info prev up next book cdrom email home

Surface of Revolution

A surface of revolution is a Surface generated by rotating a 2-D Curve about an axis. The resulting surface therefore always has azimuthal symmetry. Examples of surfaces of revolution include the Apple, Cone (excluding the base), Conical Frustum (excluding the ends), Cylinder (excluding the ends), Darwin-de Sitter Spheroid, Gabriel's Horn, Hyperboloid, Lemon, Oblate Spheroid, Paraboloid, Prolate Spheroid, Pseudosphere, Sphere, Spheroid, and Torus (and its generalization, the Toroid).

The standard parameterization of a surface of revolution is given by

$\displaystyle x(u,v)$ $\textstyle =$ $\displaystyle \phi(v)\cos u$ (1)
$\displaystyle y(u,v)$ $\textstyle =$ $\displaystyle \phi(v)\sin u$ (2)
$\displaystyle z(u,v)$ $\textstyle =$ $\displaystyle \psi(v).$ (3)

For a curve so parameterized, the first Fundamental Form has
$\displaystyle E$ $\textstyle =$ $\displaystyle \psi^2$ (4)
$\displaystyle F$ $\textstyle =$ $\displaystyle 0$ (5)
$\displaystyle G$ $\textstyle =$ $\displaystyle \phi'^2+\psi'^2.$ (6)

Wherever $\phi$ and $\phi'^2+\psi'^2$ are nonzero, then the surface is regular and the second Fundamental Form has
$\displaystyle e$ $\textstyle =$ $\displaystyle -{\vert\phi\vert\psi'\over\sqrt{\phi'^2+\psi'^2}}$ (7)
$\displaystyle f$ $\textstyle =$ $\displaystyle 0$ (8)
$\displaystyle g$ $\textstyle =$ $\displaystyle {\mathop{\rm sgn}\nolimits (\phi)(\phi''\psi'-\phi'\psi'')\over\sqrt{\phi'^2+\psi'^2}}.$ (9)

Furthermore, the unit Normal Vector is
\hat {\bf N}(u,v)={\mathop{\rm sgn}\nolimits (\phi)\over\sqr...
...left[{\matrix{\phi'\cos u\cr \psi'\sin u\cr \phi'\cr}}\right],
\end{displaymath} (10)

and the Principal Curvatures are
$\displaystyle \kappa_1$ $\textstyle =$ $\displaystyle {g\over G}={\mathop{\rm sgn}\nolimits (\phi)(\phi''\psi'-\phi'\psi'')\over(\phi'^2+\psi'^2)^{3/2}}$ (11)
$\displaystyle \kappa_2$ $\textstyle =$ $\displaystyle {e\over E}=-{\psi'\over\vert\phi\vert\sqrt{\phi'^2+\psi'^2}}.$ (12)

The Gaussian and Mean Curvatures are
$\displaystyle K$ $\textstyle =$ $\displaystyle {-\psi'^2\phi''+\phi'\psi'\psi''\over \phi(\phi'^2+\psi'^2)^2}$ (13)
$\displaystyle H$ $\textstyle =$ $\displaystyle {\phi(\phi''\psi'-\phi'\psi'')-\psi'(\phi'^2+\psi'^2)\over 2\vert\phi\vert(\phi'^2+\psi'^2)^{3/2}}$ (14)

(Gray 1993).

Pappus's Centroid Theorem gives the Volume of a solid of rotation as the cross-sectional Area times the distance traveled by the centroid as it is rotated.

Calculus of Variations can be used to find the curve from a point $(x_1,y_1)$ to a point $(x_2,y_2)$ which, when revolved around the x-Axis, yields a surface of smallest Surface Area $A$ (i.e., the Minimal Surface). This is equivalent to finding the Minimal Surface passing through two circular wire frames. The Area element is

dA=2\pi y\, ds = 2\pi y\sqrt{1+y'^2}\,dx,
\end{displaymath} (15)

so the Surface Area is
A=2\pi \int y\sqrt{1+y'^2}\,dx,
\end{displaymath} (16)

and the quantity we are minimizing is
\end{displaymath} (17)

This equation has $f_x=0$, so we can use the Beltrami Identity
f-y_x{\partial f\over\partial y_x} = a
\end{displaymath} (18)

to obtain
y\sqrt{1+y'^2}-y' {yy'\over\sqrt{1+y'^2}} = a
\end{displaymath} (19)

