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The sphere in dimensions 4, 5, …

The volume of a sphere of radius R in 2 dimensions is just the area of a circle which is π R2. (The symbol π is Pi which is 3.1415….)

The volume of a sphere of radius R in 3 dimensions is (4/3) π R3.

The volume of a sphere of radius R in 4 dimensions is (1/2) π2 R4.

Hold everything! How’d you get that? With a little calculus!

Ok, you see a sort of pattern here. The volume of a sphere in n dimensions has a nice form: it is some constant C(n) (involving π and some fractions) times the radius R raised to the dimension n:

V(n,R) = C(n) Rn —————– (1)

where I wrote V(n,R) for the volume of a sphere in n dimensions of radius R (as a function of these two variables).

Now, how do you get the constants C(n)?

With a bit of calculus and some ‘telescoping’ as we say in Math.

(In the cases n = 2 and n = 3, we can see that C(2) = π, and C(3) = 4π/3.)

First, let’s do the calculus. The volume of a sphere in n dimensions can be obtained by integrating the volume of a sphere in n-1 dimensions like this using the form (1):

V(n,R) = ∫ V(n-1,r) dz = ∫ C(n-1) rn-1 dz

where here r2 + z2 = R2 and your integral goes from -R to R (or you can take 2 times the integral from 0 to R). You solve the latter for r, plug it into the last integral, and compute it using the trig substitution z = R sin θ. When you do, you get

V(n,R) = 2C(n-1) Rn ∫ cosnθ dθ.

The cosine integral here goes from 0 to π/2, and it can be expressed in terms of the Gamma function Γ, so it becomes

V(n,R) = 2C(n-1) Rn π1/2 Γ((n+1)/2) / Γ((n+2)/2).

Now compare this with the form for V in equation (1), you see that the R’s cancel and you have C(n) expressed in terms of C(n-1). After telescoping and simplifying you eventually get the volume of a sphere in n dimensions to be:

V(n,R) = πn/2 Rn / Γ((n+2)/2).

Now plug in n equals 4 dimensions and you have what we said above. (Note: Γ(3) = 2 — in fact, for positive integers N the Gamma function has simple values given by Γ(N) = (N-1)!, using the factorial notation.)

How about the volume of a sphere in 5 dimensions?  It works out to

V(5,R) = (8/15) π2 R5.

One last thing that’s neat: if you take the derivative of the Volume with respect to the radius R, you get its Surface Area!  How crazy is that?!!

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  1. August 1, 2013 at 2:52 pm

    There is certainly a great deal to know about this subject.
    I really like all the points you have made.

  2. September 12, 2013 at 8:20 pm

    Tiffany ネックレス ペンダント

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