How should I solve this integral with changing parameters?












4












$begingroup$


I can't solve this. How should I proceed?




$$iint_De^{largefrac{y-x}{y+x}}mathrm dxmathrm dy$$




$D$ is the triangle with these coordinates $(0,0), (0,2), (2,0)$ and I've changed the parameters this way $u=y-x$ and $v= y+x$ and the Jacobian is $-1/2$ but I have problem finding the range of $u$ and $v$ to calculate the integral






definite-integrals







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    4












    $begingroup$


    I can't solve this. How should I proceed?




    $$iint_De^{largefrac{y-x}{y+x}}mathrm dxmathrm dy$$




    $D$ is the triangle with these coordinates $(0,0), (0,2), (2,0)$ and I've changed the parameters this way $u=y-x$ and $v= y+x$ and the Jacobian is $-1/2$ but I have problem finding the range of $u$ and $v$ to calculate the integral






    definite-integrals







    share|cite|improve this question









    New contributor




    khoshrang is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
    Check out our Code of Conduct.







    $endgroup$















      4












      4








      4





      $begingroup$


      I can't solve this. How should I proceed?




      $$iint_De^{largefrac{y-x}{y+x}}mathrm dxmathrm dy$$




      $D$ is the triangle with these coordinates $(0,0), (0,2), (2,0)$ and I've changed the parameters this way $u=y-x$ and $v= y+x$ and the Jacobian is $-1/2$ but I have problem finding the range of $u$ and $v$ to calculate the integral






      definite-integrals







      share|cite|improve this question









      New contributor




      khoshrang is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.







      $endgroup$




      I can't solve this. How should I proceed?




      $$iint_De^{largefrac{y-x}{y+x}}mathrm dxmathrm dy$$




      $D$ is the triangle with these coordinates $(0,0), (0,2), (2,0)$ and I've changed the parameters this way $u=y-x$ and $v= y+x$ and the Jacobian is $-1/2$ but I have problem finding the range of $u$ and $v$ to calculate the integral






      definite-integrals




      definite-integrals






      share|cite|improve this question









      New contributor




      khoshrang is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.











      share|cite|improve this question









      New contributor




      khoshrang is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.









      share|cite|improve this question




      share|cite|improve this question








      edited 1 hour ago









      mrtaurho

      5,74551540




      5,74551540






      New contributor




      khoshrang is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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      asked 1 hour ago









      khoshrangkhoshrang

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      New contributor





      khoshrang is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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      Check out our Code of Conduct.






















          3 Answers
          3






          active

          oldest

          votes


















          6












          $begingroup$

          Start by making a drawing of your domain. You can see that the $v$ is along the diagonal in the first quadrant, and $u$ is along the diagonal in the second quadrant. You can also see that the line between $(0,2)$ and $(2,0)$ is parallel to $u$, and intersects $v$ axis at $v=1$. So $v$ varies between $0$ and $1$ and $u$ varies between $-v$ and $v$.






          share|cite|improve this answer









          $endgroup$





















            5












            $begingroup$

            Your domain is $$D={(x,y):0<x<2,,,0<y<2,,,x+y<2}$$



            Using change of variables $(x,y)to(u,v)$ with $$u=frac{x-y}{x+y},,,v=x+y$$



            The region is now $$R={(u,v):-1<u<1,,0<v<2}$$



            Therefore,



            begin{align}
            iint_D expleft({-frac{x-y}{x+y}}right),mathrm{d}x,mathrm{d}y&=frac{1}{2}iint_R ve^{-u},mathrm{d}u,mathrm{d}v
            \\&=frac{1}{2}int_0^2v,mathrm{d}vint_{-1}^1 e^{-u},mathrm{d}u
            end{align}






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              Finally the $v$ appears, but with no explanation at all?
              $endgroup$
              – NickD
              33 mins ago










            • $begingroup$
              @NickD I have left the calculation of the jacobian to the OP.
              $endgroup$
              – StubbornAtom
              30 mins ago










            • $begingroup$
              That's fine, but you should at least mention it, if only for the sake of future readers.
              $endgroup$
              – NickD
              25 mins ago










            • $begingroup$
              It is pretty clear as it is, I think.
              $endgroup$
              – StubbornAtom
              24 mins ago



















