Center of Mass of a Hemisphere using Cylindrical coordinates.?

It suffices to find the center of mass of x^2 + y^2 + z^2 = 5^2 with z ≥ 0

with density δ(x,y,z) = 7z.

The equation of the upper hemisphere can be rewritten as z = √(25 - x^2 - y^2)

Via cylindrical coordinates, we have

m = ∫∫∫ δ(x,y,z) dV

....= ∫∫∫ 7z dV

....= ∫(θ = 0 to 2π) ∫(r = 0 to 5) ∫(z = 0 to √(25 - r^2)) 7z * (r dz dr dθ)

....= 2π ∫(r = 0 to 5) (7/2) rz^2 {for z = 0 to √(25 - r^2)} dr

....= 7π ∫(r = 0 to 5) (25r - r^3) dr

....= 7π (25r^2/2 - r^4/4) {for r = 0 to 5}

....= 4375π/4.

Myz = ∫∫∫ x * δ(x,y,z) dV

....= ∫∫∫ 7xz dV

....= ∫(θ = 0 to 2π) ∫(r = 0 to 5) ∫(z = 0 to √(25 - r^2)) 7(r cos θ) z * (r dz dr dθ)

....= ∫(θ = 0 to 2π) cos θ dθ * ∫(r = 0 to 5) ∫(z = 0 to √(25 - r^2)) 7r^2 z dz dr

....= 0.

Mxz = ∫∫∫ y * δ(x,y,z) dV

....= ∫∫∫ 7yz dV

....= ∫(θ = 0 to 2π) ∫(r = 0 to 5) ∫(z = 0 to √(25 - r^2)) 7(r sin θ) z * (r dz dr dθ)

....= ∫(θ = 0 to 2π) sin θ dθ * ∫(r = 0 to 5) ∫(z = 0 to √(25 - r^2)) 7r^2 z dz dr

....= 0.

Mxy = ∫∫∫ z * δ(x,y,z) dV

....= ∫∫∫ 7z^2 dV

....= ∫(θ = 0 to 2π) ∫(r = 0 to 5) ∫(z = 0 to √(25 - r^2)) 7z^2 * (r dz dr dθ)

....= 2π * ∫(r = 0 to 5) (7/3)rz^3 {for z = 0 to √(25 - r^2)} dr

....= (14π/3) * ∫(r = 0 to 5) r(25 - r^2)^(3/2) dr

....= (14π/3) * (-1/5)(25 - r^2)^(5/2) {for r = 0 to 5}

....= 8750π/3.

So, the center of mass equals

(0, 0, (8750π/3)/(4375π/4)) = (0, 0, 8/3).

I hope this helps!

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RE:

Center of Mass of a Hemisphere using Cylindrical coordinates.?

Use cylindrical coordinates to find the center of mass of a solid hemisphere with a radius of 5 if the density at any point it equal to 7 times the distance from its base. Assume the base lies flat on the xy plane.

Let H represent this hemisphere.

The distance from the base is z, so the density function is p(x,y,z) = 7z.

Note that if (x, y, z) is on the hemisphere, then so is (-x, y, z).

Furthermore, p(x, y, z) = p(-x, y, z).

So by symmetry of both the region and the density function, the x-coordinate of the center of mass is 0.

Note that if (x, y, z) is on the hemisphere, then so is (x, -y, z).

Furthermore, p(x, y, z) = p(x, -y, z).

So by symmetry of both the region and the density function, the y-coordinate of the center of mass is 0.

So we just need the z-coordinate of the center of mass.

The hemisphere H can be described by the inequalities

0 <= r <= 5, 0 <= theta <= 2pi, 0 <= z <= sqrt(25 - r^2).

z-coordinate of center of mass

= triple integral over H of zp(x,y,z) dV / triple integral over H of p(x,y,z) dV

= triple integral over H of 7z^2 dV / triple integral over H of 7z dV

= triple integral over H of z^2 dV / triple integral over H of z dV

= int 0 to 2pi int 0 to 5 int 0 to sqrt(25 - r^2) of rz^2 dz dr d theta /

int 0 to 2pi int 0 to 5 int 0 to sqrt(25 - r^2) of rz dz dr d theta

= int 0 to 2pi int 0 to 5 of (1/3)r(25 - r^2)^(3/2) dr d theta /

int 0 to 2pi int 0 to 5 of (1/2)r(25 - r^2) dr d theta

= int 0 to 2pi of [(-1/15)(25 - r^2)^(5/2) from r=0 to r=5] d theta /

int 0 to 2pi of [(1/2)(25r^2/2 - r^4/4) from r=0 to r=5] d theta

= int 0 to 2pi of (625/3) d theta / int 0 to 2pi of (625/8) d theta

= [2pi(625/3)] / [2pi(625/8)]

= 8/3.

The center of mass is (0, 0, 8/3).

Lord bless you today!