Apply the divergence theorem

Apply the divergence theorem to compute ∫∫s F·dS for the vector field F(x,y,z)= over the complete boundary S of the solid paraboloid {(x,y,z): x^2+y^2 ≤ z ≤ 1} with outward normal.

Here's what I did:
I found the divergence to be 3x^2 + 3y^2, so by the divergence theorem the integral should be
∫∫∫ 3x^2 +3y^2 dV. Then I switched the integral to ∫(θ=0 to 2pi)∫(r=0 to1)∫(z=0 to 1) 3r^2 r dz dr dθ.
which gives the answer 3pi/2.

My teacher got pi/2 as the answer so I am really confused. Can someone please tell me what went wrong?

You are integrating over a paraboloid, not a cylinder.
The bounds for z should be z = r^2 to z = 1.

So, we have
∫(θ = 0 to 2π)∫(r = 0 to 1)∫(z = r^2 to 1) 3r^2 * r dz dr dθ
= 2π ∫(r = 0 to 1) 3r^3 (1 - r^2) dr
= 2π ∫(r = 0 to 1) (3r^3 - 3r^5) dr
= 2π (3/4 - 3/6) - 0
= π/2.

I hope this helps!