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Constructing a unit normal vector

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

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  • mr pink red style avatar for user John
    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'...
    (15 votes)
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    • blobby green style avatar for user André
      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}
      (20 votes)
  • area 52 yellow style avatar for user Timothy Zhou
    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?
    (8 votes)
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    • leaf blue style avatar for user Stefen
      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:
      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.
      (12 votes)
  • old spice man green style avatar for user John Nolen
    What's the difference between the way Sal computed the unit normal and the unit normal obtained by calculating grad(S)/|grad(S)| ?
    (9 votes)
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  • blobby green style avatar for user Hunter Stufflebeam
    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
    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.
    you would get

    ∬ |∂r(s) X ∂r(t)|

    which might just be

    ∬ |dS|

    ?? 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.
    (6 votes)
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  • leaf green style avatar for user Steve Dick
    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.
    (5 votes)
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  • blobby green style avatar for user fredd.mike
    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?
    (4 votes)
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  • male robot hal style avatar for user Lucas Nakashita
    Do you have any videos about surface vectors for the sum of two vectors?
    (4 votes)
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  • leaf green style avatar for user Benjamin.S.Kjaer
    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?
    (3 votes)
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    • mr pants teal style avatar for user arpithaprasad
      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.
      (2 votes)
  • blobby green style avatar for user Niclas Samuelsson
    Are partials always parallel to each other?
    (3 votes)
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    • leaf red style avatar for user My Real Name
      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.

      I hope that was helpful!
      (2 votes)
  • orange juice squid orange style avatar for user FishHead
    Huh, I thought the cross product was the area of the parallelogram created by the two vectors. What am I missing here?
    (2 votes)
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    • blobby green style avatar for user ixoyzzz
      So this is a couple months late, but I'll answer anyway. It's actually the magnitude of the cross product that gives the area of the parallelogram. This means that the cross product, by itself, is a vector. Specifically, the cross product of two vectors gives a third vector that is perpendicular to both previous vectors. Think about it in terms of 3D space, like the hand that Sal drew on screen.

      Hence why it would be useful here: the vectors given by the partial derivatives are essentially both 'tangent' to the surface at a given point. A vector that would be perpendicular to both of those vectors would be perpendicular, or normal, to the surface at that point. That's why we take the cross product of the partial derivatives.
      (4 votes)

