Volume of Spherical Cap
Date: 02/06/98 at 01:21:06 From: Sean Arkin Subject: Volume of spherical cap I am trying to find the volume of a cap of a sphere with radius of 5. The cap has a height of 3 - it is as if the top of the sphere, 3 meters from the top, was severed from the rest of the sphere. I am looking into a solution that involves taking upper and lower bounds of this cap's volume through cylinders which contain the cap, and then smaller and smaller cylinders which contain portions of this cap. Is this a good method? Could you help me with the solution - is there a general equation for such a "cap"? Any help appreciated, Sean
Date: 02/06/98 at 16:03:49 From: Doctor Rob Subject: Re: Volume of spherical cap Sean, Yes, there is a formula: V = (1/3)*Pi*h^2*(3*r-h), where r is the radius of the sphere, and h is the heighht of the cap. In your case, r = 5, h = 3, so the volume is 36*Pi cubic units. Your method will work, but it is somewhat tedious. The best method is to use the integral calculus to find this volume: V = Integral Pi*(r^2 - x^2) dx from x = r - h to x = r, = Pi*[r^2*x - x^3/3] evaluated from x = r - h to x = r, = Pi*[r^3 - r^3/3 - r^2*(r-h) + (r-h)^3/3], = Pi*[r^3 - r^3/3 - r^3 + r^2*h + r^3/3 - r^2*h + r*h^2 - h^3/3], = Pi*[r*h^2 - h^3/3], = (1/3)*Pi*h^2*(3*r-h). If you don't know integral calculus you must use a limit of a union of cylinders of height h/n and radius sqrt[r^2-(r-h+k*h/n)^2], with 0 <= k <= n-1, as a lower bound, and a union of cylinders of height h/n and radius sqrt[r^2-(r-h+[k+1]*h/n)^2, with 0 <= k <= n-1, as an upper bound. Since the difference between the upper and lower bounds goes to zero as n goes to infinity, they approach a common limit, V. To evaluate these sums in closed form, you will have to know that 0 + 1 + 2 + ... + (n-1) = n*(n-1)/2, 1 + 2 + 3 + ... + n = n*(n+1)/2, 0^2 + 1^2 + 2^2 + ... + (n-1)^2 = (n-1)*n*(2*n-1)/6, 1^2 + 2^2 + 3^2 + ... + n^2 = n*(n+1)*(2*n+1)/6. -Doctor Rob, The Math Forum Check out our web site! http://mathforum.org/dr.math/
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