Question

A lamina occupies the part of the disk x2 + y2 ≤ 16 in the first quadrant. Find the center of mass of the lamina if the density at any point is proportional to the square of its distance from the origin.

Answer 1

Answer: The center of mass =

(8192/(1280π), 8192/(1280π)).

Step-by-step explanation:

Here, δ(x,y) =k(x^2 + y^2) for some constant k.

So, m = ∫∫ δ(x,y) dA

..........= ∫(θ = 0 to π/2) ∫(r = 0 to 4) kr^2 × (r dr dθ), via polar coordinates

..........= (π/2) × (k/4)r^4 {for r = 0 to 4}

..........= 256πk/8.

My = ∫∫ x δ(x,y) dA

......= ∫(θ = 0 to π/2) ∫(r = 0 to 4) (r cos θ) × kr^2 × (r dr dθ), via polar coordinates

......= ∫(θ = 0 to π/2) cos θ dθ × ∫(r = 0 to 4) kr^4 dr 

......= sin θ {for θ = 0 to π/2} × (1/5)kr^5 {for r = 0 to 4}

......= 1024k/5.

Mx = ∫∫ y δ(x,y) dA

......= ∫(θ = 0 to π/2) ∫(r = 0 to 4) (r sin θ) × kr^2 × (r dr dθ), via polar coordinates

......= ∫(θ = 0 to π/2) sin θ dθ × ∫(r = 0 to 4) kr^4 dr 

......= -cos θ {for θ = 0 to π/2} × (1/5)kr^5 {for r = 0 to 4}

......= 1024k/5.

Hence, the center of mass is (My/m, Mx/m) =

My/m = 1024/5 ×8/256

The same for Mx/m the density at any point is proportional to the square of its distance from the origin

(8192/(1280π), 8192/(1280π)).