Video transcript

Now that we hopefully have a conceptual understanding... ...of what a surface integral like this COULD represent, ...I want to think about how we can actually construct... ...a unit vector... ...a unit normal vector, at any point on the surface. And to do that, I will assume... ...that our surface can be parametrized... ...by the position vector function, r... ...and r is a function of two parameters. It's a function of u, and it is a function of v. You give me a u and a v and... ...that will essentially specify... ...a point on this two-dimensional surface right over here. It could be bent, so it kind of exists in three-dimensional space... But a u and a v will specify a given point on this surface. Now, let's think about what the directions of r... ...the partial of r with respect to... ...the partial of r with respect to u looks like... ...and what the direction of the partial of r... ...the partial of r with respect to v looks like. So let's say that we're at some... ...we're at some point. We're at some point, (u,v). So for some (u,v), if you'd find the position vector... ...it takes us to that point on the surface right over there. Now let's say that we increment u just a little bit. And as we increment u just a little bit, ...we're going to get to another point on our surface, ...and let's say that other point on the surface... ...is right over there. So what would r... What would this r_u vector look like? Well its magnitude is essentially going to be.. ...dependent on how fast it's happening, ...how fast we're moving towards that point, ...but its direction is going to be in that direction. It's going to be towards that point. It's going be along the surface. We're going from one point on the surface to another. It's essentially going to be tangent to the surface at that point. And I could draw a little bit bigger. It would look something like that. r... r_u. So I just zoomed in right over here. Now let's go back to this point. And now let's make v a little bit bigger. And let's say if we make v a little bit bigger, ...we go to this point right over here. So then our position vector, r, would point to this point. And so what would r_v look like? Well its magnitude, once again, would be dependent on... ...how fast we're going there, but the direction is what's interesting. The direction would also be tangential to the surface. We're going from one point on the surface to another... ...as we change v. So r_v might look something like that. And they're not necessarily... These two aren't necessarily perpendicular to each other. In fact, the way I drew them, they're not perpendicular. So r_v is like this, but they're both tangential to the plane. They're both essentially telling us, right at that point, ...what is the tangent? What is the slope in that... ...in the u direction, or in the v direction? Now, this is... When you have two... When you have two vectors that are... ...that are tangential to the plane, ...and they're not the same vector, these are actually... ...already specifying... ...these are already kind of determining a plane. And so you can imagine a plane that looks something like this. If you took linear combinations of these two things, ...you would get a plane that both of these would lie on. Now, we've done this before, but I'll re-visit it. What happens when I take the cross product... ...of r_u and r_v? What happens when I take the cross product? Well first, this is going to give us another vector. It's going to give us a vector... ...a vector that is perpendicular to both... ...to r_u AND r_v. Or another way to think about it is... ...this plane, that these... ...when you take the cross product... ...this plane is essentially a tangential plane... ...to the surface. And if something is going to be perpendicular... ...to both of these characters, ...it's going to have to be normal to them... ...or, it's definitely going to be perpendicular... ...to both of them, but it's going to be normal... ...to this plane. Which is essentially going to be... ...perpendicular to the surface itself. So this right over here... ...is going to be A normal vector. This is... I'll write it... Well, let me just write it this way. This is A normal vector. I'm not saying THE unit normal... I'm not saying THE normal vector, 'cause you have... ...you could have different normal vectors of... ...different magnitudes. This is A normal vector, when you take the cross product. And we can even think about what direction it's pointing in. And so when you have "something" cross "something else"... ...the easiest way I remember how to do it is... ...you point your left thumb... Oh, sorry. You point your RIGHT thumb... ...in the direction of the first vector... So, in this case, r_u. So let me see if I can... ...if I can draw this. I'm literally looking at my hand and trying to draw it. So you put your right thumb... -- so this is a right-hand rule, essentially -- ...in the direction of the first vector... ...and then you put your index finger in the direction of... ...the second vector... ...right over here. So this is the second vector. So that's the direction of my index finger. So my index finger is going to look something like... ...that. And then you bend... ...you bend your middle finger inward... ...and that will tell you the direction of the cross product. So if I bend my middle finger inward, ...it will look something... ...it will look something like this. And then of course, my other two fingers are just going to be... ...folded in like that, and they're not really relevant. But my other two fingers and my hand looks like that. And so that tells us the direction. The direction is going to be like that. It's going to be upward-facing. That's important, because you have normal vectors. One could... Or there's two directions of "normalcy," I guess you could say. One is going out like that... ...outwards... ...or I guess... ...in the upward direction... ...one would be going downwards, ...or going -- I guess you could say -- into the surface. But the way I have set it up right now, ...this would be going outwards. It would be A... It would be A normal vector... ...to the surface. Now, in order to go from A normal vector... ...to the UNIT normal vector, ...we just have to normalize it. We just have to divide this... ...by its magnitude. So now we have our drumroll. The unit vector... And it's going to essentially be... It's going to be a function of u... It's going to be... ...a function of u and v. You give me a u or a v... ...and I'll give you a... ...that unit normal vector. It is going to be equal to... ...the partial of r... ...the partial of r with respect to u... ...crossed with... ...the partial of r with respect to v... That just... Now that gives us A normal vector, but it hasn't been normalized. So we want to divide... ...by the magnitude... We want to divide by the magnitude... ...of the exact same thing. r_u crossed with r_v. And we're done! We have constructed a unit normal vector. And in future videos, we'll actually do this with concrete examples.