Divergence theorem: compute triple integral over a paraboloid between two planes 
Let $S$ be the surface of equation $z=x^2 + y^2$, which is between the planes $z=1$ and $z=x+y+3$. Let $F = (x^2,y,z)$. Compute the flux of $F$ across $S$

I think I should use divergence theorem, hence I only need to compute $$\int \int \int_V \operatorname{div}(F)dxdydz$$ but I don't know how to describe the volume enclosed by $S$.
I'd like to integrate this way:
$$\int \int_{D} \Big[\int_{1}^{x+y+3}2x + 2 dz\Big] dxdy$$
But I don't know how to define the domain $D$ where the coordinates $x$ and $y$ are. I think I should the two conditions on the planes, but I can't. Any help is highly appreciated.
 A: If you are applying divergence theorem, you will have to close the  surface with disks in planes $z=1$ and $z = x+y+3$. You will need to separately calculate flux through the disks and subtract from the flux through the closed surface.
Now good thing is that within the paraboloid domain, plane $z = x + y + 3$ is throughout above plane $z=1$. Also note that the intersection of the given circular paraboloid and the given plane can be seen in $XY$ plane as -
$x^2+y^2 = x + y + 3 \implies (x-0.5)^2 + (y-0.5)^2 = 3.5$
So at the minimum, it makes sense to use cylindrical coordinates with
$x = 0.5 + r \cos \theta, y = 0.5 + r \sin\theta, 0 \leq r \leq \sqrt{3.5}$
$1 \leq z \leq 4 + 0.5 \cos\theta + 0.5 \sin\theta, \ 0 \leq \theta \leq 2\pi$
So the integral to find flux becomes
$\displaystyle \int_0^{2\pi} \int_0^{\sqrt {3.5}} \int_1^{4+r\cos\theta+r\sin\theta} r(3+ 2r\cos\theta) \ dz \ dr \ d\theta$
Now please find flux through disks at $z=1$ and $z=x+y+3$ and subtract from the above. That should give you the flux through paraboloid surface.

Just as a side note: if you want to simplify the limits of integration further, you can in fact rotate the paraboloid by $\frac{\pi}{4}$. There is no change to Jacobian just like in the first case. It will simplify the equation of the plane $z = x + y + 3$.
$x = 0.5 + r \cos (\frac{\pi}{4} + \theta), y = 0.5 + r \sin(\frac{\pi}{4} + \theta), 0 \leq r \leq \sqrt{3.5}$
Equation of plane becomes $z = x+y+3 = 4 + r \cos(\frac{\pi}{4} + \theta) + r \sin(\frac{\pi}{4} + \theta) = 4 + r \sqrt2 \cos\theta$.
