Centre of mass of region using density (multivariable calculus)











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A lamina (two–dimensional plate) occupies the region inside the circle $x^2 + y^2 = 2y$, but outside the circle $x^2 +y^2 = 1$. Find the centre of mass if the density = $frac{k}{r}$ (inversely proportional to its distance from the origin). The formulae for centre of mass for $x,y$ are below: enter image description here



enter image description here



the graph of the two circles is below ($x^2 + y^2 = 2y$ is $x^2 + (y - 1)^2 = 1$ ):
enter image description here



I don't understand how to find the boundaries of region D. I know that it has to be in the form of polar co-ords $(r, theta)$ but I can't figure it out. Plz help!






calculus







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  • 1




    By symmetry the $x$-coordinate of the centre of mass is $0$. For the $y$-coordinate, we will need to integrate. Polar coordinates sound good. The point of intersection in the first quadrant is $(sqrt{3}/2,1/2)$ so for the integral we will go $theta=pi/6$ to $theta=5pi/6$, or better $pi/6$ to $pi/2$ then double.
    – André Nicolas
    May 2 '15 at 3:28















up vote
1
down vote

favorite
1












A lamina (two–dimensional plate) occupies the region inside the circle $x^2 + y^2 = 2y$, but outside the circle $x^2 +y^2 = 1$. Find the centre of mass if the density = $frac{k}{r}$ (inversely proportional to its distance from the origin). The formulae for centre of mass for $x,y$ are below: enter image description here



enter image description here



the graph of the two circles is below ($x^2 + y^2 = 2y$ is $x^2 + (y - 1)^2 = 1$ ):
enter image description here



I don't understand how to find the boundaries of region D. I know that it has to be in the form of polar co-ords $(r, theta)$ but I can't figure it out. Plz help!






calculus







share|cite|improve this question




















  • 1




    By symmetry the $x$-coordinate of the centre of mass is $0$. For the $y$-coordinate, we will need to integrate. Polar coordinates sound good. The point of intersection in the first quadrant is $(sqrt{3}/2,1/2)$ so for the integral we will go $theta=pi/6$ to $theta=5pi/6$, or better $pi/6$ to $pi/2$ then double.
    – André Nicolas
    May 2 '15 at 3:28













up vote
1
down vote

favorite
1









up vote
1
down vote

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1





A lamina (two–dimensional plate) occupies the region inside the circle $x^2 + y^2 = 2y$, but outside the circle $x^2 +y^2 = 1$. Find the centre of mass if the density = $frac{k}{r}$ (inversely proportional to its distance from the origin). The formulae for centre of mass for $x,y$ are below: enter image description here



enter image description here



the graph of the two circles is below ($x^2 + y^2 = 2y$ is $x^2 + (y - 1)^2 = 1$ ):
enter image description here



I don't understand how to find the boundaries of region D. I know that it has to be in the form of polar co-ords $(r, theta)$ but I can't figure it out. Plz help!






calculus







share|cite|improve this question















A lamina (two–dimensional plate) occupies the region inside the circle $x^2 + y^2 = 2y$, but outside the circle $x^2 +y^2 = 1$. Find the centre of mass if the density = $frac{k}{r}$ (inversely proportional to its distance from the origin). The formulae for centre of mass for $x,y$ are below: enter image description here



enter image description here



the graph of the two circles is below ($x^2 + y^2 = 2y$ is $x^2 + (y - 1)^2 = 1$ ):
enter image description here



I don't understand how to find the boundaries of region D. I know that it has to be in the form of polar co-ords $(r, theta)$ but I can't figure it out. Plz help!






calculus




calculus






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edited May 2 '15 at 5:21









alkabary

4,0791538




4,0791538










asked May 2 '15 at 3:11









user236133

62




62








  • 1




    By symmetry the $x$-coordinate of the centre of mass is $0$. For the $y$-coordinate, we will need to integrate. Polar coordinates sound good. The point of intersection in the first quadrant is $(sqrt{3}/2,1/2)$ so for the integral we will go $theta=pi/6$ to $theta=5pi/6$, or better $pi/6$ to $pi/2$ then double.
    – André Nicolas
    May 2 '15 at 3:28














  • 1




    By symmetry the $x$-coordinate of the centre of mass is $0$. For the $y$-coordinate, we will need to integrate. Polar coordinates sound good. The point of intersection in the first quadrant is $(sqrt{3}/2,1/2)$ so for the integral we will go $theta=pi/6$ to $theta=5pi/6$, or better $pi/6$ to $pi/2$ then double.
    – André Nicolas
    May 2 '15 at 3:28








1




1




By symmetry the $x$-coordinate of the centre of mass is $0$. For the $y$-coordinate, we will need to integrate. Polar coordinates sound good. The point of intersection in the first quadrant is $(sqrt{3}/2,1/2)$ so for the integral we will go $theta=pi/6$ to $theta=5pi/6$, or better $pi/6$ to $pi/2$ then double.
– André Nicolas
May 2 '15 at 3:28




