$R=\{z^2-4z+y^2 \le 0,0 \le x \le 1\}$
with $F=(x\sqrt{y^2+z^2},-z,y)$
So it's a shifted cylinder : $(z-2)^2+y^2=4$
$$ \left\{ \begin{aligned} x&=x\\ y&=2\sin\theta\\ z&=2+2\cos\theta \end{aligned} \right. $$
It's a closed surface so I can use the Divergence Theorem :
$$div(F)=\sqrt{y^2+z^2}=r.$$
The flux outgoing the surface is : $\int_{0}^{\pi}\int_{0}^{\sqrt{8+8\cos\theta}}\int_{0}^{1}r^2dx\,dr\,d\theta$
$\int_{0}^{\pi}\int_{0}^{\sqrt{8+8\cos\theta}}\int_{0}^{1}r^2dx\,dr\,d\theta$=$\int_{0}^{\pi}\int_{0}^{\sqrt{8+8\cos\theta}}r^2\,dr\,d\theta$=$\frac{1}{3}$$\int_{0}^{\pi}(8+8\cos\theta)^{\frac{3}{2}}d\theta=?$
How can I solve this ? (I tried a lot of substitution)
Or, is there an easier setting for the integral?
With the use of $\;\cos\theta=2\cos^2(\theta/2)-1\;$ the integral becomes $$I={1\over 3}\int_{0}^{\pi}\left(16\cos^2(\theta/2)\right)^{3/2}\,d\theta.$$ Note that $\cos(\theta/2)\geq 0$ if $\theta\in [0,\pi].$ Thus is $\left(\cos^2(\theta/2)\right)^{3/2}=\cos^3(\theta/2),$ and we solve $$I={1\over3}\int_{0}^{\pi}64\cos^3(\theta/2)\,d\theta={1\over3}\int_{0}^{\pi}64\big(1-\sin^2(\theta/2)\big)\cos(\theta/2)\,d\theta$$ Can you go from this?