Lateral Area of Oblique Cones

Date: 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