Verify divergence theorem of paraboloid beneath a plane Verify the divergence theorem for the vector field $\mathbf F
=3x^2y^2\mathbf i +y\mathbf  j -6xy^2z\mathbf  k$ for the volume bounded by the paraboloid $z=x^2+y^2$ and $z=2y$ .
I tried to compute the right hand side and I found $\operatorname{div}(F) =1$ . Hence, it will only be the volume bounded by that region and I got the limits and when I computed 
the value $\pi/2$. The problem is computing $\hat n$ in the left hand side. 

When we have a graph $z = f(x, y)$, the area element can be computed as $d\mathbf S = \nabla(z - f(x, y)) dx dy$. Taking outward normals, we'll have
$$\mathbf F \cdot d\mathbf S_{z = 2 y} =
\mathbf F \cdot (0, -2, 1) \,dx dy =
(-2 y - 12 x y^3) \,dx dy, \\
\mathbf F \cdot d\mathbf S_{z = x^2 + y^2} =
\mathbf F \cdot (2 x, 2 y, -1) \,dx dy=
(6 x^3 y^2 + 2 y^2 + 6 x y^2 (x^2 + y^2)) \,dx dy.$$
To evaluate the integrals, we can change to cylindrical coordinates $(x, y, z) = (r \cos t, 1 + r \sin t, z)$ and use the fact that $\int_0^{2 \pi} \sin^m x \cos^n x \,dx$ is non-zero only if both $m$ and $n$ are even:
$$\int_0^{2 \pi} (-2 y - 12 x y^3) \,dt =
\int_0^{2 \pi} (-2 y) dt = -4 \pi, \\
\int_0^{2 \pi} (6 x^3 y^2 + 2 y^2 + 6 x y^2 (x^2 + y^2)) \,dt =
\int_0^{2 \pi} 2 y^2 dt =
2 \pi (2 + r^2).$$
It remains to verify that
$$\int_0^1 2 \pi r^2 \,r dr =
\int_0^1 \int_0^{2 \pi} \int_{r^2 + 1 + 2 r \sin t}^{2 + 2 r \sin t} r \,dz dt dr.$$
