Surface integral using Gauss theorem The exercise consists in determining
$$\iint_Y \mathbf{F} \cdot \mathbf{N} \ \mathrm{d}S$$
where
$$\mathbf{F}= (x^2 yz + x \sin z , x^2 + y(1 - \sin z ), x + y - xy z^2)$$
and $Y$ is the part of the conical surface $4x^2 + (y-1)^2 = z^2$ which lies between $z = 1$ and $z = 2$. The normal vector points away from the $z$-axis.
I have tried to solve it using Gauss theorem, and I think this is the correct approach. Of course, the surface is not closed, and I need to add a "lid" and "bottom" to $Y$ to use Gauss theorem in the first place. Let $\sigma$ denote the lid, $\gamma$ the bottom and $K$ the whole surface. This yields
$$\iint_Y \mathbf{F} \cdot \mathbf{N}  \ \mathrm{d}S + \iint_{\sigma} \mathbf{F} \cdot \mathbf{N} \ \mathrm{d}S + \iint_{\gamma} \mathbf{F} \cdot \mathbf{N} \ \mathrm{d}S =  \iiint_K \nabla \cdot \mathbf{F} \ \mathrm{d}x\mathrm{d}y\mathrm{d}z$$
where $\nabla \cdot \mathbf{F}$ is the divergence, which can be determined and it is $\nabla\cdot\mathbf{F} = 1$. This produces the triple integral 
$$\iiint_{4x^2 + (y-1)^2 \leq z^2} \ \mathrm{d}x\mathrm{d}y\mathrm{d}z$$
This is where I get stuck. 
I am unsure about how to solve the triple integral, and how to further solve the surface integrals for the lid and the bottom surface.

$$\iiint_{ \{ (x,y,z) \in \mathbb{R}^3: 4x^2 + (y-1)^2 = z^2, 1 \leq z \leq 2 \} } \mathrm{d}x \mathrm{d}y \mathrm{d}z =\\
= \int_{z=1}^{z=2} \left( \int_{ \{ (x,y) \in \mathbb{R}^2: 4x^2 + (y-1)^2 = z^2 \} } \mathrm{d}x \mathrm{d}y \right) \mathrm{d}z =\\ = \int_{z=1}^{z=2}\pi\cdot\frac{z}{2}\cdot z\ dz=\frac{7\pi}{6}$$
Top ($z=2$)
$$4x^2 + (y-1)^2= 2 \Leftrightarrow \frac{x^2}{\displaystyle \frac{1}{2}}+\frac{(y-1)^2}{2} = 1$$
so a parametrization is 
$$\sigma: [0,1] \times [0,2\pi] \to \left( \frac{1}{\sqrt{2}}r\cos\theta,1+\sqrt{2}r\sin\theta,2 \right) \in \mathbb{R}^3$$
and a normal vector is 
$$\mathbf{\nu} = \frac{ \partial \sigma }{\partial r} \times \frac{ \partial \sigma}{\partial \theta} = (0,0,r)$$
thus 
$$\iint_{\mathrm{top}} \mathbf{F} \cdot \mathbf{\nu} \ \mathrm{d}x \mathrm{d}y = \iint_{ \{ (x,y) \in \mathbb{R}^2: 4x^2 + (y-1)^2 \leq 2 \} } \mathbf{F} \cdot \mathbf{\nu} \ \mathrm{d}x \mathrm{d}y =\\
= \iint_{ \{ (r,\theta): 0 \leq r \leq 1, 0 \leq \theta \leq 2 \pi \} } F ( \sigma( r,\theta )) \cdot \nu(r,\theta) \cdot r \ \mathrm{d}r \mathrm{d} \theta =\\
= \int_{ \theta=0 }^{ \theta = 2\pi } \left[ \int_{r=0}^{r=1} \left( -\frac{3}{\sqrt{2}} r^3 \cos\theta + (\sqrt{2}-4) r^3 \sin \theta + r^2 \right) \mathrm{d}r \right] \mathrm{d}\theta = \frac{2\pi}{3}$$
Bottom ($z=1$)
$$4x^2+(y-1)^2 = 1 \Leftrightarrow \frac{x^2}{\displaystyle \frac{1}{4}} + (y-1)^2 = 1$$
so a parametrization is 
$$\sigma:[0,1] \times [0,2\pi] \to \left( \displaystyle \frac{1}{2}r \cos \theta,1-r \sin \theta,1 \right) \in \mathbb{R}^3$$
and a normal vector is
$$\mathrm{\nu} = \frac{\partial\sigma}{\partial r} \times \frac{ \partial \sigma}{\partial \theta} = \left( 0,0,-\frac{r}{2} \right)$$
thus
$$\int_{\mathrm{bottom}} \mathbf{F} \cdot \mathrm{\nu} \ \mathrm{d}x \mathrm{d}y = \iint_{ \{ (x,y) \in \mathbb{R}^2:4x^2+(y-1)^2\leq 1 \} } \mathbf{F} \cdot \mathrm{\nu} \ \mathrm{d}x \mathrm{d}y=\\
= \iint_{ \{ (r,\theta): 0 \leq r \leq 1, 0 \leq \theta \leq 2\pi \} } F( \sigma(r,\theta)) \cdot \nu (r,\theta) \cdot r \ \mathrm{d}r \mathrm{d} \theta =\\
= \int_{\theta=0}^{\theta=2\pi} \left[ \int_{r=0}^{r=1} \left( -\frac{r^2}{2} + \frac{r^3}{2} \sin \theta - \frac{r^4}{4}\sin\theta\cos\theta \right) \mathrm{d}r \right] \mathrm{d}\theta = -\frac{2\pi}{6}$$
Thus 
$$\iint_{Y} \mathrm{F} \cdot \mathrm{\nu} \ \mathrm{d}x \mathrm{d}y = \iiint \nabla \cdot \mathrm{F} \ \mathrm{d}x \mathrm{d}y \mathrm{d}z - \iint_{\mathrm{top}} \mathrm{F} \cdot \mathrm{\nu} \ \mathrm{d}x \mathrm{d}y - \iint_{\mathrm{bottom}} \mathrm{F} \cdot \mathrm{\nu} \ \mathrm{d}x \mathrm{d}y=\\
= \frac{7\pi}{6} - \frac{2\pi}{3} + \frac{2\pi}{6} = \frac{5\pi}{6}$$
