Parameterization of a torus
$begingroup$
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
edited Dec 16 '15 at 19:27
Mike Miller
36.8k470137
asked Dec 16 '15 at 19:15
KenP25KenP25
112
$endgroup$
add a comment |
$begingroup$
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
edited Dec 16 '15 at 19:27
Mike Miller
36.8k470137
asked Dec 16 '15 at 19:15
KenP25KenP25
112
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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
edited Dec 16 '15 at 19:27
Mike Miller
36.8k470137
asked Dec 16 '15 at 19:15
KenP25KenP25
112
$endgroup$
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
edited Dec 16 '15 at 19:27
Mike Miller
36.8k470137
asked Dec 16 '15 at 19:15
KenP25KenP25
112
edited Dec 16 '15 at 19:27
Mike Miller
36.8k470137
asked Dec 16 '15 at 19:15
KenP25KenP25
112
edited Dec 16 '15 at 19:27
Mike Miller
36.8k470137
edited Dec 16 '15 at 19:27
Mike Miller
36.8k470137
edited Dec 16 '15 at 19:27
Mike Miller
36.8k470137
36.8k470137
asked Dec 16 '15 at 19:15
KenP25KenP25
112
asked Dec 16 '15 at 19:15
KenP25KenP25
112
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
add a comment |
$begingroup$
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
add a comment |
1 Answer
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oldest
votes
$begingroup$
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.
answered Dec 16 '15 at 22:51
EduEdu
1,3861619
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add a comment |
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1 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.
answered Dec 16 '15 at 22:51
EduEdu
1,3861619
$endgroup$
add a comment |
$begingroup$
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.
answered Dec 16 '15 at 22:51
EduEdu
1,3861619
$endgroup$
add a comment |
$begingroup$
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.
answered Dec 16 '15 at 22:51
EduEdu
1,3861619
$endgroup$
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.
answered Dec 16 '15 at 22:51
EduEdu
1,3861619
edited Sep 28 '16 at 10:48
answered Dec 16 '15 at 22:51
EduEdu
1,3861619
answered Dec 16 '15 at 22:51
EduEdu
1,3861619
answered Dec 16 '15 at 22:51
EduEdu
1,3861619
1,3861619
add a comment |
add a comment |
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$begingroup$
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