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Calc 3

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The actual span of the base of the dome is 143 feet.
1) Use cylindrical coordinates to write the surface of the dome as a function of the
distance from the center of the base; that is find z = f (r) .
2) Use your function to find the height of the dome; that is what is f (0) ?
3) Find the volume of the inside of the dome.
4) Assume the density is one, that is
P(r,"theta",z)=1 :
a. Find the moment around the x-y plane.
b. Use 3) and a) above to find the center of mass for the dome.
5) The density of the sides decreases as the dome gets higher. Assume that the
density is P(r,"theta",z)=k∗(f(0)−z) where k is a constant.
a. Find the mass of the dome.
b. Find the moment around the x-y plane.
c. Find the center of mass of the dome.

  • Calc 3 - ,

    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)

    3. Volume
    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
    a. Moment
    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.
    =∫ ρzdV
    =∫ ρz*πr²dz
    =ρ ∫ z*π(R²-z²)dz

    4b. Centre of mass, ρ=1
    Centre of mass, z̄

    5. &rho(r,θ,z)=k(R-z)

    5a. mass
    =∫ ρdV
    =∫ k(R-z)*πr²dz

    5b. Moment
    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

    = moment/mass
    = 7R/25

    7R/25 = 0.28R < 3R/8 = 0.375R
    This checks since the density is higher near the bottom.

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