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Volume of a Conical Wedge

Date: 02/19/2004 at 08:00:04
From: Kara
Subject: Volume of a conical wedge

What is the formula for the volume of the shape that remains when a 
cone is sliced parallel to the height of the cone, perpendicular to 
the base, not on center, and when the chord created by the slice on
the base circle is less than the diameter?

Date: 02/19/2004 at 12:36:12
From: Doctor Rob
Subject: Re: Volume of a conical wedge

Thanks for writing to Ask Dr. Math, Kara!

I don't know a formula, but you can find one by using integral 

Let the cone have height H and radius of the base R.  Let the chord 
have length C, and be at a distance of P = sqrt(R^2 - C^2/4) from the 
center of the base.  Let T = arcsin([C/R]/2) be half the central angle 
subtended by the chord.  Then sin(T) = (C/R)/2, cos(T) = P/R, and 
tan(T) = (C/P)/2.

Set up a cylindrical coordinate system (r,t,z) with the apex of the
cone at the origin, and the axis along the z-axis.  Then the equation
of the surface of the cone is

   r^2 = (z*R/H)^2,
     z = +/- H*r/R,

and if r >= 0 and we're looking at the part of the cone with z <= 0,
then the operative equation is

   z = -H*r/R.

The equation of the plane of the base is z = -H.  Let the equation of 
the slice be r = P/cos(t).  (Here we're assuming that the slice is 
less than half the cone, so T < Pi/2.)  Then the volume is given by 
the triple integral

            T        R               -H*r/R
V = INTEGRAL INTEGRAL        INTEGRAL      dz r dr dt.
           -T        P/cos(t)        -H

By using the symmetry of the wedge about the plane t = 0, we can write

              T        R               -H*r/R
V = 2*INTEGRAL INTEGRAL        INTEGRAL      dz r dr dt,
              0        P/cos(t)        -H

              T        R          z=-H*r/R
  = 2*INTEGRAL INTEGRAL        [z]        *r dr dt,
              0        P/cos(t)   z=-H

              T        R
  = 2*INTEGRAL INTEGRAL        [-H*r/R -(-H)]*r dr dt
              0        P/cos(t)

                T        R 
  = 2*H*INTEGRAL INTEGRAL        r - r^2/R dr dt,
                0        P/cos(t)

                T                  r=R 
  = 2*H*INTEGRAL  [r^2/2-r^3/(3*R)]          dt,
                0                  r=P/cos(t)

  = 2*H*INTEGRAL [R^2/2-R^2/3] - [P^2/2/cos^2(t)-P^3/(3*R*cos^3[t])] 

  = 2*H*INTEGRAL R^2/6 - P^2*sec^2(t)/2 + P^3/(3*R)*sec^3(t) dt,

                T                       T
  = 2*H*INTEGRAL R^2/6 dt - 2*H*INTEGRAL P^2/2*sec^2(t) dt +
                0                       0

      2*H*INTEGRAL P^3/(3*R)*sec^3(t) dt,

                    T                   T
  = H*R^2/3*INTEGRAL dt - H*P^2*INTEGRAL sec^2(t) dt +
                    0                   0

      2*H*P^3/(3*R)*INTEGRAL sec^3(t) dt,

               T                T
  = H*R^2/3*[t] - H*P^2*[tan(t)] +
               0                0

      2*H*P^3/(3*R)*[(1/2)*sec(t)*tan(t)+(1/2)*ln(sec[t]+tan[t])] ,

  = H*R^2*(1/3*T-[P/R]^2*tan[T]+

recalling that P/R = cos(T),

V = H*R^2*(1/3*T-cos^2[T]*tan[T]+

V = (1/3)*H*R^2*(T - 2*cos[T]*sin[T] + cos^3[T]*ln[sec(T)+tan(T)]).

As T approaches Pi/2, this approaches Pi*R^2*H/6, the volume of half a
cone, which is correct.  If T = 0, V = 0, also a correct value.

You can rewrite this in terms of either C or P instead of T using
C = 2*sqrt(P^2 - R^2), P = sqrt(R^2 - C^2/4), sin(T) = (C/R)/2,
cos(T) = P/R, tan(T) = (C/P)/2, and sec(T) = R/P.

You should check the above computation for errors.  I have been known 
to make a few in such lengthy derivations!

If the wedge you want is MORE than half a cone, compute the volume of 
the whole cone, (1/3)*Pi*R^2*H, and subtract the volume of the OTHER 
wedge cut off by the plane, which is less than half a cone.  That you 
can compute using the above formula.

Feel free to reply if I can help further with this question.

- Doctor Rob, The Math Forum 
Associated Topics:
College Calculus
College Conic Sections/Circles

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