By symmetry the $x$-coordinate of the centre of mass is $0$. For the $y$-coordinate, we will need to integrate. Polar coordinates sound good. The point of intersection in the first quadrant is $(sqrt{3}/2,1/2)$ so for the integral we will go $theta=pi/6$ to $theta=5pi/6$, or better $pi/6$ to $pi/2$ then double.
– André Nicolas
May 2 '15 at 3:28










2 Answers
2






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For $m$, we have that



$$begin{align}
m&=int_{pi/6}^{5pi/6} int_1^{2sin phi} frac{k}{rho}rho drho dphi\\
&=kint_{pi/6}^{5pi/6} (2sin phi -1)dphi\\
& =(2sqrt{3}-2pi/3)k
end{align}$$



For the term $M_x$, we have that



$$begin{align}
M_x&=int_{pi/6}^{5pi/6} int_1^{2sin phi} frac{krho cos phi}{rho}rho drho dphi\\
&=kint_{pi/6}^{5pi/6} cos phi int_1^{2sin phi} rho drho \\
&=kint_{pi/6}^{5pi/6}cos phi left(frac12 (4sin^2phi -1)right)dphi\\
&=0
end{align}$$



For the term $M_y$, we have that



$$begin{align}
M_y&=int_{pi/6}^{5pi/6} int_1^{2sin phi} frac{krho sin phi}{rho}rho drho dphi\\
&=kint_{pi/6}^{5pi/6} sin phi int_1^{2sin phi} rho drho \\
&=kint_{pi/6}^{5pi/6}sin phi left(frac12 (4sin^2phi -1)right)dphi\\
&=sqrt{3}k
end{align}$$






share|cite|improve this answer




























    up vote
    0
    down vote













    You drew a good picture of the region of integration, though it might
    be helpful to you if you make your drawing larger.
    On a larger drawing it's easier to add details like these:



    enter image description here



    The possible values of $theta$ for integration will range from the angle
    of the ray $overrightarrow{OA}$ to the angle of $overrightarrow{OB}$.
    We know that the length $OA = OB = 1$, and it should be clear by symmetry
    that the $y$-coordinate of $A$ is $frac12,$ and the same for $B$.
    The ray $overrightarrow{OA}$ therefore has $theta = arcsin frac12,$
    and you should recognize that $arcsin frac12 = fracpi6.$
    The ray $overrightarrow{OB}$ is in a symmetric position
    with $theta = pi - fracpi6.$



    Within that range of $theta$ values,
    consider an arbitrary line $overleftrightarrow{OP}$.
    The radius of any point on that line within in the region
    of integration is at least $1$,
    but no greater than the distance $OP$.
    That distance is $r$ for a point with polar coordinates
    $(r,theta)$ on the circle $x^2 + y^2 = 2y.$



    Subsituting $x^2 + y^2 = r^2$ and $y = rsintheta,$
    the equation of the circle is
    $r^2 = 2 rsintheta,$
    or more simply $r = 2 sintheta.$
    You should now have all you need to set up the boundaries of the
    double integral.






    share|cite|improve this answer





















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      2 Answers
      2






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      2 Answers
      2






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      up vote
      0
      down vote













      For $m$, we have that



      $$begin{align}
      m&=int_{pi/6}^{5pi/6} int_1^{2sin phi} frac{k}{rho}rho drho dphi\\
      &=kint_{pi/6}^{5pi/6} (2sin phi -1)dphi\\
      & =(2sqrt{3}-2pi/3)k
      end{align}$$



      For the term $M_x$, we have that



      $$begin{align}
      M_x&=int_{pi/6}^{5pi/6} int_1^{2sin phi} frac{krho cos phi}{rho}rho drho dphi\\
      &=kint_{pi/6}^{5pi/6} cos phi int_1^{2sin phi} rho drho \\
      &=kint_{pi/6}^{5pi/6}cos phi left(frac12 (4sin^2phi -1)right)dphi\\
      &=0
      end{align}$$



      For the term $M_y$, we have that



      $$begin{align}
      M_y&=int_{pi/6}^{5pi/6} int_1^{2sin phi} frac{krho sin phi}{rho}rho drho dphi\\
      &=kint_{pi/6}^{5pi/6} sin phi int_1^{2sin phi} rho drho \\
      &=kint_{pi/6}^{5pi/6}sin phi left(frac12 (4sin^2phi -1)right)dphi\\
      &=sqrt{3}k
      end{align}$$






      share|cite|improve this answer

























        up vote
        0
        down vote













        For $m$, we have that



        $$begin{align}
        m&=int_{pi/6}^{5pi/6} int_1^{2sin phi} frac{k}{rho}rho drho dphi\\
        &=kint_{pi/6}^{5pi/6} (2sin phi -1)dphi\\
        & =(2sqrt{3}-2pi/3)k
        end{align}$$



