How do I compute the double integral of $xy(x^2+y^2)$ with $-3le x^2-y^2le3$ and $1le xyle 4$?

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How do I compute the double integral of
$$f(x,y)=xy(x^2+y^2)$$ with
$$-3le x^2-y^2le3 \[6pt]
1le xyle 4$$



My approach is to find the points of intersection of the regions and use that to get limits of integration. Is there an easier method, or some kind of substitution?




integration definite-integrals




share|cite|improve this question

























    up vote
    1
    down vote

    favorite












    How do I compute the double integral of
    $$f(x,y)=xy(x^2+y^2)$$ with
    $$-3le x^2-y^2le3 \[6pt]
    1le xyle 4$$



    My approach is to find the points of intersection of the regions and use that to get limits of integration. Is there an easier method, or some kind of substitution?




    integration definite-integrals




    share|cite|improve this question























      up vote
      1
      down vote

      favorite









      up vote
      1
      down vote

      favorite











      How do I compute the double integral of
      $$f(x,y)=xy(x^2+y^2)$$ with
      $$-3le x^2-y^2le3 \[6pt]
      1le xyle 4$$



      My approach is to find the points of intersection of the regions and use that to get limits of integration. Is there an easier method, or some kind of substitution?




      integration definite-integrals




      share|cite|improve this question













      How do I compute the double integral of
      $$f(x,y)=xy(x^2+y^2)$$ with
      $$-3le x^2-y^2le3 \[6pt]
      1le xyle 4$$



      My approach is to find the points of intersection of the regions and use that to get limits of integration. Is there an easier method, or some kind of substitution?





      integration definite-integrals





      share|cite|improve this question












      share|cite|improve this question




      share|cite|improve this question








      edited Jul 20 at 10:22









      Blue

      43.7k868141




      43.7k868141









      asked Jul 20 at 9:40









      Piyush Divyanakar

      3,258122




      3,258122




















          2 Answers
          2






          active

          oldest

          votes

















          up vote
          2
          down vote



          accepted










          HINT



          Let



          • $xy=u$

          • $x^2-y^2=v$

          and



          $$du,dv=beginvmatrixy&x\2x&-2yendvmatrixdx,dy=|-2(x^2+y^2)|dx,dy=2(x^2+y^2)dx,dy\implies dx,dy=frac2x^2+y^2du,dv$$






          share|cite|improve this answer





















          • damm. thanks man. I got it
            – Piyush Divyanakar
            Jul 20 at 10:18










          • @PiyushDivyanakar Well done. You are welcome! Bye
            – gimusi
            Jul 20 at 10:19










          • Why is the determinant involved in the solution of $dudv$?
            – Cornman
            Jul 20 at 10:36






          • 1




            @Cornman it is the Jacobian for the change of coordinates. In that case we see that the $x^2+y^2$ term cancels out.
            – gimusi
            Jul 20 at 11:05

















          up vote
          1
          down vote













          a different approach would be
          $$x=r cos(phi), y=r sin(phi)$$
          and observing that
          $$x^2-y^2=r^2 cos(2phi), 2xy=r^2 sin(2phi)$$



          --- rk






          share|cite|improve this answer





















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            2 Answers
            2






            active

            oldest

            votes








            2 Answers
            2






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes








            up vote
            2
            down vote



            accepted










            HINT



            Let



            • $xy=u$

            • $x^2-y^2=v$

            and



            $$du,dv=beginvmatrixy&x\2x&-2yendvmatrixdx,dy=|-2(x^2+y^2)|dx,dy=2(x^2+y^2)dx,dy\implies dx,dy=frac2x^2+y^2du,dv$$






            share|cite|improve this answer





















            • damm. thanks man. I got it
              – Piyush Divyanakar
              Jul 20 at 10:18










            • @PiyushDivyanakar Well done. You are welcome! Bye
              – gimusi
              Jul 20 at 10:19










            • Why is the determinant involved in the solution of $dudv$?
              – Cornman
              Jul 20 at 10:36






            • 1




              @Cornman it is the Jacobian for the change of coordinates. In that case we see that the $x^2+y^2$ term cancels out.
              – gimusi
              Jul 20 at 11:05














            up vote
            2
            down vote



            accepted










            HINT



            Let



            • $xy=u$

            • $x^2-y^2=v$

            and



            $$du,dv=beginvmatrixy&x\2x&-2yendvmatrixdx,dy=|-2(x^2+y^2)|dx,dy=2(x^2+y^2)dx,dy\implies dx,dy=frac2x^2+y^2du,dv$$






            share|cite|improve this answer





















            • damm. thanks man. I got it
              – Piyush Divyanakar
              Jul 20 at 10:18










