Parameterization of a torus












1












$begingroup$


Given that the parameterization of a torus is given by:



$x(theta,phi) = (R + rcos(theta))cos(phi)$



$y(theta,phi) = (R + rcos(theta))sin(phi)$



$z(theta,phi) = rsin(theta)$



and the equation of a torus in Cartesian coordinates is given by:



$(R - sqrt{x^2 + y^2})^2 + z^2 = r^2$



Where $R$ represents the major radius and $r$ the minor radius.



How does one show that the parameterization equations satisfy the Cartesian coordinates? I've tried plugging in and using trig identities, but keep getting stuck.



Also, how would I calculate the volume of a general torus using a triple integral in Cartesian coordinates?






multivariable-calculus trigonometry analytic-geometry parametric







share|cite|improve this question











$endgroup$












  • $begingroup$
    Once you have the parameterization, you can use the divergence theorem to find the volume. All you need is a vector function whose div is 1.
    $endgroup$
    – TacoTesseract
    Dec 24 '15 at 15:32
















1












$begingroup$


Given that the parameterization of a torus is given by:



$x(theta,phi) = (R + rcos(theta))cos(phi)$



$y(theta,phi) = (R + rcos(theta))sin(phi)$



$z(theta,phi) = rsin(theta)$



and the equation of a torus in Cartesian coordinates is given by:



$(R - sqrt{x^2 + y^2})^2 + z^2 = r^2$



Where $R$ represents the major radius and $r$ the minor radius.



How does one show that the parameterization equations satisfy the Cartesian coordinates? I've tried plugging in and using trig identities, but keep getting stuck.



Also, how would I calculate the volume of a general torus using a triple integral in Cartesian coordinates?






multivariable-calculus trigonometry analytic-geometry parametric







share|cite|improve this question











$endgroup$












  • $begingroup$
    Once you have the parameterization, you can use the divergence theorem to find the volume. All you need is a vector function whose div is 1.
    $endgroup$
    – TacoTesseract
    Dec 24 '15 at 15:32














1












1








1





$begingroup$


Given that the parameterization of a torus is given by:



$x(theta,phi) = (R + rcos(theta))cos(phi)$



$y(theta,phi) = (R + rcos(theta))sin(phi)$



$z(theta,phi) = rsin(theta)$



and the equation of a torus in Cartesian coordinates is given by:



$(R - sqrt{x^2 + y^2})^2 + z^2 = r^2$



Where $R$ represents the major radius and $r$ the minor radius.



How does one show that the parameterization equations satisfy the Cartesian coordinates? I've tried plugging in and using trig identities, but keep getting stuck.



Also, how would I calculate the volume of a general torus using a triple integral in Cartesian coordinates?






multivariable-calculus trigonometry analytic-geometry parametric







share|cite|improve this question











$endgroup$




Given that the parameterization of a torus is given by:



$x(theta,phi) = (R + rcos(theta))cos(phi)$



$y(theta,phi) = (R + rcos(theta))sin(phi)$



$z(theta,phi) = rsin(theta)$



and the equation of a torus in Cartesian coordinates is given by:



$(R - sqrt{x^2 + y^2})^2 + z^2 = r^2$



Where $R$ represents the major radius and $r$ the minor radius.



How does one show that the parameterization equations satisfy the Cartesian coordinates? I've tried plugging in and using trig identities, but keep getting stuck.



Also, how would I calculate the volume of a general torus using a triple integral in Cartesian coordinates?






multivariable-calculus trigonometry analytic-geometry parametric




multivariable-calculus trigonometry analytic-geometry parametric






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 16 '15 at 19:27









Mike Miller

36.8k470137




36.8k470137










asked Dec 16 '15 at 19:15









KenP25KenP25

112




112












  • $begingroup$
    Once you have the parameterization, you can use the divergence theorem to find the volume. All you need is a vector function whose div is 1.
    $endgroup$
    – TacoTesseract
    Dec 24 '15 at 15:32


















