Flux through region : paraboloid and sphere I have this region : 
$D=\{(x,y,z)\mid y^2+z^2\le 3|x|,(x-2)^2+y^2+z^2 \le 4\}$
with vector field $\mathbf F=(-2x,-2y,xy)$

I can use the divergence theorem : $\mathrm{div}(\mathbf F)=-4$
Attempt :
Let's see where they intersect :
$(x-2)^2+3x=4 \implies x^2-x=0 \implies x=0,1$
In order to find the radius of integration I can plug $x=1$ into $y^2+z^2\le 3x$. so radius is $r=\sqrt{3}$
Now It's time to get the $x$ range :
$(x-2)^2+y^2+z^2 = 4 \implies x^2-4x+y^2+z^2=0 \implies x_{1,2}=\frac{4\pm\sqrt{16-4(y^2+z^2)}}{2}$
I take the positive one.
$x\in\left[\frac{y^2+z^2}{3},2+2\sqrt{4-y^2-z^2}\right]$
Integration cylindrical coordinates :
$x\in\left[\frac{r^2}{3},2+2\sqrt{4-r^2}\right]$
$\theta\in[0,2\pi]$
$r\in[0,\sqrt{3}]$
The final integral is :
$$\text{Flux} = -4\int_{0}^{2\pi}\int_{0}^{\sqrt{3}}\int_{\frac{r^2}{3}}^{2+2\sqrt{4-r^2}}r\,\mathrm dx\,\mathrm dr\,\mathrm d\theta=...=-\frac{166}{3}\pi$$
Is my Integral set-up right ? 
UPDATE :
I think I can also split the region into two parts : the first one is the paraboloid with $x \in [0,1]$ and the second one in the sphere with $x \in [1,4]$
$\int_{0}^{2\pi}\int_{0}^{\sqrt{3}}\int_{\frac{r^2}{3}}^{1}rdxdrd\theta=\frac{3}{2}\pi$
$\int_{1}^{4}(\sqrt{4x-4x^2})^2\pi dx=9\pi$
so Flux = $-4(\frac{3}{2}\pi+9\pi)=-42\pi$
?
 A: I’ll skip the divergence theorem and concentrate my answer on a computation of $\int_ D (-4) dV$.
It is not clear for me why your solutions of the equalities assure that the proposed cylindrical coordinates of the integration correspond to $D$ (and below we’ll see that the proposed bounds are wrong). But I can provide you a more simple integral setup by the following parametrization of $D$. Introduce cylindrical coordinates $y=r\cos\theta$, $z=r\sin\theta$, $r\ge 0$, $0\le\theta<2\pi$. Then $$D=\{(r,x,\theta): r^2\le\min \{3|x|, 4x-x^2\}\}.$$  If $(r,x,\theta)\in D$ then since $4x-x^2\ge 0$ we have $0\le x\le 4$. Solving inquality $3x\le 4x-x^2$, we see that $3x\le 4x-x^2$ provided $0\le x\le 1$ and  $3x\ge 4x-x^2$ provided $1\le x\le 4$. Thus (I am not perfectly sure it the calculations in the following paragraph. To make them rigorous I need to recall to myself the respective analysis theorems. But I expect that they are correct). 
$$\int_ D (-4) dV=$$ $$-4\int_0^{2\pi}\left(\int_0^1\int_0^{3x} r dr dx + \int_1^4\int_0^{4x-x^2} r dr dx\right)
 d\theta=$$ $$-8\pi\left(\int_0^1\int_0^{3x}  r dr dx + \int_1^4\int_0^{4x-x^2}  r dr dx\right)=$$
$$-8\pi\left(\int_0^1 \frac 12 r^2\Big|_0^{3x}  dx + \int_1^4 \frac 12 r^2\Big|_0^{4x-x^2}  dx\right)=$$
$$-8\pi\left(\int_0^1 \frac {9x^2}2  dx + \int_1^4 \frac {(4x-x^2)^2}2  dx\right)=$$ $$\dots=$$ $$-\frac {672}5\pi.$$
The final value I calculated by MathCad. 
Now concerning the proposed bounds. The upper bound $r\le \sqrt{3}$ is wrong, because, for instance, when $x=2$ then the bound $r^2\le 3|x|$ becomes $r^2\le 6$ and the bound $r^2\le 4x-x^2$ becomes $r^2\le 4$, so a value $r=2$ is possible. Also the bound $r^2\le 4x-x^2$ implies $|x-2|\le \sqrt{4-r^2}$ that is $2-\sqrt{4-r^2}\le x\le 2+\sqrt{4-r^2}$, but not only a bound $x\le 2+\sqrt{4-r^2}$.
