The exercise consists of determining $\iint_Y \mathbf{F} \cdot \mathbf{N} dS$ where $\mathbf{F} = (2 - x^2yz + y, x^2yz + ye^z, y^2 + 2z - e^z)$ and $Y$ is the part of the surface $x=2-(y-1)^2 - (z-2)^2$ where $x \geq 0$. The normal points away from the origin. If I interpret it correctly, $Y$ is a compact plane.
I suspect that I am supposed to use Gauss theorem, given that the $div\ \mathbf{F} = -2xyz + x^2z + 2$ and that $Y$ is compact. This simplifies the integral $\iint _Y \mathbf{F} \cdot \mathbf{N} dS = \iiint_{x=2-(y-1)^2 - (z-2)^2, x \geq 0} -2xyz + x^2z + 2 dx dy dz$ where $K$ is the interior of the surface $Y$. However, I am unsure how to how to proceed with the integral. Any input is much appreciated.
The surface represents a paraboloid with circular cross-section and vertex $(2,1,2)$, as you can see by rewriting it this way: $\frac{(y-1)^2}{2-x}+\frac{(z-2)^2}{2-x}=1$. Now, you can rewrite the triple integral you wrote as $\int_{x=0}^{x=2}(\iint_{\{(x,y)\in\mathbb{R^2}:(y-1)^2+(z-2)^2\leq 2-x\}}(x^2z-2xyz+2)dydz)dx.$ The innermost integral is better done in polar coordinates in the y-z plane and is equal to $\int_{\theta=0}^{\theta=2\pi}(\int_{r=0}^{r=\sqrt{2-x}}[x^2(r\sin\theta)-2xr^2\sin\theta\cos\theta+2]rdr)d\theta=2\pi(2-x)$ (since $\int_{0}^{2\pi}\sin(x)dx=\int_{0}^{2\pi}\sin(x)\cos(x)dx=0$) and substituting we get $\int_{x=0}^{x=2}2\pi(2-x)dx=4\pi.$