Lateral Area of Oblique ConesDate: 05/27/98 at 11:31:31 From: Chris Elbert Subject: Lateral area of oblique cones A colleague of mine asked if it was possible to find a formula for the lateral/surface areas of oblique cylinders and cones. I was able to derive a formula for oblique cylinders rather easily, but the oblique cone has eluded me. This is in part because if the base is circular, then cross-sections perpendicular to the base are elliptical, and also because the "slant height" doesn't change in a simple way as you move around the perimeter (so I have been unable to write a function I can intergrate). I have never seen this formula and my curiosity is now enormous to see what formula, if any, exists for this. Any insight you can give would be appreciated. Date: 06/01/98 at 12:10:04 From: Doctor Rob Subject: Re: Lateral area of oblique cones I set this up as follows. Let the center of the circular base be the point (0,0,0), the base be in the xy-plane, and of radius r. Let the vertex be the point (0,d,h), so that the height of the cone is h, and the slope of the axis of the cone is d/h. Parameterize the circle by: x = r*cos(t) y = r*sin(t) z = 0 Then at any point on the circle, the distance from the vertex is: sqrt(x^2+(y-d)^2+(z-h)^2) = sqrt([r*cos(t)]^2 + [r*sin(t)-d]^2 + h^2) The infinitesimal arc length is sin(A)*r*dt, where A is the angle made at a point on the circle between the tangent line and the line to the vertex. The direction vector of the point-to-vertex line is (-r*cos[t],d-r*sin[t],h), and that of the tangent line is (-sin[t],cos[t],0). The dot product of these vectors divided by their lengths will give you cos(A), and then you can find sin(A) from that. The area of the triangle with that height and base is half the distance times the arc length. We need to integrate that product with respect to t from 0 to 2*Pi, and, after some manipulation, we get: 2*Pi S = INTEGRAL sqrt([r-d*sin(t)]^2 + h^2)*r/2 dt. 0 This has the correct value S = Pi*r*s (where s is the slant-height) when d = 0 and the cone is right (because then A = Pi/2 for all points on the circle, so sin(A) = 1). Also, if d <= r and h approaches 0, this gives the correct value S = Pi*r^2. When I fed this to Mathematica(TM), it told me that this is an elliptic integral, so not expressible in closed form in terms of more familiar functions of calculus. No wonder you couldn't either derive the formula or write a function you could integrate! Perhaps the fact that this integral cannot be done in closed form is the reason nobody ever includes this case when discussing the lateral surface area of cones. -Doctor Rob, The Math Forum http://mathforum.org/dr.math/ Date: 03/25/2003 at 08:12:33 From: David W. Cantrell Subject: Lateral Area of Oblique Cones I am happy to report that I have finally managed to obtain a closed form, in terms of complete elliptic integrals, for the lateral surface area of an oblique circular cone. Letting s = Sqrt[(h^2+(r-d)^2)*(h^2+(r+d)^2)], parameter m = 1/2*(1 - (h^2+r^2-d^2)/s), and characteristic n = 1/2*(1 - (h^2+r^2+d^2)/s), the area is (#) S = 2*r*Sqrt[s]*(EllipticE[m]-EllipticK[m]+(1-n)*EllipticPi[n,m]) Note: To avoid confusion regarding notation of elliptic integrals, I have adopted the conventions used in Mathematica. These are documented at http://functions.wolfram.com/EllipticIntegrals/. Formula (#) works as desired except in the special case when h = 0 and d = r. However, if we consider (#) to be "extended by limit" in that case, then we do get pi*r^2 as required. Can any computer algebra system correctly evaluate the definite integral from t = 0 to t = 2*pi of Sqrt(h^2+(r-d*sin(t))^2)*r/2 dt? If so, I would be favorably impressed. Is anyone aware of (#), or perhaps some other closed form for the surface area, having been given before? I would be grateful for any references. Finally, for those wishing to implement (#), here are two computation checks: For h=5, r=3, d=2, the lateral surface area is 56.150873... For h=1, r=3, d=4, the lateral surface area is 33.857968... David Cantrell |
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