How do I find the upward pointing unit normals for $U$ and $D$?

[NoName3]

Let the surface $S \subset \mathbb{R}^3$ be the graph of the function $f: \mathbb{R}^2 \to \mathbb{R}$, $f(x,y)= \sqrt{x^2+y^2+1}$. Let $U$ be the finite region of $S$ below the plane $z = 2$ and let $D$ be the finite region of the plane $z = 2$ for which $x^2 + y^2 \le 3$. Let $n_U (x, y, z)$ and $n_D(x, y, z)$ be the upward pointing unit normals to $U$ and $D$ respectively. How do I find $n_U$ and $n_D$? I did the following for $n_U$ but it doesn't match the answer:

Let $\displaystyle g(x, y, z) = z-\sqrt{x^2+y^2+1}$ then a normal to the surface is $\displaystyle \nabla g = -\frac{x}{\sqrt{x^2+y^2+1}}\vec{i}-\frac{y}{\sqrt{x^2+y^2+1}}+\vec{k}$ thus a unit normal to the surface is $$ n_U=\displaystyle \frac{-x \vec{i}-y \vec{j}+\sqrt{x^2+y^2+1} \vec{k}}{\sqrt{2x^2+2y^2+1}}$$

 

[I like Serena]

Let the surface $S \subset \mathbb{R}^3$ be the graph of the function $f: \mathbb{R}^2 \to \mathbb{R}$, $f(x,y)= \sqrt{x^2+y^2+1}$. Let $U$ be the finite region of $S$ below the plane $z = 2$ and let $D$ be the finite region of the plane $z = 2$ for which $x^2 + y^2 \le 3$. Let $n_U (x, y, z)$ and $n_D(x, y, z)$ be the upward pointing unit normals to $U$ and $D$ respectively. How do I find $n_U$ and $n_D$? I did the following for $n_U$ but it doesn't match the answer:

Let $\displaystyle g(x, y, z) = z-\sqrt{x^2+y^2+1}$ then a normal to the surface is $\displaystyle \nabla g = -\frac{x}{\sqrt{x^2+y^2+1}}\vec{i}-\frac{y}{\sqrt{x^2+y^2+1}}+\vec{k}$ thus a unit normal to the surface is $$ n_U=\displaystyle \frac{-x \vec{i}-y \vec{j}+\sqrt{x^2+y^2+1} \vec{k}}{\sqrt{2x^2+2y^2+1}}$$

Hi NN! ;)

Looks fine to me.
What's the answer supposed to be?

 

[NoName3]

Hi NN! ;)

Looks fine to me.
What's the answer supposed to be?

Hi, I like Serena. ;) Thanks for the reply.

It was supposed to be $\displaystyle n_U=\displaystyle \frac{-x \vec{i}-y \vec{j}+\sqrt{x^2+y^2+1} \vec{k}}{\sqrt{x^2+y^2+2}}$

They probably misplaced the $2$ then.

Why is the answer to $n_D$ a constant vector? EDIT: I see because $z =2$.

I've noticed that while calculating $n_U$ I could have easily got $-n_U$ instead.
Is it correct to say that $-n_U$ is the downward pointing normal to $S$ and $U$?

 

[I like Serena]

Yep. All correct. (Nod)