            -1












            $begingroup$

            Computer algebra gives (for the general case):



            $$e left(x^2 text{Ei}left(-frac{2 x}{x+y}right)-frac{y^2 text{Ei}left(frac{2
            y}{x+y}right)}{e^2}right)+frac{1}{2} e^{1-frac{2 x}{x+y}} (x+y)^2$$



            over your specified region:



            $$e-frac{1}{e}$$






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              yes the final answer is right due to my answer sheet but i should solve this by hand not computer any idea of how should i solve this on paper?
              $endgroup$
              – khoshrang
              1 hour ago












            • $begingroup$
              (-1) because this is not an answer to the question. See the other excellent responses!
              $endgroup$
              – Jimmy Sabater
              9 mins ago











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            3 Answers
            3






            active

            oldest

            votes








            3 Answers
            3






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            6












            $begingroup$

            Start by making a drawing of your domain. You can see that the $v$ is along the diagonal in the first quadrant, and $u$ is along the diagonal in the second quadrant. You can also see that the line between $(0,2)$ and $(2,0)$ is parallel to $u$, and intersects $v$ axis at $v=1$. So $v$ varies between $0$ and $1$ and $u$ varies between $-v$ and $v$.






            share|cite|improve this answer









            $endgroup$


















              6












              $begingroup$

              Start by making a drawing of your domain. You can see that the $v$ is along the diagonal in the first quadrant, and $u$ is along the diagonal in the second quadrant. You can also see that the line between $(0,2)$ and $(2,0)$ is parallel to $u$, and intersects $v$ axis at $v=1$. So $v$ varies between $0$ and $1$ and $u$ varies between $-v$ and $v$.






              share|cite|improve this answer









              $endgroup$
















                6












                6








                6





                $begingroup$

                Start by making a drawing of your domain. You can see that the $v$ is along the diagonal in the first quadrant, and $u$ is along the diagonal in the second quadrant. You can also see that the line between $(0,2)$ and $(2,0)$ is parallel to $u$, and intersects $v$ axis at $v=1$. So $v$ varies between $0$ and $1$ and $u$ varies between $-v$ and $v$.






                share|cite|improve this answer









                $endgroup$



                Start by making a drawing of your domain. You can see that the $v$ is along the diagonal in the first quadrant, and $u$ is along the diagonal in the second quadrant. You can also see that the line between $(0,2)$ and $(2,0)$ is parallel to $u$, and intersects $v$ axis at $v=1$. So $v$ varies between $0$ and $1$ and $u$ varies between $-v$ and $v$.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered 55 mins ago









                AndreiAndrei

                13.1k21230




                13.1k21230























                    5












                    $begingroup$

                    Your domain is $$D={(x,y):0<x<2,,,0<y<2,,,x+y<2}$$



                    Using change of variables $(x,y)to(u,v)$ with $$u=frac{x-y}{x+y},,,v=x+y$$



                    The region is now $$R={(u,v):-1<u<1,,0<v<2}$$



                    Therefore,



                    begin{align}
                    iint_D expleft({-frac{x-y}{x+y}}right),mathrm{d}x,mathrm{d}y&=frac{1}{2}iint_R ve^{-u},mathrm{d}u,mathrm{d}v
                    \\&=frac{1}{2}int_0^2v,mathrm{d}vint_{-1}^1 e^{-u},mathrm{d}u
                    end{align}






                    share|cite|improve this answer











                    $endgroup$













                    • $begingroup$
                      Finally the $v$ appears, but with no explanation at all?
                      $endgroup$
                      – NickD
                      33 mins ago










                    • $begingroup$
                      @NickD I have left the calculation of the jacobian to the OP.
                      $endgroup$
                      – StubbornAtom
                      30 mins ago










                    • $begingroup$
                      That's fine, but you should at least mention it, if only for the sake of future readers.
                      $endgroup$
                      – NickD
                      25 mins ago










                    • $begingroup$
                      It is pretty clear as it is, I think.
                      $endgroup$
                      – StubbornAtom
                      24 mins ago
















                    5












                    $begingroup$

                    Your domain is $$D={(x,y):0<x<2,,,0<y<2,,,x+y<2}$$



                    Using change of variables $(x,y)to(u,v)$ with $$u=frac{x-y}{x+y},,,v=x+y$$



                    The region is now $$R={(u,v):-1<u<1,,0<v<2}$$



                    Therefore,



                    begin{align}
                    iint_D expleft({-frac{x-y}{x+y}}right),mathrm{d}x,mathrm{d}y&=frac{1}{2}iint_R ve^{-u},mathrm{d}u,mathrm{d}v
                    \\&=frac{1}{2}int_0^2v,mathrm{d}vint_{-1}^1 e^{-u},mathrm{d}u
                    end{align}






                    share|cite|improve this answer











                    $endgroup$













                    • $begingroup$
                      Finally the $v$ appears, but with no explanation at all?
                      $endgroup$
                      – NickD
                      33 mins ago