\end{displaymath} (20)

\end{displaymath} (21)

\end{displaymath} (22)

{y^2\over a}-1=y'^2
\end{displaymath} (23)

{dx\over dy}={1\over y'} = {a\over\sqrt{y^2-a^2}}
\end{displaymath} (24)

x=a\int {dy\over \sqrt{y^2-a^2}}= a\cosh^{-1}\left({y\over a}\right)+b
\end{displaymath} (25)

y=a\cosh\left({x-b\over a}\right),
\end{displaymath} (26)

which is called a Catenary, and the surface generated by rotating it is called a Catenoid. The two constants $a$ and $b$ are determined from the two implicit equations
$\displaystyle y_1$ $\textstyle =$ $\displaystyle a\cosh\left({x_1-b\over a}\right)$ (27)
$\displaystyle y_2$ $\textstyle =$ $\displaystyle a\cosh\left({x_2-b\over a}\right),$ (28)

which cannot be solved analytically.

\begin{figure}\begin{center}\BoxedEPSF{CatenoidXAxis.epsf scaled 800}\end{center}\end{figure}

The general case is somewhat more complicated than this solution suggests. To see this, consider the Minimal Surface between two rings of equal Radius $y_0$. Without loss of generality, take the origin at the midpoint of the two rings. Then the two endpoints are located at $(-x_0,y_0)$ and $(x_0,y_0)$, and

y_0 = a\cosh\left({-x_0-b\over a}\right)= a\cosh\left({x_0-b\over a}\right).
\end{displaymath} (29)

But $\cosh(-x)=\cosh(x)$, so
\cosh\left({-x_0-b\over a}\right)= \cosh\left({-x_0+b\over a}\right).
\end{displaymath} (30)

Inverting each side
\end{displaymath} (31)

so $b=0$ (as it must by symmetry, since we have chosen the origin between the two rings), and the equation of the Minimal Surface reduces to
y=a\cosh\left({x\over a}\right).
\end{displaymath} (32)

At the endpoints
y_0=a\cosh\left({x_0\over a}\right),
\end{displaymath} (33)

but for certain values of $x_0$ and $y_0$, this equation has no solutions. The physical interpretation of this fact is that the surface breaks and forms circular disks in each ring to minimize Area. Calculus of Variations cannot be used to find such discontinuous solutions (known in this case as Goldschmidt Solutions). The minimal surfaces for several choices of endpoints are shown above. The first two cases are Catenoids, while the third case is a Goldschmidt Solution.

To find the maximum value of $x_0/y_0$ at which Catenary solutions can be obtained, let $p\equiv{1/a}$. Then (31) gives

y_0 p=\cosh(p x_0).
\end{displaymath} (34)

Now, denote the maximum value of $x_0$ as $x_0^*$. Then it will be true that $dx_0/dp = 0$. Take $d/dp$ of (34),
y_0=\sinh(px_0)\left({x_0+p{dx_0\over dp}}\right).
\end{displaymath} (35)

Now set $dx_0/dp = 0$
y_0=x_0 \sinh(px_0^*).
\end{displaymath} (36)

From (34),
\end{displaymath} (37)

Take (37) $\div$ (36),
px_0^* = {\rm coth}(px_0^*).
\end{displaymath} (38)

Defining $u\equiv p{x_0}^*$,
u=\coth u.
\end{displaymath} (39)

This has solution $u = 1.1996789403\ldots$. From (36), $y_0p=\cosh u$. Divide this by (39) to obtain $y_0/x_0 = \sinh
u$, so the maximum possible value of $x_0/y_0$ is
{x_0\over y_0} =\mathop{\rm csch}\nolimits u= 0.6627434193\ldots.
\end{displaymath} (40)

Therefore, only Goldschmidt ring solutions exist for $x_0/y_0>0.6627\ldots$.

