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

In summary, the conversation discusses finding the upward pointing unit normals to two finite regions, $U$ and $D$, on a surface $S$. The normal for $U$ is calculated using the function $g(x,y,z) = z-\sqrt{x^2+y^2+1}$ and is given by $\displaystyle n_U=\displaystyle \frac{-x \vec{i}-y \vec{j}+\sqrt{x^2+y^2+1} \vec{k}}{\sqrt{2x^2+2y^2+1}}$. The answer provided for $n_U$ was incorrect due to a misplaced $2$. The normal for $D$ is a constant vector due to the given condition $
  • #1
NoName3
25
0
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 $$ n_U=\displaystyle \frac{-x \vec{i}-y \vec{j}+\sqrt{x^2+y^2+1} \vec{k}}{\sqrt{2x^2+2y^2+1}}$$
 
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  • #2
NoName said:
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?
 
  • #3
I like Serena said:
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 $U$?
 
Last edited:
  • #4
Yep. All correct. (Nod)
 

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

What is an upward pointing unit normal?

An upward pointing unit normal is a vector that is perpendicular to a surface and points in the direction of the surface's outward normal. It has a magnitude of 1, making it a unit vector.

Why is the upward pointing unit normal important in science?

The upward pointing unit normal is important in science because it helps us determine the direction of flow of fluids, the orientation of surfaces in space, and the direction of force acting on objects. It is also used in calculations involving surface integrals and vector calculus.

How is the upward pointing unit normal calculated?

The upward pointing unit normal is calculated by taking the cross product of two tangent vectors on the surface. The resulting vector is then normalized to have a magnitude of 1, giving us the unit normal vector.

What is the difference between an upward pointing unit normal and a downward pointing unit normal?

An upward pointing unit normal and a downward pointing unit normal are simply opposites of each other. They have the same magnitude of 1, but point in opposite directions. The choice of which one to use depends on the context and convention of the problem being solved.

How is the upward pointing unit normal used in fluid mechanics?

In fluid mechanics, the upward pointing unit normal is used to determine the direction of fluid flow. It is also used in the Navier-Stokes equations to calculate the forces acting on a fluid element. The unit normal is also used to define the boundary conditions for fluid flow problems.

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