Parameterization of a torus
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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:
$(R - sqrtx^2 + y^2)^2 + z^2 = r^2$
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
34.1k365129
asked Dec 16 '15 at 19:15
KenP25
112
add a comment |Â
up vote
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down vote
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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:
$(R - sqrtx^2 + y^2)^2 + z^2 = r^2$
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
34.1k365129
asked Dec 16 '15 at 19:15
KenP25
112
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.
– TacoTesseract
Dec 24 '15 at 15:32
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
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:
$(R - sqrtx^2 + y^2)^2 + z^2 = r^2$
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
34.1k365129
asked Dec 16 '15 at 19:15
KenP25
112
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:
$(R - sqrtx^2 + y^2)^2 + z^2 = r^2$
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
34.1k365129
asked Dec 16 '15 at 19:15
KenP25
112
edited Dec 16 '15 at 19:27
Mike Miller
34.1k365129
edited Dec 16 '15 at 19:27
Mike Miller
34.1k365129
edited Dec 16 '15 at 19:27
Mike Miller
34.1k365129
34.1k365129
asked Dec 16 '15 at 19:15
KenP25
112
asked Dec 16 '15 at 19:15
KenP25
112
asked Dec 16 '15 at 19:15
KenP25
112
112
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.
– TacoTesseract
Dec 24 '15 at 15:32
add a comment |Â
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.
– TacoTesseract
Dec 24 '15 at 15:32
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.
– TacoTesseract
Dec 24 '15 at 15:32
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.
– TacoTesseract
Dec 24 '15 at 15:32
add a comment |Â
1 Answer
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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
$$left(R-sqrtx^2+y^2right)^2+r^2sin^2theta=r^2,$$
which is
$$left(R-sqrtx^2+y^2right)^2=r^2(1-sin^2theta).$$
The last expression can also be written as
$$left(sqrtx^2+y^2-Rright)^2=r^2cos^2theta.$$
Clear away squares and rearrange to obtain
$$sqrtx^2+y^2=R+rcostheta,$$
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
Edu
1,3811519
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
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
$$left(R-sqrtx^2+y^2right)^2+r^2sin^2theta=r^2,$$
which is
$$left(R-sqrtx^2+y^2right)^2=r^2(1-sin^2theta).$$
The last expression can also be written as
$$left(sqrtx^2+y^2-Rright)^2=r^2cos^2theta.$$
Clear away squares and rearrange to obtain
$$sqrtx^2+y^2=R+rcostheta,$$
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
Edu
1,3811519
add a comment |Â
up vote
0
down vote
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
$$left(R-sqrtx^2+y^2right)^2+r^2sin^2theta=r^2,$$
which is
$$left(R-sqrtx^2+y^2right)^2=r^2(1-sin^2theta).$$
The last expression can also be written as
$$left(sqrtx^2+y^2-Rright)^2=r^2cos^2theta.$$
Clear away squares and rearrange to obtain
$$sqrtx^2+y^2=R+rcostheta,$$
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
Edu
1,3811519
add a comment |Â
up vote
0
down vote
up vote
0
down vote
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
$$left(R-sqrtx^2+y^2right)^2+r^2sin^2theta=r^2,$$
which is
$$left(R-sqrtx^2+y^2right)^2=r^2(1-sin^2theta).$$
The last expression can also be written as
$$left(sqrtx^2+y^2-Rright)^2=r^2cos^2theta.$$
Clear away squares and rearrange to obtain
$$sqrtx^2+y^2=R+rcostheta,$$
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
Edu
1,3811519
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
$$left(R-sqrtx^2+y^2right)^2+r^2sin^2theta=r^2,$$
which is
$$left(R-sqrtx^2+y^2right)^2=r^2(1-sin^2theta).$$
The last expression can also be written as
$$left(sqrtx^2+y^2-Rright)^2=r^2cos^2theta.$$
Clear away squares and rearrange to obtain
$$sqrtx^2+y^2=R+rcostheta,$$
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
Edu
1,3811519
edited Sep 28 '16 at 10:48
answered Dec 16 '15 at 22:51
Edu
1,3811519
answered Dec 16 '15 at 22:51
Edu
1,3811519
answered Dec 16 '15 at 22:51
Edu
1,3811519
1,3811519
add a comment |Â
add a comment |Â
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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.
– TacoTesseract
Dec 24 '15 at 15:32