how to determine the bounds of this integral?












4















Let $ G:= { (x,y) in mathbb{R}^2 mid x^2 + 4y^2 >1, x^2+y^2 <4 }$



I want to determine $ int_G x^2+y^2 d(x,y)$.




So, its the area between a circle with radius 2 und an ellipse with semi-axis 1 in $x$-direction and semi-axis $frac{1}{2}$ in $y$-direction, right?



How can I determine the bounds? How can I use polar coordinates to transform this?






real-analysis integration







share|cite|improve this question





























    4















    Let $ G:= { (x,y) in mathbb{R}^2 mid x^2 + 4y^2 >1, x^2+y^2 <4 }$



    I want to determine $ int_G x^2+y^2 d(x,y)$.




    So, its the area between a circle with radius 2 und an ellipse with semi-axis 1 in $x$-direction and semi-axis $frac{1}{2}$ in $y$-direction, right?



    How can I determine the bounds? How can I use polar coordinates to transform this?






    real-analysis integration







    share|cite|improve this question



























      4












      4








      4


      1






      Let $ G:= { (x,y) in mathbb{R}^2 mid x^2 + 4y^2 >1, x^2+y^2 <4 }$



      I want to determine $ int_G x^2+y^2 d(x,y)$.




      So, its the area between a circle with radius 2 und an ellipse with semi-axis 1 in $x$-direction and semi-axis $frac{1}{2}$ in $y$-direction, right?



      How can I determine the bounds? How can I use polar coordinates to transform this?






      real-analysis integration







      share|cite|improve this question
















      Let $ G:= { (x,y) in mathbb{R}^2 mid x^2 + 4y^2 >1, x^2+y^2 <4 }$



      I want to determine $ int_G x^2+y^2 d(x,y)$.




      So, its the area between a circle with radius 2 und an ellipse with semi-axis 1 in $x$-direction and semi-axis $frac{1}{2}$ in $y$-direction, right?



      How can I determine the bounds? How can I use polar coordinates to transform this?






      real-analysis integration




      real-analysis integration






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Nov 21 '18 at 18:39









      Bernard

      118k639112




      118k639112










      asked Nov 21 '18 at 18:26









      constant94constant94

      1018




      1018






















          1 Answer
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          Let
          $$ x = r cos theta, quad y = r sin theta. $$
          Then since $x^2+y^2 < 4,$ we see that $r < 2$, and since $x^2+4y^2 > 1$, we see that
          $$ r^2cos^2theta+4r^2sin^2theta = r^2+3r^2sin^2theta > 1 quad iff quad r > frac{1}{sqrt{1+3sin^2theta}}. $$
          Thus
          begin{align}
          int_G x^2+y^2 , mathrm d(x,y) &= int_0^{2pi}int_{frac{1}{sqrt{1+3sin^2theta}}}^2 r^2 cdot r , mathrm dr , mathrm dtheta.\
          end{align}
          Can you proceed?






          share|cite|improve this answer





















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            4














            Let
            $$ x = r cos theta, quad y = r sin theta. $$
            Then since $x^2+y^2 < 4,$ we see that $r < 2$, and since $x^2+4y^2 > 1$, we see that
            $$ r^2cos^2theta+4r^2sin^2theta = r^2+3r^2sin^2theta > 1 quad iff quad r > frac{1}{sqrt{1+3sin^2theta}}. $$
            Thus
            begin{align}
            int_G x^2+y^2 , mathrm d(x,y) &= int_0^{2pi}int_{frac{1}{sqrt{1+3sin^2theta}}}^2 r^2 cdot r , mathrm dr , mathrm dtheta.\
            end{align}
            Can you proceed?






            share|cite|improve this answer


























              4














              Let
              $$ x = r cos theta, quad y = r sin theta. $$
              Then since $x^2+y^2 < 4,$ we see that $r < 2$, and since $x^2+4y^2 > 1$, we see that
              $$ r^2cos^2theta+4r^2sin^2theta = r^2+3r^2sin^2theta > 1 quad iff quad r > frac{1}{sqrt{1+3sin^2theta}}. $$
              Thus
              begin{align}
              int_G x^2+y^2 , mathrm d(x,y) &= int_0^{2pi}int_{frac{1}{sqrt{1+3sin^2theta}}}^2 r^2 cdot r , mathrm dr , mathrm dtheta.\
              end{align}
              Can you proceed?






              share|cite|improve this answer
























                4












                4








                4






                Let
                $$ x = r cos theta, quad y = r sin theta. $$
                Then since $x^2+y^2 < 4,$ we see that $r < 2$, and since $x^2+4y^2 > 1$, we see that
                $$ r^2cos^2theta+4r^2sin^2theta = r^2+3r^2sin^2theta > 1 quad iff quad r > frac{1}{sqrt{1+3sin^2theta}}. $$
                Thus
                begin{align}
                int_G x^2+y^2 , mathrm d(x,y) &= int_0^{2pi}int_{frac{1}{sqrt{1+3sin^2theta}}}^2 r^2 cdot r , mathrm dr , mathrm dtheta.\
                end{align}
                Can you proceed?






                share|cite|improve this answer












                Let
                $$ x = r cos theta, quad y = r sin theta. $$
                Then since $x^2+y^2 < 4,$ we see that $r < 2$, and since $x^2+4y^2 > 1$, we see that
                $$ r^2cos^2theta+4r^2sin^2theta = r^2+3r^2sin^2theta > 1 quad iff quad r > frac{1}{sqrt{1+3sin^2theta}}. $$
                Thus
                begin{align}
                int_G x^2+y^2 , mathrm d(x,y) &= int_0^{2pi}int_{frac{1}{sqrt{1+3sin^2theta}}}^2 r^2 cdot r , mathrm dr , mathrm dtheta.\
                end{align}
                Can you proceed?







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Nov 21 '18 at 18:36









                MisterRiemannMisterRiemann

                5,8291624




                5,8291624






























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