Finding center of mass of surface of sphere contained within cone.

[s3a]

Homework Statement

Problem (also attached as TheProblem.jpg):
Find the center of mass of the surface of the sphere x^2 + y^2 + z^2 = a^2 contained within the cone z tanγ = sqrt(x^2 + y^2), 0 < γ < π/2 a constant, if the density is proportional to the distance from the z axis.

Hint: R_cm = ∫∫_S δR dS / ∫∫_S δ dS, where δ is the density.

Solution:
The solution is attached as TheSolution.jpg.

Homework Equations

Spherical coordinates:
x = p sinϕ cosθ, y = p sinϕ sinθ, z = pcosθ
∫∫_E∫ f(x,y,z) dV = ∫∫_E∫ f(p sinϕ cosθ, p sinϕ sinθ, p cosθ) p^2 sinϕ dp dθ dϕ

The Attempt at a Solution

The first thing I'm stuck on is knowing what δ and dS are. How do I determine what those are (without “cheating” and looking at the solution)?

Any input would be greatly appreciated!

 

[LCKurtz]

Homework Statement

Problem (also attached as TheProblem.jpg):
Find the center of mass of the surface of the sphere x^2 + y^2 + z^2 = a^2 contained within the cone z tanγ = sqrt(x^2 + y^2), 0 < γ < π/2 a constant, if the density is proportional to the distance from the z axis.

Hint: R_cm = ∫∫_S δR dS / ∫∫_S δ dS, where δ is the density.

Solution:
The solution is attached as TheSolution.jpg.

Homework Equations

Spherical coordinates:
x = p sinϕ cosθ, y = p sinϕ sinθ, z = pcosθ
∫∫_E∫ f(x,y,z) dV = ∫∫_E∫ f(p sinϕ cosθ, p sinϕ sinθ, p cosθ) p^2 sinϕ dp dθ dϕ

That isn't a relevant equation. Surface integrals are double integrals, not triple integrals. The radius of the sphere given as $a$.

The Attempt at a Solution

The first thing I'm stuck on is knowing what δ and dS are. How do I determine what those are (without “cheating” and looking at the solution)?

Surely your text gives $dS$ for spherical coordinates. As for $\delta$, what is the point on the $z$ axis nearest to $(x,y,z)$? What is the distance from that point to $(x,y,z)$? You could answer that in rectangular coordinates and change it to spherical, or write it directly from the figure in spherical coordinates as the solution does.