The Surface Area of the minimal Catenoid surface is given by

A=2(2\pi)\int_0^{x_0} y\sqrt{1+y'^2}\,dx,
\end{displaymath} (41)

but since
$\displaystyle y$ $\textstyle =$ $\displaystyle \sqrt{1+y'^2}\,a$ (42)
$\displaystyle y$ $\textstyle =$ $\displaystyle a\cosh\left({x\over a}\right),$ (43)

$\displaystyle A$ $\textstyle =$ $\displaystyle {4\pi\over a}\int_0^{x_0} y^2\,dx = 4\pi a\int_0^{x_0} \cosh^2\left({x\over a}\right)\,dx$  
  $\textstyle =$ $\displaystyle 4\pi a \int_0^{x_0} {\textstyle{1\over 2}}\left[{\cosh\left({2x\over a}\right)+1}\right]\, dx$  
  $\textstyle =$ $\displaystyle 2\pi a\left[{\int_0^{x_0}\cosh\left({2x\over a}\right)\, dx + \int_0^{x_0} dx}\right]$  
  $\textstyle =$ $\displaystyle 2\pi a\left[{{a\over 2}\sinh\left({2x\over a}\right)+ x}\right]_0^{x_0}$  
  $\textstyle =$ $\displaystyle \pi a^2\left[{\sinh\left({2x\over a}\right)+ {2x\over a}}\right]_0^{x_0}$  
  $\textstyle =$ $\displaystyle \pi a^2\left[{\sinh\left({2x_0\over a}\right)+ {2x_0\over a}}\right].$ (44)

\begin{figure}\begin{center}\BoxedEPSF{CatenoidSolution.epsf scaled 800}\end{center}\end{figure}

Some caution is needed in solving (33) for $a$. If we take $x_0={1/2}$ and $y_0=1$ then (33) becomes

1=a\cosh\left({1\over 2a}\right),
\end{displaymath} (45)

which has two solutions: $a_1=0.2350\ldots$ (``deep''), and $a_2=0.8483\ldots$ (``flat''). However, upon plugging these into (44) with $x_0={1/2}$, we find $A_1=6.8456\ldots$ and $A_2=5.9917\ldots$. So $A_1$ is not, in fact, a local minimum, and $A_2$ is the only true minimal solution.

The Surface Area of the Catenoid solution equals that of the Goldschmidt Solution when (44) equals the Area of two disks,

\pi a^2\left[{\sinh\left({2x_0\over a}\right)+{2x_0\over a}}\right]= 2\pi {y_0}^2
\end{displaymath} (46)

a^2\left[{2\sinh\left({x_0\over a}\right)\cosh\left({x_0\over a}\right)+{2x_0\over a}}\right]-2{y_0}^2=0
\end{displaymath} (47)

a^2\left[{\cosh\left({x_0\over a}\right)\sqrt{\cosh^2\left({x_0\over a}\right)-1}+{x_0\over a}}\right]-{y_0}^2=0.
\end{displaymath} (48)

Plugging in
{y_0\over a}=\cosh\left({x_0\over a}\right),
\end{displaymath} (49)

{y_0\over a}\sqrt{\left({y_0\over a}\right)^2-1}+\cosh^{-1}\left({y_0\over a}\right)-\left({y_0\over a}\right)^2= 0.
\end{displaymath} (50)

u\equiv {y_0\over a}
\end{displaymath} (51)

\end{displaymath} (52)

This has a solution $u=1.2113614259$. The value of $x_0/y_0$ for which
A_{\rm catenary} = A_{\rm 2\ disks}
\end{displaymath} (53)

is therefore
{x_0\over y_0} = {{x_0\over a}\over {y_0\over a}} = {\cosh^{...
{y_0\over a}} = {\cosh^{-1} u\over u} = 0.5276973967.
\end{displaymath} (54)

For $x_0/y_0 \in (0.52770, 0.6627)$, the Catenary solution has larger Area than the two disks, so it exists only as a Relative Minimum.

