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# Introduction to the surface integral

## Video transcript

In the last video, we finished off with these two results. We started off just thinking about what it means to take the partial derivative of vector-valued function, and I got to these kind of, you might call them, bizarre results. You know, what was the whole point in getting here, Sal? And the whole point is so that I can give you the tools you need to understand what a surface integral is. So let's just think about, let's draw the st plane, and then see how it gets transformed into this surface r. So let's do that. Let's say that is the t-axis, and let's say that that is the s-axis, and let's say that our vector-valued function, our positioned vector-valued function, is defined from s's between a and b, I'm just picking arbitrary boundaries, and between t being equal to c and d. So the area under question, if you take any t and any s in this rectangle right here, it will be mapped to part of that surface. And if you map each of these points, you will eventually get the surface r. So let me draw r in 3 dimensions. A surface in 3D. So that is our x-axis, that is our y-axis, and then that is the z-axis. And just as a bit of a reminder, it might look something like this. If we were to, this point right here, where s is equal to a, and t is equal to c, remember, we're going to draw the surface indicated by the position vector function s, r of s and t. So this point right here, when s is a and t is c, maybe it maps to, I'm just, you know, that point! Right there. When you take a and c, and you put it into this thing over here, you're just going to get the vector that points at that. So you could say, it'll give you a position vector that'll point right at that position, right there. And then, let's say that this line, right here, if we were to hold s constant at a, and if we were to just vary t from c to d, maybe that looks something like this. I'm just drawing some arbitrary contour there. Maybe if we hold t constant at c, and vary s from a to b, maybe that'll look at something like that. I don't know. I'm just trying to show you an example. So this point right here would correspond to that point right there, when you put it into the vector-valued function r, you would get a vector that points to that point, just like that. And this point right here in purple, when you evaluate r of s and t, it'll give you a vector that points right there, to that point over there, and we could do a couple of other points, just to get an idea of what the surface looks like, although I'm trying to keep things as general as possible. So maybe, let me do it in this bluish color. This, if we hold p at d and vary s from a to b, we're going to start here. This is when t is d and s is a. And when you vary it, maybe you get something like that. I don't know. So this point right here would correspond to a vector that points to that point, right there, and then finally this line, or this, if we hold s at b and vary t between c and d, we're going to go from that point, to that point. So it's going to look something like this-- oh, sorry, we're going to go from this point to that point. We're holding s at b, varying t from c to d, maybe it looks something like that. So our surface, we went from this nice rectangular area in the ts plane, and it gets transformed into this wacky-looking surface. We could even draw some other things right here. Let's say we get some arbitrary value. Let me pick a new color, I'll do it in white. Or a new noncolor. And let's say if we hold s at that constant value and we vary t, maybe that will look for something like this. Maybe that we'll look something like, well, I don't know. Maybe it'll look something like that. So you get an idea of what the surface might look like. Now, given this, I want to think about what these quantities are. And then when we visualize what these quantities are, we'll be able to kind of use these results of the last videos to do something that I think will be useful. So let's say that we pick arbitrary s and t. So this is the point, let me just we pick it right here. That is the point s, t. s comma t. If you were to put those values in here maybe it maps to, and I want to make sure I'm consistent with everything I've drawn, maybe it maps to this point right here. Maybe it maps to that point right there. So this point right here, that is r of s and t. For a particular s and t. I mean, I could put little subscripts here, but I want to be general. I could call this a, well, I already used a and b. I could call this x and y, this would be r of x and y. It would map to that point, right there. So that right there, or that right there. Now let's see what happens if we take, if we move just in the s direction. So this is we could do that as s. Now let us move forward by a differential, by a super small amount of s. So this right here, let's call that a s plus a super small differential in s. That's right there. So that point. Let me do that in a better color, in this yellow. So that point right there is the point s plus my differential of s. I could write delta s, but I wanted a super small change in s, comma t. And what is that going to get mapped to? Well, if