        For the term $M_x$, we have that



        $$begin{align}
        M_x&=int_{pi/6}^{5pi/6} int_1^{2sin phi} frac{krho cos phi}{rho}rho drho dphi\\
        &=kint_{pi/6}^{5pi/6} cos phi int_1^{2sin phi} rho drho \\
        &=kint_{pi/6}^{5pi/6}cos phi left(frac12 (4sin^2phi -1)right)dphi\\
        &=0
        end{align}$$



        For the term $M_y$, we have that



        $$begin{align}
        M_y&=int_{pi/6}^{5pi/6} int_1^{2sin phi} frac{krho sin phi}{rho}rho drho dphi\\
        &=kint_{pi/6}^{5pi/6} sin phi int_1^{2sin phi} rho drho \\
        &=kint_{pi/6}^{5pi/6}sin phi left(frac12 (4sin^2phi -1)right)dphi\\
        &=sqrt{3}k
        end{align}$$






        share|cite|improve this answer























          up vote
          0
          down vote










          up vote
          0
          down vote









          For $m$, we have that



          $$begin{align}
          m&=int_{pi/6}^{5pi/6} int_1^{2sin phi} frac{k}{rho}rho drho dphi\\
          &=kint_{pi/6}^{5pi/6} (2sin phi -1)dphi\\
          & =(2sqrt{3}-2pi/3)k
          end{align}$$



          For the term $M_x$, we have that



          $$begin{align}
          M_x&=int_{pi/6}^{5pi/6} int_1^{2sin phi} frac{krho cos phi}{rho}rho drho dphi\\
          &=kint_{pi/6}^{5pi/6} cos phi int_1^{2sin phi} rho drho \\
          &=kint_{pi/6}^{5pi/6}cos phi left(frac12 (4sin^2phi -1)right)dphi\\
          &=0
          end{align}$$



          For the term $M_y$, we have that



          $$begin{align}
          M_y&=int_{pi/6}^{5pi/6} int_1^{2sin phi} frac{krho sin phi}{rho}rho drho dphi\\
          &=kint_{pi/6}^{5pi/6} sin phi int_1^{2sin phi} rho drho \\
          &=kint_{pi/6}^{5pi/6}sin phi left(frac12 (4sin^2phi -1)right)dphi\\
          &=sqrt{3}k
          end{align}$$






          share|cite|improve this answer












          For $m$, we have that



          $$begin{align}
          m&=int_{pi/6}^{5pi/6} int_1^{2sin phi} frac{k}{rho}rho drho dphi\\
          &=kint_{pi/6}^{5pi/6} (2sin phi -1)dphi\\
          & =(2sqrt{3}-2pi/3)k
          end{align}$$



          For the term $M_x$, we have that



          $$begin{align}
          M_x&=int_{pi/6}^{5pi/6} int_1^{2sin phi} frac{krho cos phi}{rho}rho drho dphi\\
          &=kint_{pi/6}^{5pi/6} cos phi int_1^{2sin phi} rho drho \\
          &=kint_{pi/6}^{5pi/6}cos phi left(frac12 (4sin^2phi -1)right)dphi\\
          &=0
          end{align}$$



          For the term $M_y$, we have that



          $$begin{align}
          M_y&=int_{pi/6}^{5pi/6} int_1^{2sin phi} frac{krho sin phi}{rho}rho drho dphi\\
          &=kint_{pi/6}^{5pi/6} sin phi int_1^{2sin phi} rho drho \\
          &=kint_{pi/6}^{5pi/6}sin phi left(frac12 (4sin^2phi -1)right)dphi\\
          &=sqrt{3}k
          end{align}$$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered May 2 '15 at 3:58









          Mark Viola

          130k1273170




          130k1273170






















              up vote
              0
              down vote













              You drew a good picture of the region of integration, though it might
              be helpful to you if you make your drawing larger.
              On a larger drawing it's easier to add details like these:



              enter image description here



              The possible values of $theta$ for integration will range from the angle
              of the ray $overrightarrow{OA}$ to the angle of $overrightarrow{OB}$.
              We know that the length $OA = OB = 1$, and it should be clear by symmetry
              that the $y$-coordinate of $A$ is $frac12,$ and the same for $B$.
              The ray $overrightarrow{OA}$ therefore has $theta = arcsin frac12,$
              and you should recognize that $arcsin frac12 = fracpi6.$
              The ray $overrightarrow{OB}$ is in a symmetric position
              with $theta = pi - fracpi6.$