  • $begingroup$
    Once you have the parameterization, you can use the divergence theorem to find the volume. All you need is a vector function whose div is 1.
    $endgroup$
    – TacoTesseract
    Dec 24 '15 at 15:32
















$begingroup$
Once you have the parameterization, you can use the divergence theorem to find the volume. All you need is a vector function whose div is 1.
$endgroup$
– TacoTesseract
Dec 24 '15 at 15:32




$begingroup$
Once you have the parameterization, you can use the divergence theorem to find the volume. All you need is a vector function whose div is 1.
$endgroup$
– TacoTesseract
Dec 24 '15 at 15:32










1 Answer
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oldest

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$begingroup$

Parametrization, as wikipedia defines it, is the process of deciding and defining the parameters necessary for a complete or relevant specification of a model or geometric object. Let's do this for the torus you described.



One approach is to set $;z=rsintheta;$ (take a point on the surface of the torus and think how the $z$ coordinate is, in terms of the angle $;theta$).



Now, since you know the equation of the torus, you can plug in $;z=rsintheta;$ to get
$$left(R-sqrt{x^2+y^2}right)^2+r^2sin^2theta=r^2,$$
which is
$$left(R-sqrt{x^2+y^2}right)^2=r^2(1-sin^2theta).$$



The last expression can also be written as
$$left(sqrt{x^2+y^2}-Rright)^2=r^2cos^2theta.$$



Clear away squares and rearrange to obtain
$$sqrt{x^2+y^2}=R+rcostheta,$$
and from this you get
$$;x^2+y^2=(R+rcostheta)^2.$$



You can observe that (for a constant $;theta;$) this equation defines a circle of radius $;R+rcostheta,;$ so you may set now $;x=(R+rcostheta)cosphi;$ and $;y=(R+rcostheta)sinphi;$.
Now, you have your complete parametrization for the torus.



For the volume of the torus I recommend you to (again) think geometrically. Simplify the torus in something that you know how to deal with (in terms of measuring it). Then you might figure out how to use that calculation to get the whole volume.






share|cite|improve this answer











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    $begingroup$

    Parametrization, as wikipedia defines it, is the process of deciding and defining the parameters necessary for a complete or relevant specification of a model or geometric object. Let's do this for the torus you described.



    One approach is to set $;z=rsintheta;$ (take a point on the surface of the torus and think how the $z$ coordinate is, in terms of the angle $;theta$).



    Now, since you know the equation of the torus, you can plug in $;z=rsintheta;$ to get
    $$left(R-sqrt{x^2+y^2}right)^2+r^2sin^2theta=r^2,$$
    which is
    $$left(R-sqrt{x^2+y^2}right)^2=r^2(1-sin^2theta).$$



    The last expression can also be written as
    $$left(sqrt{x^2+y^2}-Rright)^2=r^2cos^2theta.$$



    Clear away squares and rearrange to obtain
    $$sqrt{x^2+y^2}=R+rcostheta,$$
    and from this you get
    $$;x^2+y^2=(R+rcostheta)^2.$$



    You can observe that (for a constant $;theta;$) this equation defines a circle of radius $;R+rcostheta,;$ so you may set now $;x=(R+rcostheta)cosphi;$ and $;y=(R+rcostheta)sinphi;$.
    Now, you have your complete parametrization for the torus.



    For the volume of the torus I recommend you to (again) think geometrically. Simplify the torus in something that you know how to deal with (in terms of measuring it). Then you might figure out how to use that calculation to get the whole volume.






    share|cite|improve this answer











    $endgroup$


















      0












      $begingroup$

      Parametrization, as wikipedia defines it, is the process of deciding and defining the parameters necessary for a complete or relevant specification of a model or geometric object. Let's do this for the torus you described.



      One approach is to set $;z=rsintheta;$ (take a point on the surface of the torus and think how the $z$ coordinate is, in terms of the angle $;theta$).



