Vector representation of a surface integral
Different ways of representing a flux integral. Created by Sal Khan.
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- at6:00when Sal starts explaining that dr_u is just a 'differential' and not a 'partial derivative,' I am very confused by his explanation.(13 votes)
- The derivative is the ratio of the change in r (dr) with the change in u (du). The differential, r (dr) is simply the change in r. The derivative or partial, as is the case here, would be like Miles per Hour. The differential is just Miles traveled without without the concerned for the time it took to travel that distance.(15 votes)
- When Sal replaces the dS with |r_u X r_v| du dv, is he doing a change of basis there with the absolute value of the cross product vector being the Jacobian?(9 votes)
- Yes, as he explained explained earlier in the intro to surface integral video, when you do coordinate substitution for dS then the Jacobian is the cross-product of the two differential vectors r_u and r_v.
The intuition for this is that the magnitude of the cross product of the vectors is the area of a parallelogram. We take the infinite sum of these parallelograms (by taking the infinite sum of du and dv) and we get the surface area!(6 votes)
- Do you have any specific examples of calculating flux?
Very helpful article with two examples(4 votes)
- At8:21Sal is talking about dr_u and dr_v but in the integral its just the vectors r_u and r_v.... are these equivalent?(5 votes)
- No, they are NOT equivalent. r_u = ∂r/∂u, and r_v = ∂r/∂v in the integral, these are the derivatives with respect to parameters u or v.
At4:30in the video, the ∂r_u and ∂r_v came from the expression: (∂r/∂u)du × (∂r/∂v)dv. He simplifies it by saying that du cancels with ∂u, and dv cancels with ∂v. But if you do that, then you end up with something that looks like: ∂r×∂r. He just tacked on _u and _v so that you could tell them apart.(1 vote)
- At8:54, shouldn't it be over the region R? Also, at1:55, what if the magnitude of the cross product is 0? How can we cancel them out?(1 vote)
- If the magnitude was zero then you wouldn't have a normal. It was be zero giving us an integral of zero and the integral of zero is zero.(1 vote)
- If we have a polynomial to define the surface, how would this change the problem?(1 vote)
- I have watched the surface integral videos and the 3D flux videos. In Flux video at5:35he cancels out partial dx with dx however in the surface integrals he does not cancel them out to derive the formula. Can someone please explain how this thing works??
Thanks in advance(1 vote)
- Is there a surface integral that measures the amount that the vector field flows along the surface like we had in line integrals?(1 vote)
- At7:55sal says " you can think of the cross product of the partial differentials as a UNIT NORMAL VECTOR" . How could these represent a UNIT normal? I can see that they would be a tiny normal vector but NOT a unit normal vector. Please explain. Thanks(1 vote)
- omg he didnt show us how to find the n, no one knows what a random u means(0 votes)
In the last video, we figured out how to construct a unit normal vector to a surface. And so now we can use that back in our original surface integral to try to simplify a little bit, or at least give us a clue how we can calculate these things. And also, think about different ways to represent this type of a surface integral. So if we just substitute what we came up as our normal vector, our unit normal vector right here, we will get-- so once again, it's the surface integral of F dot. And F dot all of this business right over here. And I'm going to write it all in white, just so it doesn't take me too much time. So the partial of r with respect to u crossed with the partial of r with respect to v over the magnitude of the same thing, partial of r with respect to u crossed with the partial of r with respect to v. And now, we've played with ds a lot. We know that the other way to write ds-- and I gave the intuition, hopefully, for that several videos ago when we first explored what a surface integral was all about. We know that ds is-- it can be represented as the magnitude of the partial of r with respect to u crossed with the partial of r with respect to v du dv. And Obviously, the du dv, it could be written as dv du. You could write it as da, a little chunk of area and the uv plane or in the uv domain. And actually, since now this integral's in terms of uv, we're no longer taking a surface integral. We're now taking a double integral over the uv domain. So you could say kind of a region in uv. So I'll say R to say that's it's a region in the uv plane that we're now thinking about. But there's probably a huge-- or there should be, or I'm guessing there's a huge simplification that's popping out at you right now. We're dividing by the magnitude of the cross product of these two vectors and then we're multiplying by the magnitude of the cross product of these two vectors. Those are just scalar quantities. You divide by something and multiply by something. Well, that's just the same thing as multiplying or dividing by 1. So these two characters cancel out, and our integral simplifies to the double integral over that region, the corresponding region in the uv plane, of F-- of our vector field F dotted with this cross product. This is going to give us a vector right over here. That's going to give us a vector. It gives us actually a normal vector. And then when you divide by its magnitude, it gives you a unit normal vector. So this, you're going to take the dot product of F with r, the partial of r with respect to u crossed with the partial of r with respect to v du dv. Let me scroll over to the right a little bit, du dv. And we'll see in the few videos from now that this is essentially how we go about actually calculating these things. If you have a parameterization, you can then get everything in terms of a double integral, in terms of uv this way. Now, the last thing I want to do is explore another way that you'll see a surface integral like this written. It all comes from, really, writing this part in a different way. But it hopefully gives you a little