              Within that range of $theta$ values,
              consider an arbitrary line $overleftrightarrow{OP}$.
              The radius of any point on that line within in the region
              of integration is at least $1$,
              but no greater than the distance $OP$.
              That distance is $r$ for a point with polar coordinates
              $(r,theta)$ on the circle $x^2 + y^2 = 2y.$



              Subsituting $x^2 + y^2 = r^2$ and $y = rsintheta,$
              the equation of the circle is
              $r^2 = 2 rsintheta,$
              or more simply $r = 2 sintheta.$
              You should now have all you need to set up the boundaries of the
              double integral.






              share|cite|improve this answer

























                up vote
                0
                down vote













                You drew a good picture of the region of integration, though it might
                be helpful to you if you make your drawing larger.
                On a larger drawing it's easier to add details like these:



                enter image description here



                The possible values of $theta$ for integration will range from the angle
                of the ray $overrightarrow{OA}$ to the angle of $overrightarrow{OB}$.
                We know that the length $OA = OB = 1$, and it should be clear by symmetry
                that the $y$-coordinate of $A$ is $frac12,$ and the same for $B$.
                The ray $overrightarrow{OA}$ therefore has $theta = arcsin frac12,$
                and you should recognize that $arcsin frac12 = fracpi6.$
                The ray $overrightarrow{OB}$ is in a symmetric position
                with $theta = pi - fracpi6.$



                Within that range of $theta$ values,
                consider an arbitrary line $overleftrightarrow{OP}$.
                The radius of any point on that line within in the region
                of integration is at least $1$,
                but no greater than the distance $OP$.
                That distance is $r$ for a point with polar coordinates
                $(r,theta)$ on the circle $x^2 + y^2 = 2y.$



                Subsituting $x^2 + y^2 = r^2$ and $y = rsintheta,$
                the equation of the circle is
                $r^2 = 2 rsintheta,$
                or more simply $r = 2 sintheta.$
                You should now have all you need to set up the boundaries of the
                double integral.






                share|cite|improve this answer























                  up vote
                  0
                  down vote










                  up vote
                  0
                  down vote









                  You drew a good picture of the region of integration, though it might
                  be helpful to you if you make your drawing larger.
                  On a larger drawing it's easier to add details like these:



                  enter image description here



                  The possible values of $theta$ for integration will range from the angle
                  of the ray $overrightarrow{OA}$ to the angle of $overrightarrow{OB}$.
                  We know that the length $OA = OB = 1$, and it should be clear by symmetry
                  that the $y$-coordinate of $A$ is $frac12,$ and the same for $B$.
                  The ray $overrightarrow{OA}$ therefore has $theta = arcsin frac12,$
                  and you should recognize that $arcsin frac12 = fracpi6.$
                  The ray $overrightarrow{OB}$ is in a symmetric position
                  with $theta = pi - fracpi6.$



                  Within that range of $theta$ values,
                  consider an arbitrary line $overleftrightarrow{OP}$.
                  The radius of any point on that line within in the region
                  of integration is at least $1$,
                  but no greater than the distance $OP$.
                  That distance is $r$ for a point with polar coordinates
                  $(r,theta)$ on the circle $x^2 + y^2 = 2y.$



                  Subsituting $x^2 + y^2 = r^2$ and $y = rsintheta,$
                  the equation of the circle is
                  $r^2 = 2 rsintheta,$
                  or more simply $r = 2 sintheta.$
                  You should now have all you need to set up the boundaries of the
                  double integral.






                  share|cite|improve this answer












                  You drew a good picture of the region of integration, though it might
                  be helpful to you if you make your drawing larger.
                  On a larger drawing it's easier to add details like these:



                  enter image description here



                  The possible values of $theta$ for integration will range from the angle
                  of the ray $overrightarrow{OA}$ to the angle of $overrightarrow{OB}$.
                  We know that the length $OA = OB = 1$, and it should be clear by symmetry
                  that the $y$-coordinate of $A$ is $frac12,$ and the same for $B$.
                  The ray $overrightarrow{OA}$ therefore has $theta = arcsin frac12,$
                  and you should recognize that $arcsin frac12 = fracpi6.$
                  The ray $overrightarrow{OB}$ is in a symmetric position
                  with $theta = pi - fracpi6.$



                  Within that range of $theta$ values,
                  consider an arbitrary line $overleftrightarrow{OP}$.
                  The radius of any point on that line within in the region
                  of integration is at least $1$,
                  but no greater than the distance $OP$.
                  That distance is $r$ for a point with polar coordinates
                  $(r,theta)$ on the circle $x^2 + y^2 = 2y.$



                  Subsituting $x^2 + y^2 = r^2$ and $y = rsintheta,$
                  the equation of the circle is
                  $r^2 = 2 rsintheta,$
                  or more simply $r = 2 sintheta.$
                  You should now have all you need to set up the boundaries of the
                  double integral.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered May 2 '15 at 20:45









                  David K

                  52.2k340115




                  52.2k340115






























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