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 $$\int_0^1 2 \pi r^2 \,r dr = \int_0^1 \int_0^{2 \pi} \int_{r^2 + 1 + 2 r \sin t}^{2 + 2 r \sin t} r \,dz dt dr.$$