                    • $begingroup$
                      @NickD I have left the calculation of the jacobian to the OP.
                      $endgroup$
                      – StubbornAtom
                      30 mins ago










                    • $begingroup$
                      That's fine, but you should at least mention it, if only for the sake of future readers.
                      $endgroup$
                      – NickD
                      25 mins ago










                    • $begingroup$
                      It is pretty clear as it is, I think.
                      $endgroup$
                      – StubbornAtom
                      24 mins ago














                    5












                    5








                    5





                    $begingroup$

                    Your domain is $$D={(x,y):0<x<2,,,0<y<2,,,x+y<2}$$



                    Using change of variables $(x,y)to(u,v)$ with $$u=frac{x-y}{x+y},,,v=x+y$$



                    The region is now $$R={(u,v):-1<u<1,,0<v<2}$$



                    Therefore,



                    begin{align}
                    iint_D expleft({-frac{x-y}{x+y}}right),mathrm{d}x,mathrm{d}y&=frac{1}{2}iint_R ve^{-u},mathrm{d}u,mathrm{d}v
                    \\&=frac{1}{2}int_0^2v,mathrm{d}vint_{-1}^1 e^{-u},mathrm{d}u
                    end{align}






                    share|cite|improve this answer











                    $endgroup$



                    Your domain is $$D={(x,y):0<x<2,,,0<y<2,,,x+y<2}$$



                    Using change of variables $(x,y)to(u,v)$ with $$u=frac{x-y}{x+y},,,v=x+y$$



                    The region is now $$R={(u,v):-1<u<1,,0<v<2}$$



                    Therefore,



                    begin{align}
                    iint_D expleft({-frac{x-y}{x+y}}right),mathrm{d}x,mathrm{d}y&=frac{1}{2}iint_R ve^{-u},mathrm{d}u,mathrm{d}v
                    \\&=frac{1}{2}int_0^2v,mathrm{d}vint_{-1}^1 e^{-u},mathrm{d}u
                    end{align}







                    share|cite|improve this answer














                    share|cite|improve this answer



                    share|cite|improve this answer








                    edited 47 mins ago

























                    answered 54 mins ago









                    StubbornAtomStubbornAtom

                    6,13311339




                    6,13311339












                    • $begingroup$
                      Finally the $v$ appears, but with no explanation at all?
                      $endgroup$
                      – NickD
                      33 mins ago










                    • $begingroup$
                      @NickD I have left the calculation of the jacobian to the OP.
                      $endgroup$
                      – StubbornAtom
                      30 mins ago










                    • $begingroup$
                      That's fine, but you should at least mention it, if only for the sake of future readers.
                      $endgroup$
                      – NickD
                      25 mins ago










                    • $begingroup$
                      It is pretty clear as it is, I think.
                      $endgroup$
                      – StubbornAtom
                      24 mins ago


















                    • $begingroup$
                      Finally the $v$ appears, but with no explanation at all?
                      $endgroup$
                      – NickD
                      33 mins ago










                    • $begingroup$
                      @NickD I have left the calculation of the jacobian to the OP.
                      $endgroup$
                      – StubbornAtom
                      30 mins ago










                    • $begingroup$
                      That's fine, but you should at least mention it, if only for the sake of future readers.
                      $endgroup$
                      – NickD
                      25 mins ago










                    • $begingroup$
                      It is pretty clear as it is, I think.
                      $endgroup$
                      – StubbornAtom
                      24 mins ago
















                    $begingroup$
                    Finally the $v$ appears, but with no explanation at all?
                    $endgroup$
                    – NickD
                    33 mins ago




                    $begingroup$
                    Finally the $v$ appears, but with no explanation at all?
                    $endgroup$
                    – NickD
                    33 mins ago












                    $begingroup$
                    @NickD I have left the calculation of the jacobian to the OP.
                    $endgroup$
                    – StubbornAtom
                    30 mins ago




                    $begingroup$
                    @NickD I have left the calculation of the jacobian to the OP.
                    $endgroup$
                    – StubbornAtom
                    30 mins ago












                    $begingroup$
                    That's fine, but you should at least mention it, if only for the sake of future readers.
                    $endgroup$
                    – NickD
                    25 mins ago




                    $begingroup$
                    That's fine, but you should at least mention it, if only for the sake of future readers.
                    $endgroup$
                    – NickD
                    25 mins ago












