A gold lamina in the shape of a bowl is denser at the bottom than top. It lies along the upperhemisphere x? + y² + z? = 1,z > 0. Calculate the mass of gold needed for the bowl as asurface given that the density at a point (x,y, z) on the surface of the hemisphere is8(х, у, 2)lamina in the shape of a bowl.= (2 – z)K kg/mm³ for constant K. Calculate the average density of this gold

The shape of the gold lamina is given by, x2+y2+z2=1, z≥0 The density at a point (x,y,z) is given by, δx,y,z=2−zK kg/mm3The mass of gold needed for the bowl can be calculated using the below formula. Mass=∫∫∫Rδx,y,zdV Here R is the region given by the gold lamina.Express the region and density in spherical coordinates. δx,y,z=2−zKSubstitute z=ρcosϕ in aboveδx,y,z=2−ρcosϕK and x2+y2+z2=1ρ2=1ρ=1So, ρ=0 to ρ=1 z≥0⇒ϕ=0 to ϕ=π/2 Thus, R=ρ,θ,ϕ|0≤ρ≤1, 0≤θ≤2π, 0≤ϕ≤π/2Calculate the mass as shown below. Mass=∫∫∫Rδx,y,zdVMass=∫∫∫R2−ρcosϕKρ2sinϕdρdθdϕMass=K∫0π/2∫02π∫012−ρcosϕρ2sinϕdρdθdϕMass=K∫0π/2∫02π∫012sinϕρ2−cosϕsinϕρ3dρdθdϕ Mass=K∫0π/2∫02π2sinϕρ33−cosϕsinϕρ4401dθdϕMass=K∫0π/223sinϕ−14cosϕsinϕdϕ∫02πdθMass=K∫0π/223sinϕ−14cosϕsinϕdϕ×2π−0Mass=2πK−23cosϕ−18sin2ϕ0π/2Mass=2πK−23cosπ/2−18sin2π/2−−23cos0−18sin20Mass=2πK−0−18−−23−0Mass=2πK×1324Mass=13π12K This is the required mass of gold needed for the bowl.Since the bowl is a hemisphere of radius 1, therefore, the volume of the bowl is given by, Volume=12×43πRadius3Volume=12×43π13Volume=23π Divide the mass by volume to calculate the average density. Average density=MassVolumeAverage density=13π12K23πAverage density=13π12K×32πAverage density=138K kg/mm3The mass of the gold needed is 13π12K kg. Average density=138K kg/mm3