Volume of a Conical WedgeDate: 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 calculus. 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) T = 2*H*INTEGRAL [R^2/2-R^2/3] - [P^2/2/cos^2(t)-P^3/(3*R*cos^3[t])] dt, 0 T = 2*H*INTEGRAL R^2/6 - P^2*sec^2(t)/2 + P^3/(3*R)*sec^3(t) dt, 0 T T = 2*H*INTEGRAL R^2/6 dt - 2*H*INTEGRAL P^2/2*sec^2(t) dt + 0 0 T 2*H*INTEGRAL P^3/(3*R)*sec^3(t) dt, 0 T T = H*R^2/3*INTEGRAL dt - H*P^2*INTEGRAL sec^2(t) dt + 0 0 T 2*H*P^3/(3*R)*INTEGRAL sec^3(t) dt, 0 T T = H*R^2/3*[t] - H*P^2*[tan(t)] + 0 0 T 2*H*P^3/(3*R)*[(1/2)*sec(t)*tan(t)+(1/2)*ln(sec[t]+tan[t])] , 0 = H*R^2*(1/3*T-[P/R]^2*tan[T]+ [P/R]^3/3*[sec(T)*tan(T)+ln(sec[T]+tan[T])]), recalling that P/R = cos(T), V = H*R^2*(1/3*T-cos^2[T]*tan[T]+ cos^3[T]/3*[sec(T)*tan(T)+ln(sec[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 http://mathforum.org/dr.math/ |
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