Derivation of Formula for Surface Area of TorusDate: 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/ |
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