How to proceed with this integral?











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Let $ G:= left{ (x,y) in mathbb{R}^2 : 0 < y,: x^2 + frac{y^2}{9} <1: ,: x^2+y^2 > 1 right} $.



I want to calculate this integral:
$ displaystyleint_G x^2,dxdy $.




I want to try with polar coordinates:
so I set $ (x,y) = (rcosphi,rsinphi)$, but
I am not sure how to get the right boundaries for $phi $. Isn't it $ x^2 +y^2 = r^2 $ ?



Any help is very appreciated !






calculus integration definite-integrals multiple-integral







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    up vote
    3
    down vote

    favorite
    1













    Let $ G:= left{ (x,y) in mathbb{R}^2 : 0 < y,: x^2 + frac{y^2}{9} <1: ,: x^2+y^2 > 1 right} $.



    I want to calculate this integral:
    $ displaystyleint_G x^2,dxdy $.




    I want to try with polar coordinates:
    so I set $ (x,y) = (rcosphi,rsinphi)$, but
    I am not sure how to get the right boundaries for $phi $. Isn't it $ x^2 +y^2 = r^2 $ ?



    Any help is very appreciated !






    calculus integration definite-integrals multiple-integral







    share|cite|improve this question


























      up vote
      3
      down vote

      favorite
      1









      up vote
      3
      down vote

      favorite
      1






      1






      Let $ G:= left{ (x,y) in mathbb{R}^2 : 0 < y,: x^2 + frac{y^2}{9} <1: ,: x^2+y^2 > 1 right} $.



      I want to calculate this integral:
      $ displaystyleint_G x^2,dxdy $.




      I want to try with polar coordinates:
      so I set $ (x,y) = (rcosphi,rsinphi)$, but
      I am not sure how to get the right boundaries for $phi $. Isn't it $ x^2 +y^2 = r^2 $ ?



      Any help is very appreciated !






      calculus integration definite-integrals multiple-integral







      share|cite|improve this question
















      Let $ G:= left{ (x,y) in mathbb{R}^2 : 0 < y,: x^2 + frac{y^2}{9} <1: ,: x^2+y^2 > 1 right} $.



      I want to calculate this integral:
      $ displaystyleint_G x^2,dxdy $.




      I want to try with polar coordinates:
      so I set $ (x,y) = (rcosphi,rsinphi)$, but
      I am not sure how to get the right boundaries for $phi $. Isn't it $ x^2 +y^2 = r^2 $ ?



      Any help is very appreciated !






      calculus integration definite-integrals multiple-integral




      calculus integration definite-integrals multiple-integral






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      edited Nov 22 at 8:28









      Robert Z

      92.1k1058129




      92.1k1058129










      asked Nov 22 at 8:10









      wondering1123

      10011




      10011






















          3 Answers
          3






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          up vote
          9
          down vote













          Even though I believe that the other answers are the best way to go, I think that you can use polar coordinates if you wish.



          Since $y>0$, we are integrating over the first two quadrants, so that $0 leq phi leq pi$.
          Since $x^2+y^2> 1$, we have $r > 1$, and since $x^2+y^2/9 < 1$, we see that
          $$ r^2cos^2theta + frac{r^2sin^2theta}{9} < 1 quad Rightarrow quad r < frac{3}{sqrt{8cos^2theta+1}}. $$



          Thus
          begin{align}
          int_0^pi int_{1}^{frac{3}{sqrt{8cos^2theta+1}}} r^2cos^2theta cdot r ,mathrm dr , mathrm dtheta &= frac{81}{4} int_0^pi frac{cos^2theta}{(8cos^2theta+1)^2} , mathrm dtheta - frac{1}{4} int_0^pi cos^2theta , mathrm dtheta = frac{pi}{4}.
          end{align}






          share|cite|improve this answer






























            up vote
            5
            down vote













            It is not a good idea to use polar coordinates. The integral can be written as $int_{-1}^{1}int_{sqrt{1-x^{2}}} ^{3sqrt{1-x^{2}}}x^{2}, dy, dx=int_{-1}^{1}2sqrt {1-x^{2}}x^{2}, dx$. To evaluate this put $x=sin, theta$ and use the formulas $2sin, theta cos, theta =sin, 2theta$, $2sin^{2}, 2theta =1-cos (4theta)$.






            share|cite|improve this answer




























              up vote
              5
              down vote













              Why in polar coordinates? Maybe it is easier by cartesian coordinates. Note that in $G$, $-1leq xleq 1$ and
              $$sqrt{1-x^2}<y<3sqrt{1-x^2}.$$
              Therefore
              $$int_G x^2,dxdy=int_{x=-1}^1x^2left(3sqrt{1-x^2}-sqrt{1-x^2}right)dx$$
              Can you take it from here?






              share|cite|improve this answer























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                3 Answers
                3






                active

                oldest

                votes








                3 Answers
                3






                active

                oldest

                votes









                active

                oldest

                votes






                active

                oldest

                votes








                up vote
                9
                down vote













                Even though I believe that the other answers are the best way to go, I think that you can use polar coordinates if you wish.



