Cfrgtkky

Verify the divergence theorem for the vector field $mathbf F
=3x^2y^2mathbf i +ymathbf j -6xy^2zmathbf k$ 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 $dmathbf S = nabla(z - f(x, y)) dx dy$. Taking outward normals, we'll have
$$mathbf F cdot dmathbf S_{z = 2 y} =
mathbf F cdot (0, -2, 1) ,dx dy =
(-2 y - 12 x y^3) ,dx dy, \
mathbf F cdot dmathbf S_{z = x^2 + y^2} =
mathbf F cdot (2 x, 2 y, -1) ,dx dy=
(6 x^3 y^2 + 2 y^2 + 6 x y^2 (x^2 + y^2)) ,dx dy.$$
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
$$int_0^1 2 pi r^2 ,r dr =
int_0^1 int_0^{2 pi} int_{r^2 + 1 + 2 r sin t}^{2 + 2 r sin t} r ,dz dt dr.$$