Evaluate $\int_m(\nabla \times F)\cdot NdS$ Evaluate $\int_m(\nabla \times F)\cdot N\,dS$ where $F=(e^x+z^2,\sin(yz),x^3)$ and $M=\{(x,y,z):y=\frac{x^2}{2}+z^2-4,y\leq 0\}$ and $N$ points outwards 
So my plan is:


*

*find the unit normal of $M$

*find the area element ($dS$) of $M$ by projecting $M$ onto $XZ$ plane

*find the limits of integration and calculate the integral (will got to polar coordinates)


How do I find the unit normal of such a surface? 
I thought about look at a parameterisation $\phi(x,z)=(x,\frac{x^2}{2}+z^2-4,z)$ 
Then $\hat{n}=\frac{\phi_x\times\phi_z}{\|\phi_x\times\phi_z\|}=\frac{(x,-1,2z)}{\sqrt{x^2+4z^2+1}}$
now looking the projection we have $dS=\sqrt{1+x^2+4z^2}$ 
Now $\nabla \times F=(y\sin (yz),-3x^2+2z,0)$ so it will be $$\left(\left( \frac{x^2}{2} + z^2-4\right)\sin\left(\left(\frac{x^2}{2}+z^2-4\right) z\right), -3x^2+2z,0 \right)$$
And last we will have to calculate:
$$\int_0^{2\sqrt{2}}\int_o^2\left(\left(\frac{x^2}{2}+z^2-4\right) \sin\left(\left(\frac{x^2}{2}+z^2-4\right)z\right), -3x^2+2z,0\right) \cdot (x,-1,2z) \,dx\,dx?$$

HINT
A possible way, since $div(\nabla \times F)=0$, is note that
$$\int_M(\nabla \times F)\cdot NdS=-\int_E(\nabla \times F)\cdot NdS$$
where $E$ is the surface enclosed by the boundary of $M$ at $y=0$ that is
$$E=\{(x,y,z):\frac{x^2}{2}+z^2-4\le 0,y= 0\}$$
with
$$(\nabla \times F)\cdot N=(y\sin (yz),-3x^2+2z,0)\cdot (0,1,0)=-3x^2+2z$$