                Since $y>0$, we are integrating over the first two quadrants, so that $0 leq phi leq pi$.
                Since $x^2+y^2> 1$, we have $r > 1$, and since $x^2+y^2/9 < 1$, we see that
                $$ r^2cos^2theta + frac{r^2sin^2theta}{9} < 1 quad Rightarrow quad r < frac{3}{sqrt{8cos^2theta+1}}. $$



                Thus
                begin{align}
                int_0^pi int_{1}^{frac{3}{sqrt{8cos^2theta+1}}} r^2cos^2theta cdot r ,mathrm dr , mathrm dtheta &= frac{81}{4} int_0^pi frac{cos^2theta}{(8cos^2theta+1)^2} , mathrm dtheta - frac{1}{4} int_0^pi cos^2theta , mathrm dtheta = frac{pi}{4}.
                end{align}






                share|cite|improve this answer



























                  up vote
                  9
                  down vote













                  Even though I believe that the other answers are the best way to go, I think that you can use polar coordinates if you wish.



                  Since $y>0$, we are integrating over the first two quadrants, so that $0 leq phi leq pi$.
                  Since $x^2+y^2> 1$, we have $r > 1$, and since $x^2+y^2/9 < 1$, we see that
                  $$ r^2cos^2theta + frac{r^2sin^2theta}{9} < 1 quad Rightarrow quad r < frac{3}{sqrt{8cos^2theta+1}}. $$



                  Thus
                  begin{align}
                  int_0^pi int_{1}^{frac{3}{sqrt{8cos^2theta+1}}} r^2cos^2theta cdot r ,mathrm dr , mathrm dtheta &= frac{81}{4} int_0^pi frac{cos^2theta}{(8cos^2theta+1)^2} , mathrm dtheta - frac{1}{4} int_0^pi cos^2theta , mathrm dtheta = frac{pi}{4}.
                  end{align}






                  share|cite|improve this answer

























                    up vote
                    9
                    down vote










                    up vote
                    9
                    down vote









                    Even though I believe that the other answers are the best way to go, I think that you can use polar coordinates if you wish.



                    Since $y>0$, we are integrating over the first two quadrants, so that $0 leq phi leq pi$.
                    Since $x^2+y^2> 1$, we have $r > 1$, and since $x^2+y^2/9 < 1$, we see that
                    $$ r^2cos^2theta + frac{r^2sin^2theta}{9} < 1 quad Rightarrow quad r < frac{3}{sqrt{8cos^2theta+1}}. $$



                    Thus
                    begin{align}
                    int_0^pi int_{1}^{frac{3}{sqrt{8cos^2theta+1}}} r^2cos^2theta cdot r ,mathrm dr , mathrm dtheta &= frac{81}{4} int_0^pi frac{cos^2theta}{(8cos^2theta+1)^2} , mathrm dtheta - frac{1}{4} int_0^pi cos^2theta , mathrm dtheta = frac{pi}{4}.
                    end{align}






                    share|cite|improve this answer














                    Even though I believe that the other answers are the best way to go, I think that you can use polar coordinates if you wish.



                    Since $y>0$, we are integrating over the first two quadrants, so that $0 leq phi leq pi$.
                    Since $x^2+y^2> 1$, we have $r > 1$, and since $x^2+y^2/9 < 1$, we see that
                    $$ r^2cos^2theta + frac{r^2sin^2theta}{9} < 1 quad Rightarrow quad r < frac{3}{sqrt{8cos^2theta+1}}. $$



                    Thus
                    begin{align}
                    int_0^pi int_{1}^{frac{3}{sqrt{8cos^2theta+1}}} r^2cos^2theta cdot r ,mathrm dr , mathrm dtheta &= frac{81}{4} int_0^pi frac{cos^2theta}{(8cos^2theta+1)^2} , mathrm dtheta - frac{1}{4} int_0^pi cos^2theta , mathrm dtheta = frac{pi}{4}.
                    end{align}







                    share|cite|improve this answer














                    share|cite|improve this answer



                    share|cite|improve this answer








                    edited Nov 22 at 13:05

























                    answered Nov 22 at 8:37









                    MisterRiemann

                    5,7031624




                    5,7031624






















                        up vote
                        5
                        down vote













                        It is not a good idea to use polar coordinates. The integral can be written as $int_{-1}^{1}int_{sqrt{1-x^{2}}} ^{3sqrt{1-x^{2}}}x^{2}, dy, dx=int_{-1}^{1}2sqrt {1-x^{2}}x^{2}, dx$. To evaluate this put $x=sin, theta$ and use the formulas $2sin, theta cos, theta =sin, 2theta$, $2sin^{2}, 2theta =1-cos (4theta)$.






