Center of Mass of an ice cream cone using spherical coordinates?

It suffices to find the center of mass of the region between

x^2 + y^2 + z^2 = 3^2 with z ≥ 0 and the cone x^2 + y^2 = z^2

(x^2 y^2 = z^2 must be a typo) with density δ(x,y,z) = k√(x^2 + y^2 + z^2).

Using spherical coordinates, note that

x^2 + y^2 + z^2 = 3^2 with z ≥ 0 ==> ρ = 3 with φ ≤ π/2

x^2 + y^2 = z^2 ==> ρ^2 sin φ = ρ^2 cos φ ==> φ = π/4

δ(x,y,z) = k√(x^2 + y^2 + z^2) = kρ

So, we have

m = ∫∫∫ δ(x,y,z) dV

....= ∫(θ = 0 to 2π) ∫(φ = 0 to π/4) ∫(ρ = 0 to 3) (kρ) * (ρ^2 sin φ dρ dφ dθ)

....= ∫(θ = 0 to 2π) dθ * ∫(φ = 0 to π/4) sin φ dφ * ∫(ρ = 0 to 3) kρ^3 dρ

....= 2π * (1 - √2/2) * 81k/4

....= (81πk/4) (2 - √2).

Myz = ∫∫∫ x * δ(x,y,z) dV

....= ∫(θ = 0 to 2π) ∫(φ = 0 to π/4) ∫(ρ = 0 to 3) (ρ cos θ sin φ) * (kρ) * (ρ^2 sin φ dρ dφ dθ)

....= ∫(θ = 0 to 2π) cos θ dθ * ∫(φ = 0 to π/4) ∫(ρ = 0 to 3) kρ^4 sin^2(φ) dρ dφ

....= 0.

Mxz = ∫∫∫ y * δ(x,y,z) dV

....= ∫(θ = 0 to 2π) ∫(φ = 0 to π/4) ∫(ρ = 0 to 3) (ρ cos φ) * (kρ) * (ρ^2 sin φ dρ dφ dθ)

....= ∫(θ = 0 to 2π) sin θ dθ * ∫(φ = 0 to π/4) ∫(ρ = 0 to 3) kρ^4 sin^2(φ) dρ dφ

....= 0.

Mxy = ∫∫∫ z * δ(x,y,z) dV

....= ∫(θ = 0 to 2π) ∫(φ = 0 to π/4) ∫(ρ = 0 to 3) (ρ sin θ sin φ) * (kρ) * (ρ^2 sin φ dρ dφ dθ)

....= ∫(θ = 0 to 2π) dθ * ∫(φ = 0 to π/4) sin φ cos φ dφ * ∫(ρ = 0 to 3) kρ^4 dρ

....= 2π * (1/2) sin^2(φ) {for φ = 0 to π/4} * 243k/5

....= 243πk/10

So, the center of mass equals (0, 0, (243πk/10)/((81πk/4) (2 - √2)))

= (0, 0, 6/(5(2 - √2))).

I hope this helps!

not a typical ice cream cone...rho in [ 0 , 3 ] , Θ in [ 0 , 2π ] , and Φ in [ 0 , π / 4]

assuming the cone is really √ ( x² + y² ) = z

cm = mass / volume..

mass is int of { a rho^3 sin Φ drho dΘ dΦ } , where a = proportionality constant

volume is int of { rho² sin Φ drho dΘ dΦ }