The word dome suggests that it is hollow. However, question 4b suggests using the previously calculated volume to calculate centre of mass would imply that the "dome" refers to a solid, or rather a hemisphere. The remaining calculations will be based on a solid hemisphere.
Actual "span" = 143 ft.
Radius, R = 143/2=71.5 ft.
A general point on the surface of the dome in cylindrical coordinates would be P(r, θ, z), where z is the vertical axis, and the r-θ plane corresponds to the x-y plane.
z = f(r) = √(R²-r²)
2. height of dome, H
H = f(0) = √(R²-0²)
The volume can be found by integrating thin slices over the height of the dome. In order to do so, we must express the radius of the dome as a function of z, i.e.
from z=f(r)=√(R²-r²), we obtain
(Side note: the inverse of f(r) equals f(z) by the fact that the first quadrant of the dome is symmetrical with respect to the line y=x)
=2πR³/3 [z from 0 to R]
4. Uniform density, ρ(r,θ,z)=1
If we subdivide the dome/hemisphere into thin horizontal slices, the moment about the x-y plane would be the mass of each slice multiplied by the distance from the plane.
=ρ ∫ z*π(R²-z²)dz
4b. Centre of mass, ρ=1
Centre of mass, z̄
The moment can again be calculated by multiplying the elemental disk by the distance from the x-y plane:
moment = ∫ ρz dV
5c. Centre of mass
7R/25 = 0.28R < 3R/8 = 0.375R
This checks since the density is higher near the bottom.
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