Fifty Ways to Leave Your Lover—
er, Find the Volume of a Torus
Dan Drake –University of Minnesota
http://www.math.umn.edu/∼drake
1Introduction
This is almost certainly the firsttime in this history of human civilization that Paul Simon lyrics have been used in a multivariable calculus class.
In this class we’velearned a lot of differentkinds of integrals. We’lluse almost all of them in this article/document/whateverto findthe volume of a torus.
2Why Homer Simpson is a topologist
Let’sbegin by making sure we understand what a torus is and what it looks like. The quick answer is “adoughnut”,which is why this section has the title it does. However, this author is a particular fan of Bavarian creme-filleddoughnuts, which are not torus-shaped. Let’sleave the baked goods behind and make our definitiona bit more rigorous:
A torus is obtained by rotating a circle of radius a whose center is A units away from a line around that line. A picture of this is shown
below:
1
In three dimensions, a torus looks like this:
y z
Through this article, we’llassume that A >a >0. You can relax that restriction a bit and get goofy looking torii, but we won’tworry about that.
3Slip out the back, Jack
We’llstart out using Calculus I methods to figureout the volume. We can think of the torus as a surface of revolution with another surface of revolution taken out of it. The outside is like the Bavarian creme-filleddoughnut I spoke of earlier, and can be described by rotating the semicircle
y =f (x ) =A +a 2−x 2
around the x -axis. The resulting object looks this:(1)
2
As you probably learned in firstsemester calculus, we findthe volume by a single integral. The value of f (x ) is the radius of our surface; if we square f (x ), multiply it by πand by dx , we can think of the integral as findingthe total volume of many very thin disks. The integral is
a
−a π[f (x )]d x =2 a −a 2 22πA +a −x d x, (2)
but we’vecounted too much volume. We need to account for the hole in the middle, which we do √with a very similar integral:we use the semicircle y =g (x ) =A −, which looks like this when revolved around the x -axis:
We do a very similar integral to findthe volume of the hole, then subtract that. This gives us our firstway of findingthe volume of a torus:
a
−a πA + a 2−x 2 2 2 22−πA −a −x d x. (3)
3
4Make a new plan, Stan
The new plan is actually just another way of thinking of the above method. Instead of doing two integrals with surfaces of revolution, we’llthink of the torus as a stack of very thin annuli. An annulus (pluralannuli ) is a disk with another disk cut out of its center. A compact disc is a good example of an annulus.
Think of a torus as a bagel, and slice it as you normally would slice a bagel. That gives you two pieces. Now make the same kind of slice, but slice the torus/bagelinto three pieces, or four... or millions of slices. As you make more and more slices, the resulting pieces become more like perfect annuli. Here’sa picture of the situation, taken from [2],(page
262):
To findthe volume, we can just findthe area of one of those thin annuli and integrate that from −a to a . The area of an annulus with outer radius R and inner radius r is just the area of a disk of radius R minus the area of a disk with radius r , which is π(R 2−r 2).
So, what’sthe outer and inner radius for these annuli? The outer radius is along the “outer”√semicircle described in equation (1), so R =A +In a very similar way, you can figure√out that r =A −. That means that at x =a , the area of an annulus is
2 2 πA +a 2−x 2−A −a 2−x 2.
That’sthe area of a thin annulus; to get the volume of the entire torus, we integrate that over −a ≤x ≤a to get a 2 2 πA +a 2−x 2−A −a 2−x 2d x.
−a
However, you can see that this is exactly the same as (3) above. This is just another way to think about the surface-of-revolution method.
4
5No need to be coy, Roy
Next we’lluse double integrals to findthe volume. Think of a torus as a sliced bagel and put the top half on the xy plane. A double integral will findthe volume underneath the surface of the torus, and we can then multiply by two. In cartesian coordinates, it’sdifficultto express this surface, but in polar coordinates, it’svery easy:
if A −a ≤r ≤A +a f (r, θ) = 0otherwise .
All we have to do is integrate that function over appropriate bounds in cylindrical coordinates. It’sclear that θwill range from 0to 2π; you can see from the above definitionthat r will range from A −a to A +a . The integral is almost
2πA +a a −(r −A ) d r d θ.2
0A −a That’snot quite right, because in cylindrical coordinates, we have to multiply by the determinant of the Jacobian of the transformation that takes us from cartesian to cylindrical coordinates. That’sa long-winded way to say “multiplythe integrand by r ”.Once we do that, we have the second way of findingthe volume of a torus:2πA +a 2r a 2−(r −A ) 2d r d θ.
