10.9. SPHERICAL COORDINATES IN p DIMENSIONS 255

It is always the case that ρ measures the distance from the point in Rp to the origin inRp, 0. Each φ i ∈ R and the transformations will be one to one if each φ i ∈ (0,π) , and

θ ∈ (0,2π) . Denote by hp

(ρ, φ⃗ ,θ

)the above transformation.

It can be shown using math induction and geometric reasoning that these coordinatesmap ∏

p−2i=1 (0,π)× (0,2π)× (0,∞) one to one onto an open subset of Rp which is ev-

erything except for the set of measure zero Ψp (N) where N results from having someφ i equal to 0 or π or for ρ = 0 or for θ equal to either 2π or 0. Each of these are setsof Lebesgue measure zero and so their union is also a set of measure zero. You can seethat hp

(∏

p−2i=1 (0,π)× (0,2π)× (0,∞)

)omits the union of the coordinate axes except for

maybe one of them. This is not important to the integral because it is just a set of measurezero.

Theorem 10.9.1 Let y =hp

(⃗φ ,θ ,ρ

)be the spherical coordinate transformations

in Rp. Then letting A = ∏p−2i=1 (0,π)× (0,2π) , it follows h maps A× (0,∞) one to one onto

all of Rp except a set of measure zero given by hp (N) where N is the set of measure zero(Ā× [0,∞)

)\ (A× (0,∞))

Also∣∣∣detDhp

(⃗φ ,θ ,ρ

)∣∣∣ will always be of the form∣∣∣detDhp

(⃗φ ,θ ,ρ

)∣∣∣= ρp−1

Φ

(⃗φ ,θ

). (10.9)

where Φ is a continuous function of φ⃗ and θ .2 Then if f is nonnegative and Lebesguemeasurable,∫

Rpf (y)dmp =

∫hp(A)

f (y)dmp =∫

Af(hp

(⃗φ ,θ ,ρ

))ρ

p−1Φ

(⃗φ ,θ

)dmp (10.10)

Furthermore whenever f is Borel measurable and nonnegative, one can apply Fubini’stheorem and write∫

Rpf (y)dy =

∫∞

0ρ

p−1∫

Af(h(⃗

φ ,θ ,ρ))

Φ

(⃗φ ,θ

)dφ⃗dθdρ (10.11)

where here dφ⃗dθ denotes dmp−1 on A. The same formulas hold if f ∈ L1 (Rp) .

Proof: Formula 10.9 is obvious from the definition of the spherical coordinates becausein the matrix of the derivative, there will be a ρ in p− 1 columns. The first claim is alsoclear from the definition and math induction or from the geometry of the above description.It remains to verify 10.10 and 10.11. It is clear hp maps Ā× [0,∞) onto Rp. Since hp isdifferentiable, it maps sets of measure zero to sets of measure zero. Then

Rp = hp (N∪A× (0,∞)) = hp (N)∪hp (A× (0,∞)) ,

the union of a set of measure zero with hp (A× (0,∞)) . Therefore, from the change ofvariables formula,∫

Rpf (y)dmp =

∫hp(A×(0,∞))

f (y)dmp

=∫

A×(0,∞)f(hp

(⃗φ ,θ ,ρ

))ρ

p−1Φ

(⃗φ ,θ

)dmp

2Actually it is only a function of the first but this is not important in what follows.