I'm asked to calculate the flux through a manifold defined by: $$\begin{align}M = &\{(x,y,z) \in \mathbb{R}^3 \vert\:z = 4-3\,x^2-3\,y^2,\:z\in[1,4]\}\\\\&\{(x,y,z) \in \mathbb{R}^3 \vert\:x^2+y^2=1,\: z\in[0,1]\} \\\\&\{(x,y,z) \in \mathbb{R}^3 \vert\:x^2+y^2\leq1,\:z =0\}\end{align}$$ In other words, this surface:
Through a Vector-field F = $\left(\begin{array}{c}x\,y\\ -\dfrac{y^2}{2}\\ z\end{array}\right)$
Now what I ask myself: can I use Divergence Theorem in the meaning of $$\int_{\partial M}\langle F,N\rangle\,\mathrm{dS} = \int_{M_1} \text{div}\:F\,\mathrm{dV}+\int_{M_2}\text{div}\:F\,\mathrm{dV}$$
since $M$ is a closed surface but can't covered by one integral, but by 2 ($M_1:$ the paraboloid, $M_2:$ the cylinder).
I'm also having a hard time calculating the volume integral (in case I can calculate it this way) of the paraboloid: $$\int_{M_1} \text{div}\:F\,\mathrm{dV} = \int_1^4\int_0^{2\,\pi}\int_0^r \text{div}\:F\,r\,\mathrm{dr\,d\varphi\,dh} $$
Especiallay because $r$ is varying depending on $h$. My simple guess: $r(h) = h+3\,r^2$ ?
Using the divergence theorem is the right idea. You have
$$\operatorname{div}(F)=\frac{\partial(xy)}{\partial x}+\frac{\partial\left(-\frac{y^2}2\right)}{\partial y}+\frac{\partial(z)}{\partial z}=y-y+1=1$$
so the flux through $M$ is indeed equal to the volume of the region $R$ interior to $M$. You can capture this in rectangular coordinates by the integral,
$$\iiint_R\mathrm dV=\int_{-1}^1\int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}}\int_0^{4-3x^2-3y^2}\mathrm dz\,\mathrm dy\,\mathrm dx$$
or more cleanly in cylindrical coordinates by
$$\iiint_R\mathrm dV=\int_0^{2\pi}\int_0^1\int_0^{4-3r^2}r\,\mathrm dz\,\mathrm dr\,\mathrm d\theta$$