0A −a (4)
6Hop on the bus, Gus
By now you should know the basic following properties of single, double, and triple integrals:when you integrate the function “1”,you get length, area, and volume, respectively. That is,
1d x =length of interval I
I
R
S 1d A =area of the planar region R 1d V =volume of the 3-dimensional solid S.
Therefore, if asked to findthe volume of a torus, the easiest answer is that the volume is
1d V
T
5
where T is the torus 1. That answer is perfectly legitimate, but of course it isn’tvery satisfying. Let’suse the Divergence Theorem to turn this into something more interesting. Recall that the Divergence Theorem (alsocalled Gauss’Theorem, but everything is Gauss’Theorem) states that The Divergence Theorem. If S is a solid region in R 3and ∂Sis the boundary of S , then
¤F ·n d σ=div F d V ∂SS
for “nice”vector fieldsF .
We have a triple integral whose integrand is 1; let’stry to work backwards and convert that integral into a surface integral. We need a vector fieldwhose divergence is exactly 1. Here are a couple possibilities:
F (x, y, z ) =(x, 0, 0)
=(0, y, 0)
=(0, 0, z )
=(x/3, y/3, z/3)
=(x, sin(cos(z )) , log(π+|y |+9 1+y 4))
No reasonable person would ever want to work with that last vector field,but its divergence is certainly 1. I came up with the last vector fieldby just writing down a hairy-looking function of z in the y coordinate, and another hairy-looking function of y in the z coordinate. The partial derivatives will clearly be zero.
Certainly there are infinitelymany functions of y and z (thereare ℵ1of them, if you happen to know anything about transfinitecardinals), which means that the awe-inspiring power of calculus enables us to findthe volume of a torus using any one of infinitelymany vector fields—thereare INFINITELY MANY WAYS to findthe volume of a torus! Stick that in your pipe and smoke it, Paul Simon.
1Let’srecord one of those ways, using the vector fieldF (x, y, z ) =(x, y, z ):¤1(x, y, z ) ·n d σ.3
∂T(5)
7Drop offthe key, Lee
The other major theorem we learned this semester is Stokes’Theorem, which relates surface integrals and line integrals. At this point, you might tempted to think of something like this (“SF”1This isn’tquite right; a torus is actually the 2-dimensional surface, and not the solid region it encloses.
6
means a scalar-valued function, “VF”means a vector field):
triple integral of SF =surface integral of VF
surface integral of VF =line integral of VF
⇓
triple integral of SF =line integral of VF
But alas, our reach has exceeded our grasp. This will never happen unless the scalar function is identically zero and the vector fieldis conservative, and in that case, the integrals will all be zero. There’sa couple ways to explain why this is so. First, let’srecall Stoke’sTheorem:
Stokes’Theorem. Let S be a smooth surface in R 3. If F is a differentiablevector field,then
curl F d σ=F d x ,
S ∂S(DivergenceThm) (Stokes’Thm)
provided some conditions on orientation are met.
The firstreason can be summarized by saying “divcurl F =0”.If we apply the Divergence Theorem to the situation described above in Stokes’Theorem, we would have
div(curlF ) d V =curl F d σ.
T S
However, if you take any vector field,findits curl, and then findthe divergence of the resulting vector field,you’llalways get zero. So the triple integral on the left will be zero, making the double integral and in turn the single integral all zero. So there’sno chance we’llbe able to measure the volume of something using this method (unlessits volume is zero, which isn’tvery interesting). The second and more serious problem with this is the fact that the Divergence Theorem required a closed surface—onethat encloses a three-dimensional solid region. But Stokes’Theorem requires the surface to have some sort of boundary, and no surface that encloses a solid region can have a boundary:compare the surface of a sphere, which has no boundary and encloses a solid region, and a disk, which has a boundary but doesn’tenclose any sort of solid region. So we simply can’trelate the single integral in Stokes’to the triple integral in the Divergence Theorem.
Despite that unfortunate reality of calculus, we still have found a number of ways to findthe volume of a torus. We didn’tfind50ways, but Paul Simon’ssong didn’tlist 50ways to leave your lover, either!
Below are some possibly useful references for the types of integrals and theorems mentioned here.
7
References
[1]David M. Bressoud, Second year calculus , Springer-Verlag, 1991.
