finding the boundaries of these types of integrals? I did fail at searching for examples like this as I don't even know their name(if they do have a specific one).



Oh and also Happy New Year in advance!






integration multivariable-calculus multiple-integral




integration multivariable-calculus multiple-integral






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 30 '18 at 21:21









gt6989b

34.5k22456




34.5k22456










asked Dec 30 '18 at 21:01









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

134




134








  • 1




    $begingroup$
    Making a graph of this is helpful
    $endgroup$
    – Milan Stojanovic
    Dec 30 '18 at 21:11










  • $begingroup$
    If you made a graph you would see that this is symetrical , so you can just evalute integral when $x+y < 1$ and multiply by four
    $endgroup$
    – Milan Stojanovic
    Dec 30 '18 at 21:16














  • 1




    $begingroup$
    Making a graph of this is helpful
    $endgroup$
    – Milan Stojanovic
    Dec 30 '18 at 21:11










  • $begingroup$
    If you made a graph you would see that this is symetrical , so you can just evalute integral when $x+y < 1$ and multiply by four
    $endgroup$
    – Milan Stojanovic
    Dec 30 '18 at 21:16








1




1




$begingroup$
Making a graph of this is helpful
$endgroup$
– Milan Stojanovic
Dec 30 '18 at 21:11




$begingroup$
Making a graph of this is helpful
$endgroup$
– Milan Stojanovic
Dec 30 '18 at 21:11












$begingroup$
If you made a graph you would see that this is symetrical , so you can just evalute integral when $x+y < 1$ and multiply by four
$endgroup$
– Milan Stojanovic
Dec 30 '18 at 21:16




$begingroup$
If you made a graph you would see that this is symetrical , so you can just evalute integral when $x+y < 1$ and multiply by four
$endgroup$
– Milan Stojanovic
Dec 30 '18 at 21:16










3 Answers
3






active

oldest

votes


















0












$begingroup$

Sketch the region $G$.



enter image description here



It is now clear that $$G=left{(x,y)inmathbb R^2: |y|le 1-|x|,,,|x|le 1right}$$



So,



begin{align}
iint_G x^2 ,dx,dy&=int_{-1}^{1}x^2int_{-1+|x|}^{1-|x|},dy ,dx
\&=int_{-1}^1 2x^2 (1-|x|),dx
\&=2int_0^1 2x^2(1-x),dx
end{align}






share|cite|improve this answer











$endgroup$





















    0












    $begingroup$

    In double integrals of the form $int (int f(x,y),dx),dy$: the limits of the outer integral are the largest possible values of $y$ over the entire domain of integration; then, for every fixed $y$, the limits of the inner integral are the largest and smallest values of $x$ that can occur in conjunction with that particular value of $y$ (in particular, these are often functions of $y$, whereas the outer limits are always constant).



    How big and small can $y$ get in the region $G$?



    Given a fixed value of $y$, how big and small can $x$ get (as a function of $y$) so that the point $(x,y)$ is still in $G$?






    share|cite|improve this answer









    $endgroup$





















      0












      $begingroup$

      You have to solve the inequalities for the four cases:



      $xgeq0,land,ygeq0,land x+yleq1$, the triangle with vertices at $(0,0),(0,1),(1,0)$, limited by the function $y=1-x$



      $xgeq0,land,y<0,land x-yleq1$, the triangle with vertices at $(0,0),(0,-1),(1,0)$ with $y=x-1$



      $x<0,land,ygeq0,land -x+yleq1$, the triangle with vertices at $(0,0),(0,1),(-1,0)$, with $y=1+x$



      $x<0,land,y<0,land -x-yleq1$, the triangle with vertices at $(0,0),(0,-1),(-1,0)$, with $y=-x-1$



      So, $x$ runs between $0$ and $1$ or $-1$. The integral is better solved splitting it in two or four for a clearer calculation.






      share|cite|improve this answer









      $endgroup$













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        3 Answers
        3






        active

        oldest

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        3 Answers
        3






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        active

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        0












        $begingroup$

        Sketch the region $G$.