bit more intuition of what this thing is even saying. So I'm just going to rewrite. I'm going to rewrite this chunk right over here. I'm just going to rewrite that chunk. And I'm going to use slightly different notation because it will hopefully help make a little bit more sense. So the partial of r with respect to u I can write as the partial of r with respect to u. And we're taking the cross product. Let me make my u's a little bit more u-like so we confuse them with v's. And we're taking the cross product of that with the partial of r with respect to v. So very small changes in our vector-- in our parameterization right here, our position vector given a small change in v. Very small changes in the vector given a small change in u. And then we're multiplying that times du dv. We're multiplying that times du dv. Now, du and dv are just scalar quantities. They're infinitesimally small. But for the sake of this argument, you can just view-- they're not vectors, they're just scalar quantities. And so you can essentially include them-- if you have the cross product. If you have a cross b times some scalar value-- I don't know, x, you could rewrite this as x times a cross b, or you could write this as a cross x times b, because x is just a scalar value. It's just a number. So we could do the same thing over here. We can rewrite all of this business as-- and I'm going to group the du where we have the partial-- or with respect to u in the denominator. And I'll do the same thing with the v's. And so you will get the partial of r with respect to u times du, times that scalar. So that'll give us a vector. And we're going to cross that. We're going to cross that with the partial of r with respect to v dv. Now, these might look notationally like two different things, but that just comes from the necessity of when we take partial derivatives to say, oh, no, this vector function is defined-- it's a function of multiple variables and this is taking a derivative with respect to only one of them. So this is, how much does our vector change when you have a very small change in u? But this is also an infinitesimally small change in u over here, we're just using slightly different notation. So for the sake of-- and this is a little bit loosey-goosey mathematics, but it will hopefully give you the intuition for why this thing could be written in a different way. These are essentially the same quantity. So if you divide by something and multiply by something, you can cancel them out. If you divide by something and multiply by something, you can cancel them out. And all you're left with then-- all you're left with is the differential of r. And since we lost the information that it's in the u-direction, I'll write here, the differential of r in the u-direction. I don't want to get the notation confused. This is just the differential. This is just how much r changed. This is not the partial derivative of r with respect to u. This right over here is, how much does r change given per unit change, per small change in u? This just says a differential in the direction of-- I guess as u changes, this is how much that infinitely small change that just r changes. This isn't change in r with respect to change in u. And we're going to cross that. Now, we're going to cross that with the partial of r, the partial of r in the v-direction. Now, this right over here, let's just conceptualize this. And this goes back to our original visions of what a surface integral was all about. If we're on a surface-- and I'll draw surface. Let me draw another surface. I won't use the one that I had already drawn on. If we draw a surface, and for a very small change in u-- and we're not going to think about the rate. We're just thinking about kind of the change in r. You're going in that direction. So if that thing looks like this, this is actually kind of a distance moved on the surface. Because remember, this isn't the derivative. This is the differential. So it's just a small change along the surface, that's that over there. And that this is a small change when you change v. So it's also a change along the surface. When you take the cross product of these two things, you get a vector that is orthogonal. You get a vector that is normal to the surface. So it is normal to the surface and its magnitude-- and we saw this when we first learned about cross products. Its magnitude is equal to the area that is defined by these two vectors. So its magnitude is equal to area. So in a lot of ways, you can really think of it-- you really could think of it as a unit normal vector times ds. And so the way that we would, I guess notationally do this, is we can call this-- because this is kind of a ds, but it's a vector version of the ds. Over here, this is just an area right over here. This is just a scalar value. But now, we have a vector that points normally from the surface, but its magnitude is the same thing as that ds that we were just talking about. So we can call this thing right over here, we can call this ds. And the key difference here is this is a vector now. So we'll call it ds with a little vector over it to know that this thing. This isn't the scalar ds that is just concerned with the area. But when you view things this way, we just saw that this entire thing simplifies to ds. Then our whole surface integral can be rewritten. Instead of writing it like this, we can write it as the integral or the surface integral-- those integral signs were too fancy. The surface integral of F dot. And instead of saying a normal vector times the scalar quantity, that little chunk of area on the surface, we can now just call that the vector differential ds. And I want to make it clear, these are two different things. This is a vector. This is essentially what we're calling it. This right over here is a scalar times a normal vector. So these are three different ways of really representing the same thing. And in different contexts, you will see different things, depending on what the author of whoever's trying to communicate is trying to communicate. This right over here is the one that we'll use most frequently as we actually try to calculate these surface integrals.