[2]Richard E. Williamson and Hale F. Trotter, Multivariable mathematics , Prentice-Hall, 1996. Dan Drake ([email protected])
87November 2002
Fifty Ways to Leave Your Lover—
er, Find the Volume of a Torus
Dan Drake –University of Minnesota
http://www.math.umn.edu/∼drake
1Introduction
This is almost certainly the firsttime in this history of human civilization that Paul Simon lyrics have been used in a multivariable calculus class.
In this class we’velearned a lot of differentkinds of integrals. We’lluse almost all of them in this article/document/whateverto findthe volume of a torus.
2Why Homer Simpson is a topologist
Let’sbegin by making sure we understand what a torus is and what it looks like. The quick answer is “adoughnut”,which is why this section has the title it does. However, this author is a particular fan of Bavarian creme-filleddoughnuts, which are not torus-shaped. Let’sleave the baked goods behind and make our definitiona bit more rigorous:
A torus is obtained by rotating a circle of radius a whose center is A units away from a line around that line. A picture of this is shown
below:
1
In three dimensions, a torus looks like this:
y z
Through this article, we’llassume that A >a >0. You can relax that restriction a bit and get goofy looking torii, but we won’tworry about that.
3Slip out the back, Jack
We’llstart out using Calculus I methods to figureout the volume. We can think of the torus as a surface of revolution with another surface of revolution taken out of it. The outside is like the Bavarian creme-filleddoughnut I spoke of earlier, and can be described by rotating the semicircle
y =f (x ) =A +a 2−x 2
around the x -axis. The resulting object looks this:(1)
2
As you probably learned in firstsemester calculus, we findthe volume by a single integral. The value of f (x ) is the radius of our surface; if we square f (x ), multiply it by πand by dx , we can think of the integral as findingthe total volume of many very thin disks. The integral is
a
−a π[f (x )]d x =2 a −a 2 22πA +a −x d x, (2)
but we’vecounted too much volume. We need to account for the hole in the middle, which we do √with a very similar integral:we use the semicircle y =g (x ) =A −, which looks like this when revolved around the x -axis:
We do a very similar integral to findthe volume of the hole, then subtract that. This gives us our firstway of findingthe volume of a torus:
a
−a πA + a 2−x 2 2 2 22−πA −a −x d x. (3)
3
4Make a new plan, Stan
The new plan is actually just another way of thinking of the above method. Instead of doing two integrals with surfaces of revolution, we’llthink of the torus as a stack of very thin annuli. An annulus (pluralannuli ) is a disk with another disk cut out of its center. A compact disc is a good example of an annulus.
Think of a torus as a bagel, and slice it as you normally would slice a bagel. That gives you two pieces. Now make the same kind of slice, but slice the torus/bagelinto three pieces, or four... or millions of slices. As you make more and more slices, the resulting pieces become more like perfect annuli. Here’sa picture of the situation, taken from [2],(page
262):
To findthe volume, we can just findthe area of one of those thin annuli and integrate that from −a to a . The area of an annulus with outer radius R and inner radius r is just the area of a disk of radius R minus the area of a disk with radius r , which is π(R 2−r 2).
So, what’sthe outer and inner radius for these annuli? The outer radius is along the “outer”√semicircle described in equation (1), so R =A +In a very similar way, you can figure√out that r =A −. That means that at x =a , the area of an annulus is
2 2 πA +a 2−x 2−A −a 2−x 2.
That’sthe area of a thin annulus; to get the volume of the entire torus, we integrate that over −a ≤x ≤a to get a 2 2 πA +a 2−x 2−A −a 2−x 2d x.
−a
However, you can see that this is exactly the same as (3) above. This is just another way to think about the surface-of-revolution method.
4
5No need to be coy, Roy
Next we’lluse double integrals to findthe volume. Think of a torus as a sliced bagel and put the top half on the xy plane. A double integral will findthe volume underneath the surface of the torus, and we can then multiply by two. In cartesian coordinates, it’sdifficultto express this surface, but in polar coordinates, it’svery easy:
if A −a ≤r ≤A +a f (r, θ) = 0otherwise .
All we have to do is integrate that function over appropriate bounds in cylindrical coordinates. It’sclear that θwill range from 0to 2π; you can see from the above definitionthat r will range from A −a to A +a . The integral is almost
2πA +a a −(r −A ) d r d θ.2
0A −a That’snot quite right, because in cylindrical coordinates, we have to multiply by the determinant of the Jacobian of the transformation that takes us from cartesian to cylindrical coordinates. That’sa long-winded way to say “multiplythe integrand by r ”.Once we do that, we have the second way of findingthe volume of a torus:2πA +a 2r a 2−(r −A ) 2d r d θ.
