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.

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)).

it has changed from the width and the height.

step-by-step explanation: