Calculus double integral

Question

Find ∫∫Rx2dA where R={(x,y)∣16x^2+4y^2≤64}

Answer 1

Let x = 2u and y = 4v∂

Then ∂x/∂u =2 ∂x/∂v=0

∂y/∂u =0 ∂y/∂v=4

That is the Jacobian ∂(x,y) ∂/ ∂(u,v)= 8

Hence ∫∫_Rx²dA = ∫∫Ω4u²8du dv where Ω = { (u,v)∈ℜ² | u²+v²≤1} Now let u=rcosθ and v= r sinθ

Then ∫∫_Rx²dA = ∫∫Ω4u²8du dv = ∫ ^θ=2π _θ=0 ∫^r=1 _r=032r²cos²θ dr dθ= 32/3 ∫ ^θ=2π _θ=0[r³cos²θ ] |₀¹dθ=

= 32/3 ∫ ^θ=2π _θ=0[cos²θ ]dθ = 32/6 ∫ ₀^π2(1+cos2θ)dθ = 16/3[ θ +1/2 sin2θ] ₀^2π= 32π/3