4-D volume of a 4-D cone (or pyramid) and 4-D sphere?

Question

A triangle is formed when you extend the edges of a line segment to a point outside the line containing the line segment. The area is bh/2, where b is the length of the line and h is the perpendicular distance.

Similarly, a cone, pyramid, etc. is formed when you extend the edges of a circle, polygon, etc. to a point outside the plane of the figure. The volume is ah/3, where a is the area and h is the perpendicular distance.

Does this pattern continue to higher dimensions? For example, if you extend a 3-D solid to a point outside the solid's 3-D space, is the "hyper-volume" vh/4, where v is the volume of the solid and h the perpendicular distance?

Also, what about circles, spheres, etc? Letting d = diameter, the area of a circle is πd²/4. The volume of a sphere is πd³/6. Does this pattern continue, i.e. is the hypervolume of the 4-D sphere πd^4/8?

Answer 1

You are ALMOST right about this.

For the cases of cone & pyramid, you are right: that's because the extension into the next dimension is done by linear scaling, from 0 to maximum. Since the lower-dimensional "volume" scales like distance^n, when you integrate it, you will get something like (distance^(n+1))/(n+1) for the (n+1)-dimensional volume.

Where you make a mistake for the sphere is that the scaling up to go to a higher-dimensional sphere is not linear, but curvilinear. When you go from a circle to a sphere, the radii of the sequential slices do not grow linearly, or you would end up with a cone. Instead, the thickness of each slice determines how far up it has gone, and thus what its area should be.

So let's define:
V(n, R) = n-dimensional volume of an n-sphere of radius R.
We further define:
a(n) = V(n, R)/(R^n)
So
V(n, R) = a(n)*R^n

How do we go to an (n+1)-sphere?

V(n+1, R) = 2 * integral (V(n, sqrt(R^2 - z^2) dz {z = 0, R}

In other words, as I increase z, the radius of the n-dimensional "slices" has to be reduced so that the total radius is unchanged. Thus,
V(n+1, R) = 2 * integral (a(n) (R^2 - z^2)^(n/2) dz {z = 0, R}
= 2*a(n)*R^(n+1) integral( (1 - x^2)^(n/2) dx {x = 0, 1}
= 2*a(n)*R^(n+1) integral( cos^n(b) d(sin(b))
= 2*a(n)*R^(n+1) integral(cos^(n+1)(b) da {b = 0, pi/2}

Therefore, since V(n+1, R) = a(n+1) * R^(n+1)
a(n+1) = 2*a(n)*integral(cos^(n+1)(b) db {b = 0, pi/2}

Using this formula, we start with :
a(1) = 2
and get:
a(2) = (pi/2)*a(1) = pi
and
a(3) = (4/3)*a(2) = (4pi)/3

So I'm sure that if you work on the integral above, you can easily verify the explicit formula given in the wiki article on n-spheres.
(The only thing you need to know about the gamma function for this issue is that gamma(1/2) = sqrt(pi)/2.)

Answer 2

Assuming the pyramid must be a regular square pyramid, The largest square pyramid inside a cube with side length d, will have height f = d and base side e = d (such that it shares a cube's side as its base), and its apex at the center of the opposite cube side, such that its volume V[pyramid] = (1/3) e^2 f = d^3 / 3 = one-third the volume V[cube] of the cube. Thus, V[pyramid] is maximized as V[cube] is maximized. A cube inside a cylinder with radius b and height c, will have volume V[cube] such that: V[cube] = min(2b, c)^3 *** Eq. 1 Since c is inversely proportional Assuming both caps of the cylinder are such that their circular perimeters are parallel "small circles" of the sphere, (that is, for the cylinders maximum radius b[max] = 1, c = 0; and likewise, for the cylinder's maximum height c[max] = 2, b = 0). Let point o be the center of the sphere, and Let point p be on the perimeter of a circular cap. Let oq be a radius of the cylinder such that pq is perpendicular to the circular cap. Then, op = the spherical radius a = 1, pq = half the cylinder's height = c/2, and oq = the cylinder's radius b opq is a right triangle with hypotenuse op. By the Pythagorean Theorem, c^2 / 4 + b^2 = 1 => c = 2 sqrt(1 - b^2) *** Eq. 2 Since 0 < b < 1 and 0 < c < 2, we need only consider the positive branches of the sqrts. Now, b is inversely proportional to c, and both are monotonic, meaning that (by Eq. 1), V[cube] is maximized where b = c. So our Pythagorean equation can be simplified to: b^2 / 4 + b^2 = 1 => b = 2 sqrt(5) / 5 And, so V[pyramid, max] = V[cube, max] / 3 = b^3 / 3 = (2 sqrt(5) / 5)^3 / 3 = 8 sqrt(5) / 75 *** SOLUTION ≈ 5.7% of the volume of the sphere --- Since the inradius of a unit cube is 1/2 and a circumradius is sqrt(3)/2, I would imagine packing the sphere in the cube would maximize space utilization. The most awkward volume to fill seems to be the pyramid. So I would guess a maximal order would go like this (from innermost to outermost): pyramid, cylinder, cube, sphere I would guess that a minimal order would be cylinder, cube, sphere, pyramid

Answer 3

The formula for the hypervolume of a generalized cone does extend in the way you are guessing. However, the hypervolumes of spheres are trickier. For example, the volume of a 4D sphere is pi^2 r^4 /2 . Every other dimension adds a power of pi and the coefficients do not form a simple sequence either.

Answer 4

The volume of an n-dimensional hypersphere is

V = pi^(n/2) * R^n / Gamma[n/2 + 1]

where n is the number of dimensions, R is the radius, and Gamma[] is Euler's gamma function, which connects the factorial function in a smooth curve.

Gamma[x] = (x-1)!
when x is an integer. When it's not
Gamma[x] = Integral[ t^(x-1)*exp(-t), t, 0, infinity ]

Answer 5

Apparently it's not quite that simple. Lithiumdeuteride is right. See link.