                    $begingroup$
                    It is pretty clear as it is, I think.
                    $endgroup$
                    – StubbornAtom
                    24 mins ago




                    $begingroup$
                    It is pretty clear as it is, I think.
                    $endgroup$
                    – StubbornAtom
                    24 mins ago











                    -1












                    $begingroup$

                    Computer algebra gives (for the general case):



                    $$e left(x^2 text{Ei}left(-frac{2 x}{x+y}right)-frac{y^2 text{Ei}left(frac{2
                    y}{x+y}right)}{e^2}right)+frac{1}{2} e^{1-frac{2 x}{x+y}} (x+y)^2$$



                    over your specified region:



                    $$e-frac{1}{e}$$






                    share|cite|improve this answer









                    $endgroup$













                    • $begingroup$
                      yes the final answer is right due to my answer sheet but i should solve this by hand not computer any idea of how should i solve this on paper?
                      $endgroup$
                      – khoshrang
                      1 hour ago












                    • $begingroup$
                      (-1) because this is not an answer to the question. See the other excellent responses!
                      $endgroup$
                      – Jimmy Sabater
                      9 mins ago
















                    -1












                    $begingroup$

                    Computer algebra gives (for the general case):



                    $$e left(x^2 text{Ei}left(-frac{2 x}{x+y}right)-frac{y^2 text{Ei}left(frac{2
                    y}{x+y}right)}{e^2}right)+frac{1}{2} e^{1-frac{2 x}{x+y}} (x+y)^2$$



                    over your specified region:



                    $$e-frac{1}{e}$$






                    share|cite|improve this answer









                    $endgroup$













                    • $begingroup$
                      yes the final answer is right due to my answer sheet but i should solve this by hand not computer any idea of how should i solve this on paper?
                      $endgroup$
                      – khoshrang
                      1 hour ago












                    • $begingroup$
                      (-1) because this is not an answer to the question. See the other excellent responses!
                      $endgroup$
                      – Jimmy Sabater
                      9 mins ago














                    -1












                    -1








                    -1





                    $begingroup$

                    Computer algebra gives (for the general case):



                    $$e left(x^2 text{Ei}left(-frac{2 x}{x+y}right)-frac{y^2 text{Ei}left(frac{2
                    y}{x+y}right)}{e^2}right)+frac{1}{2} e^{1-frac{2 x}{x+y}} (x+y)^2$$



                    over your specified region:



                    $$e-frac{1}{e}$$






                    share|cite|improve this answer









                    $endgroup$



                    Computer algebra gives (for the general case):



                    $$e left(x^2 text{Ei}left(-frac{2 x}{x+y}right)-frac{y^2 text{Ei}left(frac{2
                    y}{x+y}right)}{e^2}right)+frac{1}{2} e^{1-frac{2 x}{x+y}} (x+y)^2$$



                    over your specified region:



                    $$e-frac{1}{e}$$







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered 1 hour ago









                    David G. StorkDavid G. Stork

                    11k41432




                    11k41432












                    • $begingroup$
                      yes the final answer is right due to my answer sheet but i should solve this by hand not computer any idea of how should i solve this on paper?
                      $endgroup$
                      – khoshrang
                      1 hour ago












                    • $begingroup$
                      (-1) because this is not an answer to the question. See the other excellent responses!
                      $endgroup$
                      – Jimmy Sabater
                      9 mins ago


















                    • $begingroup$
                      yes the final answer is right due to my answer sheet but i should solve this by hand not computer any idea of how should i solve this on paper?
                      $endgroup$
                      – khoshrang
                      1 hour ago












                    • $begingroup$
                      (-1) because this is not an answer to the question. See the other excellent responses!
                      $endgroup$
                      – Jimmy Sabater
                      9 mins ago
















                    $begingroup$
                    yes the final answer is right due to my answer sheet but i should solve this by hand not computer any idea of how should i solve this on paper?
                    $endgroup$
                    – khoshrang
                    1 hour ago






                    $begingroup$
                    yes the final answer is right due to my answer sheet but i should solve this by hand not computer any idea of how should i solve this on paper?
                    $endgroup$
                    – khoshrang
                    1 hour ago














                    $begingroup$
                    (-1) because this is not an answer to the question. See the other excellent responses!
                    $endgroup$
                    – Jimmy Sabater
                    9 mins ago




                    $begingroup$
                    (-1) because this is not an answer to the question. See the other excellent responses!
                    $endgroup$
                    – Jimmy Sabater
                    9 mins ago










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