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## Multivariable calculus

### Course: Multivariable calculus>Unit 4

Lesson 13: Flux in 3D

# Constructing a unit normal vector

Deriving a unit normal vector from the surface parametrization. Created by Sal Khan.

## Want to join the conversation?

• Why is Sal using 'u' and 'v' parameters when he can use 'x' and 'y' because we are in Cartesian coordinate system!?
It's the same thing and I get confused while using 'u' and 'v'... • In this case, the surface is parameterized as {x,y,f(x,y)}, so yes, he could use 'x' and 'y'
But he was explaining for a general case, where the parameterization would not necessarily be like that, i could be a sphere {cos u*cos v,cos u*sin v, sin u}
• Does the surface of integration have to be orientable? For instance, the Mobius strip has a half-twist in it, and it's not possible to choose a consistent direction for the normal vector. Does that mean we can't integrate on the Mobius strip? • It is possible - but you have to be very very careful - and you will need a lot more sophisticated tools than you will find here on Khan, and also, typically, at least finishing your undergraduate level math program, though you may have the young clever mind to tackle it now. Here is the basic idea behind one method:
https://en.wikipedia.org/wiki/Orientability#Orientable_double_cover
Other options are cutting and/or partitioning - but each introduces new problems (like boundary values that don't cancel cleanly) that if not taken into account and dealt with, will not yield a correct result.
• What's the difference between the way Sal computed the unit normal and the unit normal obtained by calculating grad(S)/|grad(S)| ? • When one uses the cross product of ∂r/∂s X ∂r/∂t to find a normal vector, one crosses the partials of r to ensure that since both ∂r/∂s and ∂r/∂t are tangent to the surface, the normal vector is ensured to be normal to the surface at that point, correct? When we then assured that the normal vector in question was a unit normal vector, we divided by the magnitude of ∂r/∂s X ∂r/∂t,
|∂r/∂s X ∂r/∂t|
However, when we derived the surface element dσ, we ended up using the parallelogram area defined by
|(∂r/∂s)ds X (∂r/∂t)dt|= |∂r/∂s X ∂r/∂t|dsdt.
What I do not understand is why we used the cross product of the partials of r when we defined the normal unit vector, leaving them in the limit definition of the partial derivative, but when we defined dσ we multiplied the partial derivative by ds or dt to represent
|(∂r/∂s)ds X (∂r/∂t)dt| as the differences of the vectors r(s+ds,t) and r(s,t) crossed with the difference of r(s,t+dt) and r(s,t). Shouldn't these two vector areas be the same? The best solution I can think of at the moment is that if you define the unit normal vector like we did in the cross product defining dσ, you would get

(∂r/∂s X ∂r/∂t)dsdt
------------------------
|∂r/∂s X ∂r/∂t|dsdt.

which should be equal to

(∂r/∂s X ∂r/∂t)
-----------------
|∂r/∂s X ∂r/∂t|

if you are allowed to play with the differentials like that...
That way,
(∂r/∂s X ∂r/∂t)dsdt
∬ F∙ ------------------------ (|∂r/∂s X ∂r/∂t|dsdt. )
R |∂r/∂s X ∂r/∂t|dsdt.

would still be equivalent to

∬ F∙(∂r/∂s X ∂r/∂t)dsdt
R
which is what we want.

Second question (sort of)
If you then represent that cross product in its differential form, you would get
∬ F∙(∂s(s) X ∂r(t)) --------> ∬ F∙dS
R S
If you did the same thing with

∬ |∂r/∂s X ∂r/∂t|dsdt.
R
you would get

∬ |∂r(s) X ∂r(t)|
R

which might just be

∬ |dS|
S

?? Can you do this, and does it actually make any sense?

I've looked elsewhere on the internet for derivations of theses integrals, but I cannot figure out why these two areas are represented differently, and what all that rearranging might do. • Why is Sal using the thumb for crossing two(2) vectors? I've always seen index and middle fingers used for 1st and 2nd vectors respectively and imagine curling the rest of your hand in the direction that you would turn a screwdriver from 1st vector to line up with the 2nd vector and that your thumb would point in the direction of the resulting cross product. Also be sure the order of the vectors 1 and 2 are not switched because cross product is not commutative. • OK
This answers my question about the direction of a surface. But still I'm bothered about the arbitrariness of Ru and Rv. If you cross Rv into Ru you get negative the cross of Ru into Rv how do you know the correct order. My reason for asking has to do with what direction is taken for the surface when evaluating the magnetic flux? • Do you have any videos about surface vectors for the sum of two vectors? • How do I know the Unit Normal vector is the one pointing outwards and not inwards? I guess my answer will be just the negativ of the right one if I do it wrong? • Well if you calculate the curl correctly then the normal vector will point where it should. If you want to see which way its pointing, then you should try this: Suppose you have two vectors a and b and you want to find a x b. Then, place your right palm on the first vector ( which is vector a) perpendicular to the plane containing your two vectors (note that you can do this in two ways, above the plane or below the plane). Then, curl your palm inwards towards the second vector (vector b), with your thumb up (note that this can be done in one way only, either with your hand above the plane or below it). The direction in which your thumb points gives you the direction of the curl of the two vectors.
• Are partials always parallel to each other? • Based on the way you phrased the question, the short answer is no. Say, for example, you defined a surface S with parameters X and Y. The partial of the height Z with respect to X, in most cases, will almost never be pointing the same direction as the partial of Z with respect to Y. If they did, then the parallelogram that they'd form would have no area. You can also tell that they wouldn't be parallel based on the value of the partial derivative; if two partials don't have the same value, that tells you that their "rise over run" will be different, so they wouldn't be parallel.

But I think what your question meant was whether the partials of a surface are always parallel to the surface itself. In that case, the answer is yes. Think about it in 2D: the derivative tells you the slope of a tangent line - the slope of a line parallel to a curve at a specific point. Now in 3D, the partial derivative will tell you the slope of a tangent line in some direction on a surface. So by definition the surface and its partials are always tangent to each other. 