D.  

x2 + y2 dV

dV = ρsinϕ dρ dϕ dϴ 

To turn into spherical coordinates,  

f(x,y,z)= x2 + y2 =  (ρcosϴsinϕ) 2 +(ρsinϴsinϕ) 2

= (ρsinϕ) 2 (cos2ϴ + sin2ϴ) = (ρsinϕ) 2

Since we are integrating the unit ball, 0 ≤ r ≤1 and r = ρ 

Also, we are integrating only where x ≥ 0 so -pi/2 ≤ ϴ ≤ pi/2 

Lastly, 0 ≤ ϕ ≤ pi 

ϴ = -pi/2 ϴ = -pi/2 ϕ=0 ϕ=pi   ρ=0 ρ=1(ρsinϕ) 2 * ρsinϕ dρ dϕ dϴ

ρ=0 ρ=1 ρ3sin3ϕ dρ = ρ4/4 sin3ϕ from 0 to 1 = 1/4 sin3ϕ

1/4 ϕ=0 ϕ=pi  sin3ϕ dϕ = 1/4 ϕ=0 ϕ=pi  sin ϕ(sin2ϕ) dϕ

=1/4 ϕ=0 ϕ=pi  sin ϕ(1-cos2ϕ) dϕ    let u = cos ϕ dx = du/-sin ϕ

=-1/4 ϕ=0 ϕ=pi (1-u 2)du = -1/4(cos ϕ – cos3ϕ/3) from 0 to pi

= -1/4(-1+1/3) + ¼(1-1/3) = 1/3 

And then integrating the last integral, = 1/3 ϴ from –pi/2 to pi/2, the answer becomes pi/3