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### Derivation of Formula for Surface Area of Torus

```Date: 05/05/2005 at 16:19:40
From: Shawn
Subject: surface area of a torus

Dear Dr. Math -

I saw the formula for the surface area of a torus on your web site,
and I'm wondering if you can show me how that formula was derived?

Thanks.

- Shawn

```

```
Date: 05/05/2005 at 18:03:13
From: Doctor Rick
Subject: Re: surface area of a torus

Hi, Shawn.

Let's consider a cross-section of the torus along its axis.  We see
two circles of radius r, their centers at (-R,0) and (R,0).  I will
take an element of arc length, ds = r*dtheta, on the right-hand circle
with theta defined as the angle measured at the center of the small
circle, from the positive x direction to the element under
consideration.

This arc element sweeps out an area equal to ds times the
circumference of the circle it sweeps out as the circle is rotated to
form the torus. The radius of that circle is R + r*cos(theta).  Thus
the circumference is 2*pi(R + r*cos(theta)), and the surface area
swept out is

dA = 2*pi(R + r*cos(theta))*r*dtheta

The surface area is the integral of dA for theta = 0 to 2*pi:

A = 2*pi*r*INTEGRAL((R+r*cos(theta))dtheta, 0, 2*pi)
= 2*pi*r(2*pi*R + r*INTEGRAL(cos(theta)dtheta, 0, 2*pi))
= 2*pi*r(2*pi*R + r*[sin(2*pi) - sin(0)])
= 2*pi*r*2*pi*R
= 4*pi^2*rR

How's that?  It's pretty simple after all, befitting a simple formula.

- Doctor Rick, The Math Forum
http://mathforum.org/dr.math/
```
Associated Topics:
College Higher-Dimensional Geometry

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