            • @PiyushDivyanakar Well done. You are welcome! Bye
              – gimusi
              Jul 20 at 10:19










            • Why is the determinant involved in the solution of $dudv$?
              – Cornman
              Jul 20 at 10:36






            • 1




              @Cornman it is the Jacobian for the change of coordinates. In that case we see that the $x^2+y^2$ term cancels out.
              – gimusi
              Jul 20 at 11:05












            up vote
            2
            down vote



            accepted







            up vote
            2
            down vote



            accepted






            HINT



            Let



            • $xy=u$

            • $x^2-y^2=v$

            and



            $$du,dv=beginvmatrixy&x\2x&-2yendvmatrixdx,dy=|-2(x^2+y^2)|dx,dy=2(x^2+y^2)dx,dy\implies dx,dy=frac2x^2+y^2du,dv$$






            share|cite|improve this answer













            HINT



            Let



            • $xy=u$

            • $x^2-y^2=v$

            and



            $$du,dv=beginvmatrixy&x\2x&-2yendvmatrixdx,dy=|-2(x^2+y^2)|dx,dy=2(x^2+y^2)dx,dy\implies dx,dy=frac2x^2+y^2du,dv$$







            share|cite|improve this answer













            share|cite|improve this answer



            share|cite|improve this answer











            answered Jul 20 at 10:15









            gimusi

            65.4k73584




            65.4k73584











            • damm. thanks man. I got it
              – Piyush Divyanakar
              Jul 20 at 10:18










            • @PiyushDivyanakar Well done. You are welcome! Bye
              – gimusi
              Jul 20 at 10:19










            • Why is the determinant involved in the solution of $dudv$?
              – Cornman
              Jul 20 at 10:36






            • 1




              @Cornman it is the Jacobian for the change of coordinates. In that case we see that the $x^2+y^2$ term cancels out.
              – gimusi
              Jul 20 at 11:05
















            • damm. thanks man. I got it
              – Piyush Divyanakar
              Jul 20 at 10:18










            • @PiyushDivyanakar Well done. You are welcome! Bye
              – gimusi
              Jul 20 at 10:19










            • Why is the determinant involved in the solution of $dudv$?
              – Cornman
              Jul 20 at 10:36






            • 1




              @Cornman it is the Jacobian for the change of coordinates. In that case we see that the $x^2+y^2$ term cancels out.
              – gimusi
              Jul 20 at 11:05















            damm. thanks man. I got it
            – Piyush Divyanakar
            Jul 20 at 10:18




            damm. thanks man. I got it
            – Piyush Divyanakar
            Jul 20 at 10:18












            @PiyushDivyanakar Well done. You are welcome! Bye
            – gimusi
            Jul 20 at 10:19




            @PiyushDivyanakar Well done. You are welcome! Bye
            – gimusi
            Jul 20 at 10:19












            Why is the determinant involved in the solution of $dudv$?
            – Cornman
            Jul 20 at 10:36




            Why is the determinant involved in the solution of $dudv$?
            – Cornman
            Jul 20 at 10:36




            1




            1




            @Cornman it is the Jacobian for the change of coordinates. In that case we see that the $x^2+y^2$ term cancels out.
            – gimusi
            Jul 20 at 11:05




            @Cornman it is the Jacobian for the change of coordinates. In that case we see that the $x^2+y^2$ term cancels out.
            – gimusi
            Jul 20 at 11:05










            up vote
            1
            down vote













            a different approach would be
            $$x=r cos(phi), y=r sin(phi)$$
            and observing that
            $$x^2-y^2=r^2 cos(2phi), 2xy=r^2 sin(2phi)$$



            --- rk






            share|cite|improve this answer

























              up vote
              1
              down vote













              a different approach would be
              $$x=r cos(phi), y=r sin(phi)$$
              and observing that
              $$x^2-y^2=r^2 cos(2phi), 2xy=r^2 sin(2phi)$$



              --- rk






              share|cite|improve this answer























                up vote
                1
                down vote










                up vote
                1
                down vote









                a different approach would be
                $$x=r cos(phi), y=r sin(phi)$$
                and observing that
                $$x^2-y^2=r^2 cos(2phi), 2xy=r^2 sin(2phi)$$



                --- rk






                share|cite|improve this answer













                a different approach would be
                $$x=r cos(phi), y=r sin(phi)$$
                and observing that
                $$x^2-y^2=r^2 cos(2phi), 2xy=r^2 sin(2phi)$$



                --- rk







                share|cite|improve this answer













                share|cite|improve this answer



                share|cite|improve this answer











                answered Jul 20 at 15:50









                Dr. Richard Klitzing

                7586




                7586






















                     

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