Computing the Gaussian curvature of this surface $z=e^{(-1/2)(x^2+y^2)}$. Compute the Gaussian curvature of $z=e^{(-1/2)(x^2+y^2)}$. Sketch this surface and show where $K=0 $, $K>0$, and $K<0$.
So would the easiest way to do this question be to construct a parametrization $$\mathbf{x}(u,v)=(u, v, e^{-\frac{1}{2}(u^2+v^2)} )?$$
If so, I calculated the Normal to be $$\left( \frac{u}{u^2+v^2+e^{u^2+v^2}},  \frac{v}{u^2+v^2+e^{u^2+v^2}},  1 \right).$$ Is that correct? Thanks
 A: If $g$ is the first fundamental form, and $h$ is the second fundamental form. Then we know
$$K = \frac{ \det h}{ \det g } $$
Since you have a graph $z=f(x,y)$, it's straight forward to compute the fundamental forms by definition. I'll leave $g$ to you, but $h$ is given by
$$ h = f_{xx} dx^2 + 2 f_{xy} dxdy + f_{yy} dy^2 $$
Hint: Consider $\gamma = ( x,y,f )$, by definition of the first fundamental form we have 
$$g_{ij} = ( \gamma_i , \gamma_j) $$
where the subscript denotes a derivative in $i$ and $( \cdot , \cdot)$ denotes the inner product. So we have
$$g = ( 1 + f_x^2 ) dx^2 + 2f_xf_y dxdy + ( 1 + f_y^2) dy^2 $$
A: The parametrization $$\sigma = (u,v, e^{-(x^2+y^2)/2})$$ would indeed be the simplest parametrization, but not for computational purposes.
So we can also let $x = r\cos\theta$ and $y = r\sin\theta$ and get,
$$\sigma = (r\cos\theta,r\sin\theta, e^{-r^2/2}).$$
After finding the normal vector, we get
$$\mathbf{N} = \frac{(e^{-r^2/2}r^2 \cos \theta, e^{-r^2/2}r^2 \sin\theta, r)}{|r|\sqrt{(e^{-r^2}r^2 +1)}}$$ which looks a lot like what you have got.
Find the component of the second fundamental, and we get (we will take $r > 0$ for now)
Therefore, 
$$L = \left < (0,0,(r^2 - 1)e^{\frac{-r^2}{2}}), \mathbf{N} \right > = \frac{r(r^2 - 1)e^{\frac{-r^2}{2}}}{\sqrt{(e^{-r^2}r^2 +1)}}$$
$$M =  \left < (-\sin \theta, \cos \theta,0 ), \mathbf{N} \right > = 0$$
$$N =  \left < (-r\cos \theta, -r\sin \theta,0 ), \mathbf{N} \right > = -r^3e^{\frac{-r^2}{2}}$$
$$LN - M^2 = LN = \frac{-r^4(r^2 - 1)e^{-r^2}}{ \sqrt{(e^{-r^2}r^2 +1)} }$$
On the other hand, 
$$E =  \| (\cos \theta, \sin\theta, -re^{\frac{-r^2}{2}}) \|^2 = 1 + r^2 e^{-r^2}$$
$$F = \left < (-r\sin \theta, r\cos\theta, 0),  (\cos \theta, \sin\theta, -re^{\frac{-r^2}{2}})\right >  = 0$$
$$G = \| (-r\sin \theta, r\cos\theta, 0) \|^2 = r^2$$
$$EG - F^2 = EG = r^2(1 + r^2 e^{-r^2})$$
So FINALLY,  
$$K = \frac{LN - M^2}{EG - F^2 } = \frac{LN}{EG} = \frac{1}{r^2(1 + r^2 e^{-r^2})}   \frac{-r^4(r^2 - 1)e^{-r^2}}{ \sqrt{(e^{-r^2}r^2 +1)} } = \frac{r^2(1-r^2)e^{-r^2}}{(1+r^2 e^{-r^2})^{3/2}}$$
