Answer:
the amount of metal is approximately 19.47 cm³
Explanation:
the volume enclosed by the cylinder is
V= π*R²*L
then using differentials
dV = ∂V/∂L *dL + ∂V/∂R *dR
dV = π*R²*dL + 2*π*R*L *dR
ΔV ≈ π*R²*ΔL + 2*π*R*L *ΔR
replacing values
ΔV ≈ π* (4 cm)²* (0.1 cm*2) + 2*π*R*30 cm *0.05 cm = 19.47 cm³
then the amount of metal ( in volume ) Vmetal is
V metal = V(R+dR, L+dL) - V(R, L) = ΔV ≈ 19.47 cm³