sci.math FAQ: Surface Area of Sphere

There are reader questions on this topic!
Help others by sharing your knowledge

Newsgroups: sci.math
From: alopez-o@neumann.uwaterloo.ca (Alex Lopez-Ortiz)
Subject: sci.math FAQ: Surface Area of Sphere
Summary: Part 35 of many, New version,
Message-ID: <DI76nI.EC1@undergrad.math.uwaterloo.ca>
Sender: news@undergrad.math.uwaterloo.ca (news spool owner)
Date: Fri, 17 Nov 1995 17:16:30 GMT
Reply-To: alopez-o@neumann.uwaterloo.ca

Archive-Name: sci-math-faq/surfaceSphere
Last-modified: December 8, 1994
Version: 6.2

Formula for the Surface Area of a sphere in Euclidean N -Space

   This is equivalent to the volume of the N -1 solid which comprises the
   boundary of an N -Sphere.

   The volume of a ball is the easiest formula to remember: It's r^N
   (pi^(N/2))/((N/2)!) . The only hard part is taking the factorial of a
   half-integer. The real definition is that x! = Gamma (x + 1) , but if
   you want a formula, it's:

   (1/2 + n)! = sqrt(pi) ((2n + 2)!)/((n + 1)!4^(n + 1)) To get the
   surface area, you just differentiate to get N (pi^(N/2))/((N/2)!)r^(N
   - 1) .

   There is a clever way to obtain this formula using Gaussian integrals.
   First, we note that the integral over the line of e^(-x^2) is sqrt(pi)
   . Therefore the integral over N -space of e^(-x_1^2 - x_2^2 - ... -
   x_N^2) is sqrt(pi)^n . Now we change to spherical coordinates. We get
   the integral from 0 to infinity of Vr^(N - 1)e^(-r^2) , where V is the
   surface volume of a sphere. Integrate by parts repeatedly to get the
   desired formula.

   It is possible to derive the volume of the sphere from ``first
   principles''.

     _________________________________________________________________

    alopez-o@barrow.uwaterloo.ca
    Tue Apr 04 17:26:57 EDT 1995