please solve this

Question

x^2+y^2<a^2

∫∫ x^2y^2dA

Answer 1

Use polar coordinates; r dr dθ = dx dy, x = rcosθ, y=rsinθ, cosθsinθ = 1/2sin2θ, and use I[a,b] for the integral from a to b and E[a,b] for the evaluation from a to b, 0 < θ < 2π, and 0 < r < a, we have, as sin² 2θ = 1/2 - 1/2 cos 4θ,

I[0, a] I[0, 2π] 1/4 r⁵ sin² 2θ dθ dr = I[0, a] I[0, 2π] 1/4 r⁵ ( 1/2 - 1/2 cos 4θ) dθ dr =

I[0, a] 1/4 r⁵ ( 1/2θ - 1/8 sin 4θ)I[0, 2π] dr =

I[0, a] 1/4 r⁵π dr = 1/24 r⁶π [0,a] = π a⁶/24