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/