What are the conditions for the volume formula V = (H/6)(B + 4M + T) to work?

Question

If H = height, B = area of bottom face, M = area of cross section halfway up, and T = area of top face, then the volume formula V = (H/6)(B + 4M + T) works for a variety of solids, such as a cone, a box, a fustrum, a pyramid, even a sphere or ellipsoid. When does it work or does not work? What are the conditions necessary for it to work?

Dr D, yes, it's based on Simpson's rule, which is an APPROXIMATION. Now, how about those exactly correct volume formulas for certain solids?

Right, it won't work for all solids, even those solids of revolution. The question is, what are the necessary conditions for it to work?

Thanks, everybody, for your interesting answers to this problem. I'm still trying to find exceptions to the rule, where the cross section is not necessarily a cubic function of height.

Answer 1

(H/6)(B + 4M + T)
is the Simpson's rule approximation for the required integral, using just 3 points.

Simpson's rule assumes a quadratic fit, so it works if the area of any cross section has a linear, quadratic or cubic relationship with the height at that section.

It works for a hemisphere.

I haven't tested this, but for a figure described by:
ax^3 + by^3 + cz^3 = R^3
it will not work.

*EDIT*
I would say the condition is that the area at any section
A(z) = P(z^3)
It doesn't even have to be symmetric. Once the area can be expressed as a cubic function of height, then Simpson's approximation becomes EXACT, independent of the number of points. This is because Simpson's rule tries to approximate the function using a quadratic fit, so if the function is already cubic (or of lower order), then it works exactly.
Remember to use that formula, you must have H defined somehow. So set the z axis in that direction, and A(z) = P(z^3).
Note also that the formula only needs to work with one height axis. Consider a right cylinder. It works if the z axis is defined longitudinally. It does not work if the z axis is defined transversely.

For example:
x^2 + y^2 = z - 2z^2 + 3z^3 works
The volume is 13π/12 via integration as well as the formula.

x^2 + y^2 = z^4 does not work
Between z = 0 to 1, the exact volume is π/5, but the formula gives 5π/24

*EDIT*
Just to clarify some confusion in my own head, here is a direct quote from the site below:
"Since it uses quadratic polynomials to approximate functions, Simpson's rule actually gives exact results when approximating integrals of polynomials up to cubic degree."

Bottom line: Simpson's rule gives exact integrals for cubic functions. Hence the formula works as long as in any particular axis, the cross sectional area is a polynomial function (of order 3 or less) of the height at that section.

Answer 2

Considering an area generating function G(x), the error bounds for Simpson's rule is:

ε ≤ (1/90)h^5 G[4](x*);

where G[4] is the fourth derivative with respect to x of G(x), and x* gives the local maximum for the fourth derivative.

Therefore, on the desired interval, if the fourth derivative is *everywhere zero* then, obviously, Simpson's rule gives an exact value.

(Edit: in deference to Dr. D's comment, the fourth derivative of a cubic polynomial or lower is always zero......)

In mathematical terms, this is a "sufficient" condition, but not a "necessary" one. It's still possible for the *actual error* to be zero, even if the *error bounds* are not.

Of course finding the *actual* error, would probably require integrating to find the exact volume of the solid..........which totally defeats the purpose of using such a formula in the first place.

~W.O.M.B.A.T.

Answer 3

The formula will work anytime the cross-sectional area varies with the height as a parabolic function. This includes cones, pyramids, etc., which vary as A = h², and also spheres, hemispheres, etc., which vary as A = 1 - h². (A = area, h = height.) Are there others?

* * * * *

Thinking about this a little more, it makes sense that this application of Simpson's Rule gives the exact value when the relationship between the cross-sectional area and height follow a parabolic function, since Simpson's Rule basically looks for the parabola that best fits a function. Obviously, if the function is quadratic, then the best fit is the function itself.

I suppose that the cross-sections for prisms do not vary, i.e. follow the constant function A = c for some value c. Since this equation also works for prisms, we can call this a second class of solids for which the formula works. On the other hand, we can think of a constant function as a "flat parabola", i.e. A = 0h² + c, and argue that the quadratic relationship still holds.

Answer 4

I agree with Dr. D on this (cubic functions)

Cubic term vanishes due to the central position of 4M weight.
That's why Simpsons quadratures are so popular in numerical integration (extra accuracy for nothing).

Few interesting examples in addition to the cited above include volume of popular 'Earth with cylindrical hole through the center', and volume of tetrahedron through area of its quadrilater mid-section. Both of them are only quadratic, though.

My two cents.

Answer 5

whoever wrote the equation was trying to avoid imaginary solutions, ie, solutions where the square root turns out to be of a negative number, which cannot be done. The expression sqrt(1+x)(9-x) is imaginary when x<-1 or x>9, which can be expressed as (-1<=x<=9) the writer also wanted to avoid solutions where the area is negative, as negative area has no meaning. this can be done by avoiding x<0. combining the two inequalities, we get (0<=x<=9)