There also exist solutions with a disk (of radius $r$) between the rings supported by two Catenoids of revolution. The Area is larger than that for a simple Catenoid, but it is a Relative Minimum. The equation of the Positive half of this curve is

y=c_1\cosh\left({{x\over c_1}+c_3}\right).
\end{displaymath} (55)

At $(0,r)$,
\end{displaymath} (56)

At $(x_0,y_0)$,
y_0=c_1\cosh\left({{x_0\over c_1}+c_3}\right).
\end{displaymath} (57)

The Area of the two Catenoids is
$\displaystyle A_{\rm catenoids}$ $\textstyle =$ $\displaystyle 2(2\pi)\int_0^{x_0}y\sqrt{1+y'^2}\,dx
= {4\pi\over c_1} \int_0^{x_0} y^2\,dx$  
  $\textstyle =$ $\displaystyle 4\pi c_1\int_0^{x_0} \cosh^2\left({{x\over c_1}+c_3}\right)\, dx.$ (58)

Now let $u\equiv x/c_1+c_3$, so $du=dx/c_1$
$\displaystyle A$ $\textstyle =$ $\displaystyle 4\pi{c_1}^2\int_{c_3}^{x_0/x_1+c_3} \cosh^2 u\,du$  
  $\textstyle =$ $\displaystyle 4\pi{c_1}^2 {\textstyle{1\over 2}}\int_{c_3}^{x_0/x_1+c_3} [\cosh(2u)+1]\,du$  
  $\textstyle =$ $\displaystyle 2\pi{c_1}^2\left[{{\textstyle{1\over 2}}\sinh(2u)+u}\right]_{c_3}^{x_0/x_1+c_3}$  
  $\textstyle =$ $\displaystyle 2\pi{c_1}^2\left\{{{\textstyle{1\over 2}}\sinh\left[{2\left({{x_0...
...}+c_3}\right)}\right]-{\textstyle{1\over 2}}\sinh(2c_3)+{x_0\over c_1}}\right\}$  
  $\textstyle =$ $\displaystyle \pi{c_1}^2\left\{{\sinh\left[{2\left({{x_0\over c_1}+c_3}\right)}\right]-\sinh(2c_3)+{2x_0\over c_1}}\right\}.$  

The Area of the central Disk is
A_{\rm disk} =\pi r^2 = \pi{c_1}^2\cosh^2 c_3,
\end{displaymath} (60)

so the total Area is

A = \pi{c_1}^2\left\{{\sinh\left[{2\left({{x_0\over c_1}+c_3...
...eft[{\cosh^2 c_3-\sinh(2c_3)}\right]+{2x_0\over c_1}}\right\}.
\end{displaymath} (61)

By Plateau's Laws, the Catenoids meet at an Angle of 120°, so
$\displaystyle \tan 30^\circ$ $\textstyle =$ $\displaystyle \left[{dy\over dx}\right]_{x=0} = \left[{\sinh\left({{x\over c_1}+c_3}\right)}\right]_{x=0}$  
  $\textstyle =$ $\displaystyle \sinh c_3 = {1\over\sqrt{3}}$ (62)

\end{displaymath} (63)

This means that

$\displaystyle \cosh^2 c_3-\sinh(2c_3)$ $\textstyle =$ $\displaystyle [1+\sinh^2 c_3]-2\sinh c_3\sqrt{1+\sinh^2 c_3}$  
  $\textstyle =$ $\displaystyle (1+{\textstyle{1\over 3}})-2\left({1\over\sqrt{3}}\right)\sqrt{1+{\textstyle{1\over 3}}}$  
  $\textstyle =$ $\displaystyle {4\over 3}-{2\over\sqrt{3}} {2\over\sqrt{3}} = 0,$ (64)

A = \pi{c_1}^2\left\{{\sinh\left[{2\left({{x_0\over c_1}+c_3}\right)}\right]+{2x_0\over c_1}}\right\}.
\end{displaymath} (65)

Now examine $x_0/y_0$,
{x_0\over y_0} = {{x_0\over c_1}\over {y_0\over c_1}} = {{x_...
...\over c_1}+c_3}\right)} = u\mathop{\rm sech}\nolimits (u+c_3),
\end{displaymath} (66)

where $u\equiv x_0/c_1$. Finding the maximum ratio of $x_0/y_0$ gives
{d\over du}\left({x_0\over y_0}\right)= \mathop{\rm sech}\no...
...ts (u+c_3)-u\tanh(u+c_3)\mathop{\rm sech}\nolimits (u+c_3) = 0
\end{displaymath} (67)

\end{displaymath} (68)

with $c_3=\sinh^{-1}(1/\sqrt{3}\,)$ as given above. The solution is $u=1.0799632187$, so the maximum value of $x_0/y_0$ for two Catenoids with a central disk is $y_0=0.4078241702$.

If we are interested instead in finding the curve from a point $(x_1,y_1)$ to a point $(x_2,y_2)$ which, when revolved around the y-Axis (as opposed to the x-Axis), yields a surface of smallest Surface Area $A$, we proceed as above. Note that the solution is physically equivalent to that for rotation about the x-Axis, but takes on a different mathematical form. The Area element is

dA=2\pi x\, ds = 2\pi x\sqrt{1+y'^2}\,dx
\end{displaymath} (69)

A =2\pi \int x\sqrt{1+y'^2}\, dx,
\end{displaymath} (70)

and the quantity we are minimizing is
\end{displaymath} (71)

Taking the derivatives gives
$\displaystyle {\partial f\over \partial y}$ $\textstyle =$ $\displaystyle 0$ (72)
$\displaystyle {d\over dx}{\partial f\over\partial y'}$ $\textstyle =$ $\displaystyle {d\over dx}\left({xy'\over\sqrt{1+y'^2}}\right),$ (73)

so the Euler-Lagrange Differential Equation becomes
{\partial f\over \partial y}-{d\over dx}{\partial f\over\partial y'}= {d\over dx}\left({xy'\over\sqrt{1+y'^2}}\right)= 0.
\end{displaymath} (74)

{xy'\over\sqrt{1+y'^2}} = a
\end{displaymath} (75)

\end{displaymath} (76)

\end{displaymath} (77)

{dy\over dx}={a\over\sqrt{x^2-a^2}}
\end{displaymath} (78)

y=a\int {dx\over\sqrt{x^2-a^2}}+b = a\cosh^{-1}\left({x\over a}\right)+b.
\end{displaymath} (79)

Solving for $x$ then gives
x=a\cosh\left({y-b\over a}\right),
\end{displaymath} (80)

which is the equation for a Catenary. The Surface Area of the Catenoid product by rotation is

$\displaystyle A$ $\textstyle =$ $\displaystyle 2\pi\int x\sqrt{1+y'^2}\,dx = 2\pi\int x\sqrt{1+{a^2\over x^2-a^2}}\,dx$  
  $\textstyle =$ $\displaystyle 2\pi \int{x\over\sqrt{x^2-a^2}}\sqrt{(x^2-a^2)+a^2}\,dx$  
  $\textstyle =$ $\displaystyle 2\pi\int {x^2\, dx\over\sqrt{x^2-a^2}}$  
  $\textstyle =$ $\displaystyle \left[{{x\over 2}\sqrt{x^2-a^2}+{a^2\over 2}\ln\left({x+\sqrt{x^2-a^2}\,}\right)}\right]^{x_2}_{x_1}$  
  $\textstyle =$ $\displaystyle {1\over 2}\left[{x_2\sqrt{{x_2}^2-a^2}-x_1\sqrt{{x_1}^2-a^2}+a^2\ln\left({x_2+\sqrt{{x_2}^2-a^2}\over x_1+\sqrt{{x_1}^2-a^2}\,}\right)}\right].$ (81)

Isenberg (1992, p. 80) discusses finding the Minimal Surface passing through two rings with axes offset from each other.

See also Apple, Catenoid, Cone Conical Frustum, Cylinder, Darwin-de Sitter Spheroid, Eight Surface, Gabriel's Horn, Hyperboloid, Lemon, Meridian, Oblate Spheroid, Pappus's Centroid Theorem, Paraboloid, Parallel (Surface of Revolution), Prolate Spheroid, Pseudosphere, Sinclair's Soap Film Problem, Solid of Revolution, Sphere, Spheroid, Toroid, Torus


Arfken, G. Mathematical Methods for Physicists, 3rd ed. Orlando, FL: Academic Press, pp. 931-937, 1985.

Goldstein, H. Classical Mechanics, 2nd ed. Reading, MA: Addison-Wesley, p. 42, 1980.

Gray, A. ``Surfaces of Revolution.'' Ch. 18 in Modern Differential Geometry of Curves and Surfaces. Boca Raton, FL: CRC Press, pp. 357-375, 1993.

Isenberg, C. The Science of Soap Films and Soap Bubbles. New York: Dover, pp. 79-80 and Appendix III, 1992.

info prev up next book cdrom email home

© 1996-9 Eric W. Weisstein