      Now, since you know the equation of the torus, you can plug in $;z=rsintheta;$ to get
      $$left(R-sqrt{x^2+y^2}right)^2+r^2sin^2theta=r^2,$$
      which is
      $$left(R-sqrt{x^2+y^2}right)^2=r^2(1-sin^2theta).$$



      The last expression can also be written as
      $$left(sqrt{x^2+y^2}-Rright)^2=r^2cos^2theta.$$



      Clear away squares and rearrange to obtain
      $$sqrt{x^2+y^2}=R+rcostheta,$$
      and from this you get
      $$;x^2+y^2=(R+rcostheta)^2.$$



      You can observe that (for a constant $;theta;$) this equation defines a circle of radius $;R+rcostheta,;$ so you may set now $;x=(R+rcostheta)cosphi;$ and $;y=(R+rcostheta)sinphi;$.
      Now, you have your complete parametrization for the torus.



      For the volume of the torus I recommend you to (again) think geometrically. Simplify the torus in something that you know how to deal with (in terms of measuring it). Then you might figure out how to use that calculation to get the whole volume.






      share|cite|improve this answer











      $endgroup$
















        0












        0








        0





        $begingroup$

        Parametrization, as wikipedia defines it, is the process of deciding and defining the parameters necessary for a complete or relevant specification of a model or geometric object. Let's do this for the torus you described.



        One approach is to set $;z=rsintheta;$ (take a point on the surface of the torus and think how the $z$ coordinate is, in terms of the angle $;theta$).



        Now, since you know the equation of the torus, you can plug in $;z=rsintheta;$ to get
        $$left(R-sqrt{x^2+y^2}right)^2+r^2sin^2theta=r^2,$$
        which is
        $$left(R-sqrt{x^2+y^2}right)^2=r^2(1-sin^2theta).$$



        The last expression can also be written as
        $$left(sqrt{x^2+y^2}-Rright)^2=r^2cos^2theta.$$



        Clear away squares and rearrange to obtain
        $$sqrt{x^2+y^2}=R+rcostheta,$$
        and from this you get
        $$;x^2+y^2=(R+rcostheta)^2.$$



        You can observe that (for a constant $;theta;$) this equation defines a circle of radius $;R+rcostheta,;$ so you may set now $;x=(R+rcostheta)cosphi;$ and $;y=(R+rcostheta)sinphi;$.
        Now, you have your complete parametrization for the torus.



        For the volume of the torus I recommend you to (again) think geometrically. Simplify the torus in something that you know how to deal with (in terms of measuring it). Then you might figure out how to use that calculation to get the whole volume.






        share|cite|improve this answer











        $endgroup$



        Parametrization, as wikipedia defines it, is the process of deciding and defining the parameters necessary for a complete or relevant specification of a model or geometric object. Let's do this for the torus you described.



        One approach is to set $;z=rsintheta;$ (take a point on the surface of the torus and think how the $z$ coordinate is, in terms of the angle $;theta$).



        Now, since you know the equation of the torus, you can plug in $;z=rsintheta;$ to get
        $$left(R-sqrt{x^2+y^2}right)^2+r^2sin^2theta=r^2,$$
        which is
        $$left(R-sqrt{x^2+y^2}right)^2=r^2(1-sin^2theta).$$



        The last expression can also be written as
        $$left(sqrt{x^2+y^2}-Rright)^2=r^2cos^2theta.$$



        Clear away squares and rearrange to obtain
        $$sqrt{x^2+y^2}=R+rcostheta,$$
        and from this you get
        $$;x^2+y^2=(R+rcostheta)^2.$$



        You can observe that (for a constant $;theta;$) this equation defines a circle of radius $;R+rcostheta,;$ so you may set now $;x=(R+rcostheta)cosphi;$ and $;y=(R+rcostheta)sinphi;$.
        Now, you have your complete parametrization for the torus.



        For the volume of the torus I recommend you to (again) think geometrically. Simplify the torus in something that you know how to deal with (in terms of measuring it). Then you might figure out how to use that calculation to get the whole volume.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Sep 28 '16 at 10:48

























        answered Dec 16 '15 at 22:51









        EduEdu

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