Penteract (I would prefer to consider the irregular case penteractoid or penteroid or whatever else it may be called as the regular one is a special case of this and can always be inferred from this)

Surteron bulk: 2(lbht + lbhw + lbtw + lhtw + bhtw)

Pentavolume: lbhtw

Tesserinder

Surteron bulk: 2πr(rab + rac + rbc + abc)

Pentavolume: πr

^{2}abc

Duocyldyinder

Surteron bulk: 2π

^{2}ab{ab + (a + b)h}

Pentavolume: π

^{2}a

^{2}b

^{2}h (where a and b are the two radii)

Cubspherinder

Surteron bulk: 4πr

^{2}{ab + (2/3)r(a + b)}

Pentavolume:(4/3)πr

^{3}ab

Cylspherinder

Surteron bulk: 4π

^{2}ab

^{2}{a + (2/3)b} (I'm not sure whether I computed this right, though most probably I did)

Pentavolume: (4/3)π

^{2}a

^{2}b

^{3}(where a is the radius of the circular portion and b is the radius of the spherical portion)

Glominder

Surteron bulk: π

^{2}r

^{3}(2h + r)

Pentavolume: (1/2)π

^{2}r

^{4}h

Pentasphere

Surteron bulk: (8/3)π

^{2}r

^{4}

Pentavolume: (8/15)π

^{2}r

^{5}

As for the pentasphere, this is what is given in the wiki:

The hypervolumes of a pentasphere are given by:

total edge length = 0

total surface area = 0

total surcell volume = 0

surteron bulk = π^{2}∕2 · r^{4}

pentavolume = π^{2}∕8 · r^{5}

How could this be? This surteron bulk is not even the derivative of the pentavolume w.r.t. the radius. Unless I am very much mistaken, this is true of any number of dimensions and, in addition, we have these methods (quoted from viewtopic.php?f=25&t=1891):

Klitzing wrote:You either can calculate the volume of the unit hyperball recursively:

or explicitely, but separate for even and odd numbers, by means of

- Code: Select all
`V_(2k) = pi^k / k!`

V_(2k+1) = 2 . k! . (4 pi)^k / (2k+1)!

--- rk

Whew! That finishes all the rotatera!