Normal Vectors and Surfaces.

May 22, 2010

When we do integration over surfaces, we want to “cut up our surfaces” into little boxes.  This makes sense, right?  Think about an orange.


Now, if we want to find the surface area of this orange, we have to measure “little bits of area” of it.  In order to do that, we need to peel the orange.   Once we have the peel off, let’s put it down on a table and cut it up into little one inch by one inch squares.  Then we’ll have a little bit left over, but we’ll be able to estimate this reasonably.  We can cut it up into smaller pieces and get a better estimate.  We can keep cutting and cutting smaller and smaller pieces and get better and better estimates, and this is essentially the algorithm to find the exact surface area.

But what if we couldn’t peel this orange?  What if we needed to figure out what the surface area was while the skin was still on?  We could make little “boxes” on the orange and just approximate the area of those boxes, and this is generally how we do it for real surfaces.

It gets a bit weirder when we have a vector field on the surface — that is, for every point on the surface, we have an associated vector.  What happens when we want to integrate over this?  For each “slice” of our orange, we have an associated vector magnitude and direction.  How do we talk about where every particular slice is on our surface, and how do we talk about the magnitude of each of those slices?

One particularly nice way is to think in terms of normal vectors.  Given some surface, we have our partial derivatives in the x- and y-directions


and this gives us a plane: specifically, it gives us the plane that contains those two partial derivative vectors like in the picture above.  At the point where these two partials are radiating from, there are exactly two vectors which are perpendicular to both of those partials: one pointing “up” away from the two partials, and one pointing “down.”  Neither one of these is better than the other, but we usually pick some direction and stick to it.  So, for example, the following picture shows (in red) the vector I’m referring to:


and this vector is called the vector normal to the surface at the given point.  So, this will tell us exactly where and what direction every slice on our surface will be, but it does not tell us the magnitude.

Upsetting.  But, we can see “how much” of the vector field accounts for the magnitude in the direction that we’re considering by taking the dot product between the vector and the surface.  This leads to the definition of our surface integral:

\displaystyle \int_{S}v\cdot dS

with surface S and vector field v which is a function of the points on the surface.  Notice that “dS” is going to be a small slice of our surface that we’re integrating with respect to.

Okay, so now we have this “formal” definition, we need a practical definition; one that we can actually calculate things with!  We can use our normal vectors to help us.  We just need to take the dot product of our vector field with the normal vector to see how much of the magnitude goes “through” the surface.  This gives us the new definition

\displaystyle \int_{S}(v\cdot n)dS

which is slightly more useful.

Okay, so now that we know why we care about normal vectors, how do we find them?  In three dimensions, cross products take a strange form, but they turn out to be really easy to calculate if you know how to take the determinant of a 3×3 matrix by minors.  I haven’t written about this, but it’s relatively easy to learn.  If you don’t want to learn, you can just memorize the formula below.

Cross Products

Given two vectors v and u, we have that we can write them like this:

v = (v_{x}, v_{y}, v_{z})

u = (u_{x}, u_{y}, y_{z})

where the components are directions of the vectors.

So, for example

v = (2,1,3)

is 2 units in the x-direction, 1 unit in the y-direction, and 3 units in the z-direction.  In order to take the cross product of these two vectors, we make this pseudo-matrix:

\left| \begin{array}{ccc}\ i & j & k \\ u_{x} & u_{y} & u_{z} \\ v_{x} & v_{y} & v_{z} \end{array} \right|.

where the “i, j, k” on the top are the unit vectors in the x-direction, y-direction, and z-direction respectively.  This is the standard notation for these vectors.  Now, to compute the cross product, we compute the determinant of the matrix (which I do by minors):

\left| \begin{array}{cc}\ u_{y} & u_{z} \\ v_{y} & v_{z} \end{array} \right| i - \left| \begin{array}{cc} u_{x} & u_{z} \\ v_{x} & v_{z} \end{array} \right| j + \left| \begin{array}{cc} u_{x} & u_{y} \\ v_{x} & v_{y} \end{array} \right| k.

This gives us

u\times v = (u_{y}v_{z} - u_{z}v_{y})i + (u_{z}v_{x} - u_{x}v_{z})j + (u_{x}v_{y} - u_{x}v_{y})k

which is exactly the same as

u\times v = ((u_{y}v_{z} - u_{z}v_{y}), (u_{z}v_{x} - u_{x}v_{z}), (u_{x}v_{y} - u_{x}v_{y})).

And this is it.  Let’s, as an example, compute the cross product of a few vectors.


Let’s let v = (1,2,3), u = (2,-1,2), w = (1,0,0).

  1. To compute v\times u, we simply compute the matrix above.  We obtain (7,4,-5) as our normal vector.
  2. To compute v\times w, we do the same thing.  We obtain (0,3,-2).
  3. Same thing for u\times w; we obtain (0,2,1).

Try these for yourself.


  1. Prove A\times B = -B\times A.  This statement refers to the idea that the “orientation” changes when we consider vectors in a different order.  Specifically, if we get a vector pointing “outwards” when we have A\times B, we will get one pointing “inwards” when we take B\times A.
  2. On the other hand, prove that A\cdot B = B\cdot A.  Explain what this means!
  3. Prove that X\cdot (Y + Z) = X\cdot Y + X\cdot Z.
  4. Prove A\cdot (B\times C) = B(A\cdot C) - C(A\cdot B), where subtraction is matrix subtraction.  This, and problem 3, are two important equalities that deal with cross products.
  5. Given a nonzero vector v = (v_{x}, v_{y}, v_{z}), what is \sqrt{v\cdot v}?  What does this mean?  What is v\times v?  Why do you think this is?

Leave a Reply

Fill in your details below or click an icon to log in: Logo

You are commenting using your account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s

%d bloggers like this: