# Volume and Surface Area of Torus (Doughnut)

TR Smith is a product designer and former teacher who uses math in her work every day.

A torus is one of the basic 3-dimensional curved shapes in geometry. Inner tubes of tires, doughnuts, and rings are all examples of tori (toruses) and torus-like shapes. Like a sphere, a torus has circular cross-sections when you cut it in the x, y, and z directions, but unlike a sphere a torus is characterized by a hole in the center. Another way to visualize a torus is to take a flexible cylinder and curl it so that the ends are fused.

There are several ways to calculate the volume and surface area of a torus depending on how you measure its dimensions. Imagine a doughnut with perfect circular cross-sections laying on a plate. There is the diameter of the doughnut's hole X, the diameter of the entire pastry Y, and the diameter of the vertical circular cross-sections H. Since the vertical cross-sections are circles, the thickness of the ring (the width between the hole and outer edge) is also H. Conveniently, the variables X, Y, and H are related by the expression 2H = Y - X. If you know any two of the three measurements you can find the volume and surface area of the torus.

Measurements of a torus with circular cross-sections.

## Torus Volume Formula

If the total diameter of the torus is Y and the diameter of the hole is X, then the volume equation in terms of X and Y is

V = (π^2)(Y^2 - X^2)(Y - X)/32

If you know the total diameter Y and the height H, then the volume can be computed with an equivalent formula

V = (π^2)(Y - H)(H^2)/4

And finally, if you know the hole diameter X and the height H, then the volume is given by another equivalent equation

V = (π^2)(X + H)(H^2)/4

Notice that the volume formulas have the constant π^2, rather than plain π. Another geometric formula that involves π^2 is the volume of a 4-dimensional hypersphere of radius R; the volume equation is V = (π^2)(R^4)/2.

A view from inside a torus.

## Torus Surface Area Formula

In terms of the measurements X and Y, the surface area of a torus is given by the equation

S = (π^2)(Y^2 - X^2)/8

In terms of Y and H, the surface area is

S = (π^2)(Y - H)(H/2)

And in terms of X and H, the toric surface area equation is

S = (π^2)(X + H)(H/2)

Like the toric volume equations, these surface area equations all contain the constant π^2. Another surface area formula that involves π^2 is that of the 4-dimensional hypersphere of radius R. Its surface area is given by S = 2(π^2)R^3.

## Example Problem

An inflatable inner tube is inflated so that its total diameter is 1.24 meters and its thickness is 0.33 meters. What volume of air does it contain and what is the surface area of the inner tube? For this problem we have Y = 1.24 and H = 0.33. Therefore the volume formula gives us

V = (π^2)(Y - H)(H^2)/4
= (π^2)(1.24 - 0.33)(0.33^2)/4
= 0.2445 cubic meters

The surface area formula gives us

S = (π^2)(Y - H)(H/2)
= (π^2)(1.24 - 0.33)(0.33/2)
= 1.4819 square meters

## Parametric Equations to Graph a Torus Surface

To graph the surface of a torus in 3-dimensiona xyz-coordinate space, you can use the parametric equations

x(u, v) = (K + A*cos(u))*cos(v)
y(u, v) = (K + A*cos(u))*sin(v)
z(u, v) = A*sin(u)

where K is the distance from the center of the hole to the center of the torus tube, and A is the radius of the torus tube. The parameters u and v range from 0 to 2π. This set of parametric equations produces a torus whose center lies at the origin of the coordinate space such that the z-axis passes through the hole without touching the torus.

In non-parametric coordinates, the upper half surface of a torus is given by the equation

z = f(x, y) = sqrt[ A^2 - K^2 + 2K*sqrt(x^2 + y^2) - x^2 - y^2 ]

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• FitnezzJim 2 years ago from Fredericksburg, Virginia

I noted you have a grid showing on the surface of the torus picture. It has axes that that run perpendicular and parallel with a radius from the center of the torus. The image makes for a nice Cartesian like grid that is squished on the inner radius X of the torus, and expanded on the outer radius Y.

What relationship is needed to have that grid skewed from its current orientation in such a way that the axes originating from the origin of the grid come back to the origin in a finite number of paths around the center of the torus, for both axes?

• Author

TR Smith 2 years ago from Eastern Europe

If I'm picturing your description correctly, what you'd need is a pair of toriodal spirals, one going in one direction, and the other going perpendicular to the first.

One set of curves you could use is parameterized by

x(t) = (4 + sin(20t))sin(t), y(t) = (4 + sin(20t))cos(t), z(t) = cos(20t)

and

x(t) = (4+sin(t))sin(20t), y(t) = (4+sin(t))cos(20t), z(t)=cos(t)

where the first curve is like looping string through a donut hole over and over, and the second is like winding string "parallel" to the hole.