please find this :

Question

∫∫ xdA if x>1 and x^2+y^2<2

Answer 1

Better to do this problem in polar coordinate: x = rcosφ, y = rsinφ. 0 ≤ r ≤ 2^1/2, - π/4 ≤ φ ≤ π/4, dA = rdrdφ.

So, ∫∫ xdA if x>1 and x^2+y^2<2 = ∫∫r²drdφ = ∫₀^sqrt2r²dr∫_-π/4^π/4cosφdφ = r³/3₀^sqrt2sinφ_-π/4^π/4 = 2^3/2/3·2sinπ/4 =

= 4/3

Answer 2

I[1,sqrt(2)]I[-sqrt(2-x^2),sqrt(2-x^2)]x dydx =

I[1,sqrt(2)] xy E[-sqrt(2-x^2),sqrt(2-x^2)] dx =

I[1,sqrt(2)] 2x sqrt(2-x^2)dx =

-2/3(2-x^2)^3/2 E[1,sqrt(2)] =

2/3