0A −a (4)
6Hop on the bus, Gus
By now you should know the basic following properties of single, double, and triple integrals:when you integrate the function “1”,you get length, area, and volume, respectively. That is,
1d x =length of interval I
I
R
S 1d A =area of the planar region R 1d V =volume of the 3-dimensional solid S.
Therefore, if asked to findthe volume of a torus, the easiest answer is that the volume is
1d V
T
5
where T is the torus 1. That answer is perfectly legitimate, but of course it isn’tvery satisfying. Let’suse the Divergence Theorem to turn this into something more interesting. Recall that the Divergence Theorem (alsocalled Gauss’Theorem, but everything is Gauss’Theorem) states that The Divergence Theorem. If S is a solid region in R 3and ∂Sis the boundary of S , then
¤F ·n d σ=div F d V ∂SS
for “nice”vector fieldsF .
We have a triple integral whose integrand is 1; let’stry to work backwards and convert that integral into a surface integral. We need a vector fieldwhose divergence is exactly 1. Here are a couple possibilities:
F (x, y, z ) =(x, 0, 0)
=(0, y, 0)
=(0, 0, z )
=(x/3, y/3, z/3)
=(x, sin(cos(z )) , log(π+|y |+9 1+y 4))
No reasonable person would ever want to work with that last vector field,but its divergence is certainly 1. I came up with the last vector fieldby just writing down a hairy-looking function of z in the y coordinate, and another hairy-looking function of y in the z coordinate. The partial derivatives will clearly be zero.
Certainly there are infinitelymany functions of y and z (thereare ℵ1of them, if you happen to know anything about transfinitecardinals), which means that the awe-inspiring power of calculus enables us to findthe volume of a torus using any one of infinitelymany vector fields—thereare INFINITELY MANY WAYS to findthe volume of a torus! Stick that in your pipe and smoke it, Paul Simon.
1Let’srecord one of those ways, using the vector fieldF (x, y, z ) =(x, y, z ):¤1(x, y, z ) ·n d σ.3
∂T(5)
7Drop offthe key, Lee
The other major theorem we learned this semester is Stokes’Theorem, which relates surface integrals and line integrals. At this point, you might tempted to think of something like this (“SF”1This isn’tquite right; a torus is actually the 2-dimensional surface, and not the solid region it encloses.
6
means a scalar-valued function, “VF”means a vector field):
triple integral of SF =surface integral of VF
surface integral of VF =line integral of VF
⇓
triple integral of SF =line integral of VF
But alas, our reach has exceeded our grasp. This will never happen unless the scalar function is identically zero and the vector fieldis conservative, and in that case, the integrals will all be zero. There’sa couple ways to explain why this is so. First, let’srecall Stoke’sTheorem:
Stokes’Theorem. Let S be a smooth surface in R 3. If F is a differentiablevector field,then
curl F d σ=F d x ,
S ∂S(DivergenceThm) (Stokes’Thm)
provided some conditions on orientation are met.
The firstreason can be summarized by saying “divcurl F =0”.If we apply the Divergence Theorem to the situation described above in Stokes’Theorem, we would have
div(curlF ) d V =curl F d σ.
T S
However, if you take any vector field,findits curl, and then findthe divergence of the resulting vector field,you’llalways get zero. So the triple integral on the left will be zero, making the double integral and in turn the single integral all zero. So there’sno chance we’llbe able to measure the volume of something using this method (unlessits volume is zero, which isn’tvery interesting). The second and more serious problem with this is the fact that the Divergence Theorem required a closed surface—onethat encloses a three-dimensional solid region. But Stokes’Theorem requires the surface to have some sort of boundary, and no surface that encloses a solid region can have a boundary:compare the surface of a sphere, which has no boundary and encloses a solid region, and a disk, which has a boundary but doesn’tenclose any sort of solid region. So we simply can’trelate the single integral in Stokes’to the triple integral in the Divergence Theorem.
Despite that unfortunate reality of calculus, we still have found a number of ways to findthe volume of a torus. We didn’tfind50ways, but Paul Simon’ssong didn’tlist 50ways to leave your lover, either!
Below are some possibly useful references for the types of integrals and theorems mentioned here.
7
References
[1]David M. Bressoud, Second year calculus , Springer-Verlag, 1991.
[2]Richard E. Williamson and Hale F. Trotter, Multivariable mathematics , Prentice-Hall, 1996. Dan Drake ([email protected])
87November 2002