                        share|cite|improve this answer

























                          up vote
                          5
                          down vote













                          It is not a good idea to use polar coordinates. The integral can be written as $int_{-1}^{1}int_{sqrt{1-x^{2}}} ^{3sqrt{1-x^{2}}}x^{2}, dy, dx=int_{-1}^{1}2sqrt {1-x^{2}}x^{2}, dx$. To evaluate this put $x=sin, theta$ and use the formulas $2sin, theta cos, theta =sin, 2theta$, $2sin^{2}, 2theta =1-cos (4theta)$.






                          share|cite|improve this answer























                            up vote
                            5
                            down vote










                            up vote
                            5
                            down vote









                            It is not a good idea to use polar coordinates. The integral can be written as $int_{-1}^{1}int_{sqrt{1-x^{2}}} ^{3sqrt{1-x^{2}}}x^{2}, dy, dx=int_{-1}^{1}2sqrt {1-x^{2}}x^{2}, dx$. To evaluate this put $x=sin, theta$ and use the formulas $2sin, theta cos, theta =sin, 2theta$, $2sin^{2}, 2theta =1-cos (4theta)$.






                            share|cite|improve this answer












                            It is not a good idea to use polar coordinates. The integral can be written as $int_{-1}^{1}int_{sqrt{1-x^{2}}} ^{3sqrt{1-x^{2}}}x^{2}, dy, dx=int_{-1}^{1}2sqrt {1-x^{2}}x^{2}, dx$. To evaluate this put $x=sin, theta$ and use the formulas $2sin, theta cos, theta =sin, 2theta$, $2sin^{2}, 2theta =1-cos (4theta)$.







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered Nov 22 at 8:20









                            Kavi Rama Murthy

                            46.6k31854




                            46.6k31854






















                                up vote
                                5
                                down vote













                                Why in polar coordinates? Maybe it is easier by cartesian coordinates. Note that in $G$, $-1leq xleq 1$ and
                                $$sqrt{1-x^2}<y<3sqrt{1-x^2}.$$
                                Therefore
                                $$int_G x^2,dxdy=int_{x=-1}^1x^2left(3sqrt{1-x^2}-sqrt{1-x^2}right)dx$$
                                Can you take it from here?






                                share|cite|improve this answer



























                                  up vote
                                  5
                                  down vote













                                  Why in polar coordinates? Maybe it is easier by cartesian coordinates. Note that in $G$, $-1leq xleq 1$ and
                                  $$sqrt{1-x^2}<y<3sqrt{1-x^2}.$$
                                  Therefore
                                  $$int_G x^2,dxdy=int_{x=-1}^1x^2left(3sqrt{1-x^2}-sqrt{1-x^2}right)dx$$
                                  Can you take it from here?






                                  share|cite|improve this answer

























                                    up vote
                                    5
                                    down vote










                                    up vote
                                    5
                                    down vote









                                    Why in polar coordinates? Maybe it is easier by cartesian coordinates. Note that in $G$, $-1leq xleq 1$ and
                                    $$sqrt{1-x^2}<y<3sqrt{1-x^2}.$$
                                    Therefore
                                    $$int_G x^2,dxdy=int_{x=-1}^1x^2left(3sqrt{1-x^2}-sqrt{1-x^2}right)dx$$
                                    Can you take it from here?






                                    share|cite|improve this answer














                                    Why in polar coordinates? Maybe it is easier by cartesian coordinates. Note that in $G$, $-1leq xleq 1$ and
                                    $$sqrt{1-x^2}<y<3sqrt{1-x^2}.$$
                                    Therefore
                                    $$int_G x^2,dxdy=int_{x=-1}^1x^2left(3sqrt{1-x^2}-sqrt{1-x^2}right)dx$$
                                    Can you take it from here?







                                    share|cite|improve this answer














                                    share|cite|improve this answer



                                    share|cite|improve this answer








                                    edited Nov 22 at 8:33

























                                    answered Nov 22 at 8:20









                                    Robert Z

                                    92.1k1058129




                                    92.1k1058129






























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