How to find a normal vector to a surface at a given point?

Question

Suppose we define a curve $C$ by letting $C$ be the intersection of the unit sphere $x^2+y^2+z^2=1$ and the cylinder $x^2+y^2=x$ that takes place for $z>0$.

Now I want to use Stoke's theorem to be able to compute a line integral around $C$ and as such I will need to find a unit normal vector to any point in the interior of $C$ this region we can call $S$.

I'm really struggling on how to find a unit normal vector for $S$ so that I can calculate:

$$\iint_S~(\nabla \times \vec{F})~\hat{n}~dS$$

I have got a parametrization for the curve $S$ but I don't really know if that will be of any use.

Answer 1

This results in an interesting curve:

One surface $S$ with $\partial S = C$ is that bit of the cylinder sticking out of the unit sphere. It has easy normal vectors as well.

Another choice is the part of the unit sphere within the cylinder.

Answer 2

A vector normal to the surface of the sphere $x^2 + y^2 + z^2 = 1$, $<x,y,z>$ would do it.

Or $<x/z, y/z, 1>$, or $<x/(1-x^2+y^2)^{1/2} , y/(1-x^2-y^2)^{1/2}, 1)$

$<-\partial z/\partial z,-\partial z/\partial y,1>$ is a generic approach for any surface in Cartesian coordinates.

Asked to provide more information...

It is high-school geometry that a vector from the center of a sphere will be normal to the surface. So, $<x,y,z>$ jumps out as a quick go-to...

$<x/z, y/z, 1>$ is nice because then a square unit in the x,y plane corresponds to the sq units above it.

But, if you are going to integrate in terms of $x$ and $y$ then it is handy to replace any $z$ terms as functions of $x,y$

To find the vector perpendicular to a more complicate surface... F(x,y,z) = K,

$\nabla F = <\partial F/\partial x,\partial F/\partial y,\partial F/\partial z>$ will be perpendicular to the surface.

If you can parameterize your surface in terms of $u$ and $v$ then: $\partial S = <\partial x/\partial u,\partial y/\partial u,\partial z/\partial u>\times<\partial x/\partial v,\partial y/\partial v,\partial z/\partial v>$

and finally, you can use the existing $x,y$ that you already as your parameters.

$\partial S = <\partial x/\partial x,\partial y/\partial x,\partial z/\partial x>\times<\partial x/\partial y,\partial y/\partial y,\partial z/\partial y>$ which equals $<1,0,\partial z/\partial x>\times<0,1,\partial z/\partial y> = <-\partial z/\partial y,-\partial z/\partial y,1>$

Hope this helps

Answer 3

Typically when doing this you solve for $\hat{n}\cdot dS$ all at once rather than just the normal vector. When doing this there are a few ways.

If You Know A Normal Vector
Say you already know a normal vector to the surface, $\vec{N}$. Then $$\hat{n}\cdot dS=\pm\frac{\vec{N}}{\vec{N}\cdot \hat{k}}$$ where the $\pm$ changes based on how you orient the surface.

Using the Gradient
You may have just thought the last method was counter intuitive, but you do know a normal vector to the surface! Say you have a function $g(x,y,z)=f(x,y,z)$, change it to the form $F(x,y,z)=c$ where $c$ is a constant. Then you are just looking at a level curve for the function $F$, and remember that gradients are always normal to level curves, so $$\hat{n}\cdot dS=\pm\frac{\nabla F}{F_z}$$

You Have an Explicit Function for Z
Another common case would be if you were to be given the surface in the form $z=f(x,y,z)$. Then $$\hat{n}\cdot dS=\pm<-z_x,-z_y,1>$$
You Have The Parametrization of the Surface
There is one last way that I know of that is for if you're given the parametrization of the function. Say the surface $S$ is defined by $$S:\begin{cases} x=x(u,v) \\ y=y(u,v) \\ z=z(u,v) \end{cases}$$ Or in position vector notation we have the position vector, $$\vec{r}=<x(u,v),y(u,v),z(u,v)>$$ Then $$\hat{n}\cdot dS=\pm\left(\frac{\partial \vec{r}}{\partial u}\times\frac{\partial \vec{r}}{\partial v}\right)\text{d}u\text{d}v$$

Answer 4

This surface is called Viviani's window.

There are different ways of dealing with this. The best option, I believe, is to parametrize it in cartesian coordinates: $$ x=x,\quad y=y, \quad z=\sqrt{1-x^2-y^2},\quad (x,y)\in D $$ with $$ D=\{(x,y)\;|\;x^2+y^2\le x \} $$ This way $$ \int_C \vec{F}\cdot d\vec{r}=\iint_S \nabla\times \vec{F}\cdot d\vec{S} =\iint_D \nabla\times \vec{F}(x,y)\cdot \vec{r}_x\times\vec{r}_y\; dA $$ This is where parametrizing in cartesian coordinates is handy: computing the normal vector (which is what you asked for) $\vec{r}_x\times\vec{r}_y$ is always very simple as you have $1$'s and $0$'s in your vectors: $$ \vec{r}_x\times\vec{r}_y = (1,0,\frac{-x}{\sqrt{1-x^2-y^2}})\times (0,1,\frac{-y}{\sqrt{1-x^2-y^2}}) =(\frac{x}{\sqrt{1-x^2-y^2}},\frac{y}{\sqrt{1-x^2-y^2}},1) $$

You end up with $$ \iint_D \nabla\times \vec{F}(x,y)\cdot \vec{r}_x\times\vec{r}_y\; dA=\iint_D f(x,y)\; dA $$ and from there a switch to polar coordinates and you are done: $$ D=\{(r,\theta)\;|\; -\pi/2\le \theta \le \pi/2, 0\le r\le \cos{\theta} \} $$