Evaluate the double integral ∫ [0 to 3] ∫ [ 0 to sqrt(9-x^2) for (x^2 + y^2)^(3/2).?

x = 0 ~ 3
y = 0 ~ √( 9 -x^2 )
the integral region is in the first quadrant,
that is, in polar form, θ = 0 ~ π/2

y = √( 9 -x^2 )
x^2 + y^2 = 9 = 3^2
So, r = 3

∫ [ 0 , 3 ] ∫ [ 0 , √(9-x^2) ] (x^2 + y^2)^(3/2) dy dx
= ∫ [ 0 , π/2 ] ∫ [ 0 , 3 ] ( r^2 )^(3/2) * r dr dθ
= ∫ [ 0 , π/2 ] ∫ [ 0 , 3 ] r^3 * r dr dθ
= ∫ [ 0 , π/2 ] ∫ [ 0 , 3 ] r^4 dr dθ
= ∫ [ 0 , π/2 ] { [ (1/5)r^5 ] , from r = 0 to r = 3 } dθ
= ∫ [ 0 , π/2 ] (1/5)3^5 dθ
= ∫ [ 0 , π/2 ] 48.6 dθ
= [ 48.6 θ ] , from θ = 0 , θ = π/2
= 48.6 * π/2
= 24.3 π ..... Ans