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 Math Help > Calculus > Integral > 5D Volume of a 5-Sphere

A student came to me with this problem:

Compute the volume of a 5-dimensional ball of radius 1. This is the region {(a, b, c, d, e) : a2 +b2 +c2 +d2 +e2 1}. [Hint: For each point (a0, b0, c0) inside the 3-dimensional ball, show that the intersection of the 5-dimensional ball with the plane a = a0, b = b0, c = c0 is a circle.  Integrate the area of this circle over all points inside the 3-dimensional ball].

I'll admit to trying this with rectangular coordinates to begin with...

 ∫ 1 ∫ sqrt(1-c2) ∫ sqrt(1-b2-c2) π(1-a2-b2-c2)da db dc -1 -sqrt(1-c2) -sqrt(1-b2-c2)

and this worked fine for the inner integral.  But then for the second integral, I got

(4 π (-(b sqrt(1 - b2 - c2) (-5 + 2 b2 + 5 c2))/8 + (3 (-1 + c2)2 arctan(b/sqrt(1 - b2 - c2)))/8))/3

and that's when my student suggested switching to spherical coordinates.

### Spherical coordinates

Imagine dividing the space not into little cubes measuring dx dy dz, as in rectangular coordinates, but rather into little patches of the surface of concentric spheres.  To help you imagine this, use the earth as an example, with its familiar longitude and latitude.  In spherical coordinates, longitude is θ (theta), and latitude is φ (phi), except instead of going from -π/2 to π/2, the latitude of our sphere is measured from 0 to π.  And, finally, the distance of our little patch from the origin is called ρ (rho). (note: this convention is not universal; see the Wikipedia article, Spherical coordinate system for a complete explanation.)

If f(ρ,φ,θ) represents some sort of density function within a sphere, and we want to integrate that function to find the total mass of the object, then the appropriate integral using spherical coordinates is

 ∫ 2π ∫ π ∫ 1 f(ρ,φ,θ) ρ2 sin φ dρ dφ dθ 0 0 0

The order of integration can be interchanged, if it helps; however this order seems to work best, with the inner integral being the distance from the center, then the latitude, then the longitude.  The "size" of each little patch of space is calculated this way:

the thickness, top to bottom, is dρ;
the length, North to South, is ρ dφ
and the width, East to West, is ρ sin φ dθ

Of these dimensions, perhaps the width is hardest to grasp...  Think of the circles on the globe that represent a particular latitude.  The circles nearer the poles are much smaller, so the actual width of a patch of space whose angular measure is dθ on one of these circles is less than that of a patch of space nearer the equator.  The factor that accounts for the varying circumference of the circles of different latitudes is sin φ.  Hence the volume of each patch of space is ρ2 sin φ dρ dφ dθ.

### Volume of the 5-sphere

Using the hint in the problem, consider each point of a solid sphere in 3 dimensions as representative of a plane in 5-space, with all the planes being parallel to one another.  That is, consider the point within the sphere as "fixing" the first three coordinates, with the other two allowed to vary freely.  Then, the intersection of the 5-sphere with one of the planes represented by a point ρ units from the center of the 3-sphere is a circle whose radius is 1-ρ.  The reason this is true (if you don't immediately grasp it) is that the sum of the squares of all five dimensions must be less than or equal to 1.  ρ2 is the sum of the squares of three of the dimensions, and the circle of radius 1-ρ accounts for the squares of the two "hyperdimensions".

So then the integral we need to solve is

 ∫ 2π ∫ π ∫ 1 π(1-ρ2) ρ2 sin φ dρ dφ dθ 0 0 0

The inner integral is (ρ^3 π sin(φ))/3 - (ρ^5 π sin(φ))/5.  Evaluating this from 0 to 1 gives us (2/15) π sin(φ).

the integral now becomes

 ∫ 2π ∫ π (2/15) π sin(φ) dφ dθ 0 0

Now, the integral of (2/15) π sin(φ) is simply (-2/15) π cos(φ), which evaluated from 0 to pi is (4/15) π, making our integral now

 ∫ 2π (4/15) π dθ 0

which evaluates to (8/15) π2

### Internet references

Wikipedia: Spherical coordinate system, and N-sphere

Mathworld: Hypersphere

### Related pages in this website

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