Once you have the picture right and realize that the edge of the hemisphere is z^{2}+x^{2}+y^{2} = 4, or ρ^{2} = 4, the bounds of the integral

triple integral of ρ^{2}sinφ dρdθdφ becomes a matter of finding the correct φ lower limit and lower ρ limit because θ goes all the way around (0 to 2π) and ρ has an upper limit of 2 and φ has an upper limit of π/2 (I've chosen φ as the angle that z=ρcosφ.in case your text switches theta and phi)

The sides of the cone limit φ: tan(π/2-φ) = sqrt(x^{2}+y^{2})/z from the geometry = 1/sqrt(3) from the cone equation.

The limiting lower value of ρ depends on φ and how it cuts the z=1 plane which is ρcosφ = 1 or

ρ = 1/cosφ, the lower limit for ρ.

When you do the integral you want to integrate ρ before φ. Integrating Theta will just end up multiplying by 2π.

Good Luck!π