Calculus 3 Triple Integral

Question

Find the center of mass of the solid  bounded by the paraboloid  and the plane . Assume the density is constant.

Answer 1

The projection of the intersection of the paraboloid with the plane z=1 on the xy-plane is the circle

x² + y²= 11/3

The mass m is equal to

m = ρ ∫_θ=0^θ=2π ∫_r=0 ^{r= √(11/3)} ∫^z=11_z=3r^2 rdz dr dθ =

= 121πρ / 6

M_yz = ρ ∫_θ=0^θ=2π ∫_r=0 ^{r= √(11/3)} ∫^z=11_z=3r^2 x rdz dr dθ =

M_yz = ρ ∫_θ=0^θ=2π ∫_r=0 ^{r= √(11/3)} ∫^z=11_z=3r^2 r²cosθdz dr dθ = 0

M_xz = ρ ∫_θ=0^θ=2π ∫_r=0 ^{r= √(1/3)} ∫^z=1_z=3r^2 y rdz dr dθ

M_xz = ρ ∫_θ=0^θ=2π ∫_r=0 ^{r= √(1/3)} ∫^z=1_z=3r^2 r²sinθdz dr dθ =0

M_xy = ρ ∫_θ=0^θ=2π ∫_r=0 ^{r= √(11/3)} ∫^z=11_z=3r^2 z rdz dr dθ =

M_xy = ρ ∫_θ=0^θ=2π ∫_r=0 ^{r= √(11/3)} ∫^z=11_z=3r^2 z rdz dr dθ = π(11³)ρ /9

Then the center of mass is at (x,y,z) = ( M_yz/m , M_xz/m, M_xy/m ) = ( 0, 0, 66/9 )