        enter image description here



        It is now clear that $$G=left{(x,y)inmathbb R^2: |y|le 1-|x|,,,|x|le 1right}$$



        So,



        begin{align}
        iint_G x^2 ,dx,dy&=int_{-1}^{1}x^2int_{-1+|x|}^{1-|x|},dy ,dx
        \&=int_{-1}^1 2x^2 (1-|x|),dx
        \&=2int_0^1 2x^2(1-x),dx
        end{align}






        share|cite|improve this answer











        $endgroup$


















          0












          $begingroup$

          Sketch the region $G$.



          enter image description here



          It is now clear that $$G=left{(x,y)inmathbb R^2: |y|le 1-|x|,,,|x|le 1right}$$



          So,



          begin{align}
          iint_G x^2 ,dx,dy&=int_{-1}^{1}x^2int_{-1+|x|}^{1-|x|},dy ,dx
          \&=int_{-1}^1 2x^2 (1-|x|),dx
          \&=2int_0^1 2x^2(1-x),dx
          end{align}






          share|cite|improve this answer











          $endgroup$
















            0












            0








            0





            $begingroup$

            Sketch the region $G$.



            enter image description here



            It is now clear that $$G=left{(x,y)inmathbb R^2: |y|le 1-|x|,,,|x|le 1right}$$



            So,



            begin{align}
            iint_G x^2 ,dx,dy&=int_{-1}^{1}x^2int_{-1+|x|}^{1-|x|},dy ,dx
            \&=int_{-1}^1 2x^2 (1-|x|),dx
            \&=2int_0^1 2x^2(1-x),dx
            end{align}






            share|cite|improve this answer











            $endgroup$



            Sketch the region $G$.



            enter image description here



            It is now clear that $$G=left{(x,y)inmathbb R^2: |y|le 1-|x|,,,|x|le 1right}$$



            So,



            begin{align}
            iint_G x^2 ,dx,dy&=int_{-1}^{1}x^2int_{-1+|x|}^{1-|x|},dy ,dx
            \&=int_{-1}^1 2x^2 (1-|x|),dx
            \&=2int_0^1 2x^2(1-x),dx
            end{align}







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Dec 31 '18 at 7:12

























            answered Dec 30 '18 at 22:10









            StubbornAtomStubbornAtom

            6,05811239




            6,05811239























                0












                $begingroup$

                In double integrals of the form $int (int f(x,y),dx),dy$: the limits of the outer integral are the largest possible values of $y$ over the entire domain of integration; then, for every fixed $y$, the limits of the inner integral are the largest and smallest values of $x$ that can occur in conjunction with that particular value of $y$ (in particular, these are often functions of $y$, whereas the outer limits are always constant).



                How big and small can $y$ get in the region $G$?



                Given a fixed value of $y$, how big and small can $x$ get (as a function of $y$) so that the point $(x,y)$ is still in $G$?






                share|cite|improve this answer









                $endgroup$


















                  0












                  $begingroup$

                  In double integrals of the form $int (int f(x,y),dx),dy$: the limits of the outer integral are the largest possible values of $y$ over the entire domain of integration; then, for every fixed $y$, the limits of the inner integral are the largest and smallest values of $x$ that can occur in conjunction with that particular value of $y$ (in particular, these are often functions of $y$, whereas the outer limits are always constant).



                  How big and small can $y$ get in the region $G$?



                  Given a fixed value of $y$, how big and small can $x$ get (as a function of $y$) so that the point $(x,y)$ is still in $G$?






                  share|cite|improve this answer









                  $endgroup$
















                    0












                    0








                    0





                    $begingroup$

                    In double integrals of the form $int (int f(x,y),dx),dy$: the limits of the outer integral are the largest possible values of $y$ over the entire domain of integration; then, for every fixed $y$, the limits of the inner integral are the largest and smallest values of $x$ that can occur in conjunction with that particular value of $y$ (in particular, these are often functions of $y$, whereas the outer limits are always constant).



                    How big and small can $y$ get in the region $G$?



                    Given a fixed value of $y$, how big and small can $x$ get (as a function of $y$) so that the point $(x,y)$ is still in $G$?






                    share|cite|improve this answer









                    $endgroup$



                    In double integrals of the form $int (int f(x,y),dx),dy$: the limits of the outer integral are the largest possible values of $y$ over the entire domain of integration; then, for every fixed $y$, the limits of the inner integral are the largest and smallest values of $x$ that can occur in conjunction with that particular value of $y$ (in particular, these are often functions of $y$, whereas the outer limits are always constant).



                    How big and small can $y$ get in the region $G$?



                    Given a fixed value of $y$, how big and small can $x$ get (as a function of $y$) so that the point $(x,y)$ is still in $G$?







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Dec 30 '18 at 21:23









                    Greg MartinGreg Martin

                    35.6k23464




                    35.6k23464























                        0












                        $begingroup$

                        You have to solve the inequalities for the four cases:



                        $xgeq0,land,ygeq0,land x+yleq1$, the triangle with vertices at $(0,0),(0,1),(1,0)$, limited by the function $y=1-x$



                        $xgeq0,land,y<0,land x-yleq1$, the triangle with vertices at $(0,0),(0,-1),(1,0)$ with $y=x-1$



                        $x<0,land,ygeq0,land -x+yleq1$, the triangle with vertices at $(0,0),(0,1),(-1,0)$, with $y=1+x$



                        $x<0,land,y<0,land -x-yleq1$, the triangle with vertices at $(0,0),(0,-1),(-1,0)$, with $y=-x-1$



                        So, $x$ runs between $0$ and $1$ or $-1$. The integral is better solved splitting it in two or four for a clearer calculation.






                        share|cite|improve this answer









                        $endgroup$


















                          0












                          $begingroup$

                          You have to solve the inequalities for the four cases:



                          $xgeq0,land,ygeq0,land x+yleq1$, the triangle with vertices at $(0,0),(0,1),(1,0)$, limited by the function $y=1-x$



                          $xgeq0,land,y<0,land x-yleq1$, the triangle with vertices at $(0,0),(0,-1),(1,0)$ with $y=x-1$



                          $x<0,land,ygeq0,land -x+yleq1$, the triangle with vertices at $(0,0),(0,1),(-1,0)$, with $y=1+x$



                          $x<0,land,y<0,land -x-yleq1$, the triangle with vertices at $(0,0),(0,-1),(-1,0)$, with $y=-x-1$



                          So, $x$ runs between $0$ and $1$ or $-1$. The integral is better solved splitting it in two or four for a clearer calculation.






                          share|cite|improve this answer









                          $endgroup$
















                            0












                            0








                            0





                            $begingroup$

                            You have to solve the inequalities for the four cases:



                            $xgeq0,land,ygeq0,land x+yleq1$, the triangle with vertices at $(0,0),(0,1),(1,0)$, limited by the function $y=1-x$



                            $xgeq0,land,y<0,land x-yleq1$, the triangle with vertices at $(0,0),(0,-1),(1,0)$ with $y=x-1$



                            $x<0,land,ygeq0,land -x+yleq1$, the triangle with vertices at $(0,0),(0,1),(-1,0)$, with $y=1+x$



                            $x<0,land,y<0,land -x-yleq1$, the triangle with vertices at $(0,0),(0,-1),(-1,0)$, with $y=-x-1$



                            So, $x$ runs between $0$ and $1$ or $-1$. The integral is better solved splitting it in two or four for a clearer calculation.






                            share|cite|improve this answer









                            $endgroup$



                            You have to solve the inequalities for the four cases:



                            $xgeq0,land,ygeq0,land x+yleq1$, the triangle with vertices at $(0,0),(0,1),(1,0)$, limited by the function $y=1-x$



                            $xgeq0,land,y<0,land x-yleq1$, the triangle with vertices at $(0,0),(0,-1),(1,0)$ with $y=x-1$



                            $x<0,land,ygeq0,land -x+yleq1$, the triangle with vertices at $(0,0),(0,1),(-1,0)$, with $y=1+x$



                            $x<0,land,y<0,land -x-yleq1$, the triangle with vertices at $(0,0),(0,-1),(-1,0)$, with $y=-x-1$



                            So, $x$ runs between $0$ and $1$ or $-1$. The integral is better solved splitting it in two or four for a clearer calculation.







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered Dec 30 '18 at 21:36









                            Rafa BudríaRafa Budría

                            5,8151825




                            5,8151825






























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