we apply these two point in r, we're going to get something that maybe is right over there. And I want to be very clear. This right here, that is r of s plus ds comma t. That's what that is. That's the point when we just shift s by a super small differential, this distance here, you can view as ds. It's a super small change in s. And then when we map it or transform it, or put that point into our vector-valued function, let me copy and paste the original vector-valued function, just so we have a good image of what we're talking about this whole time. Let me put it right down there. So just to be clear what's going on, when we took this little blue point right here, this s and t, and we put the s and t values here, we get a vector that points to that point on the surface, right there. When you add a ds to your s-values, you get a vector that points at that yellow point right there. So going back to the results we got in the last presentation, or the last video, what is this? r of s plus delta s, or r of s plus ds, the differential of s, t, well, that is that, right there. That is the vector that points to that position. This right here is a vector that points to this blue position. So what is the difference of those two vectors? And this is a bit of basic vector math, you might remember. The difference of these two vectors, head to tails. The difference of these two vectors is going to be this vector. If you subtract this vector that vector, you're going to get that vector, right there. A vector that looks just like that. So that's what this is equal to, that vector. And it makes sense. This blue vector plus the orange vector, this vector, right here, plus the orange vector, is equal to this vector. It makes complete sense. Heads to tails. So that's what that represents. Now let's do the same thing in the t direction. I'm running out of color. I'll go back to the pink, or maybe the magenta. So we had that s and t. Now if we go up a little bit, in that direction, let's say that that is t, so this is the point s, t plus a super small change in t, that's that point right there. This distance right there is dt, you can view it that way. If you put s and t plus dt into our vector value function, what are you to get? You're going to get a vector that maybe points to this point, right here. Maybe I'll draw it right here. Maybe it points to this point, right here. A vector that points right there. So that will be mapped to a vector that points to that position, right over there. Now by the same argument that we did on the s-side, this point, or the vector that points to that, that is r of st plus dt. That is the exact same thing as that right there, and of course, this, we already saw. This is the same thing as that over there. So what is that vector minus this blue vector? The magenta vector minus the blue vector? Well, once again, this is hopefully a bit of a review of adding vectors. It's going to be a vector that looks like this. I'll do it in white. It's going to be a vector that looks like that. And you can imagine, if you take the blue vector plus the white vector, the blue vector plus this white vector is going to equal this purple vector. So it makes sense, if the purple vector minus the blue vector is going to be equal to this white vector. So something interesting is going on here. I have these two, this is a vector that is kind of going along this parameterized surface, as we changed our s by a super small amount. And then this is a vector that is going along our surface if we change our t by a super small amount. Now, you may or may not remember this, and I've done several videos where I showed this to you. But the magnitude, if I take 2 vectors, and I take their cross product, so if I take the cross product of a and b, and I take the magnitude of the resulting vector-- remember, when you take the cross product, you get a third vector that is perpendicular to both of these. But if you were just to take the magnitude of that vector, that is equal to the area of a parallelogram, defined by a and b. What do I mean by that? Well, if that is vector a and that is vector b, that's a and that is b, if you were to just take the cross product of those two, you're going to get a third vector. You're going to get a third vector that's perpendicular to both of them, it kind of popped out of the page. That would be a cross b. But the magnitude of this, so if you just take a cross product, you're going to get a vector. But then if you take the magnitude of that vector, you're just saying, how big is that vector, how long is that vector? That's going to be the area of the parallelogram defined by a and b. And I've proved that in the linear algebra videos, maybe I'll prove it again in this. I mean it's because-- well, I won't go into that in detail. I've done it before, don't want to make this video too long. So the parallelogram defined by a and b, you just imagine a, and then take another kind of parallel version of a, is right over there, and another parallel version of b is right over there. So this is the parallelogram defined by a and b. So, going back to our surface example, if we were to take the cross product of this orange vector and this white vector, I'm going to get the surface area, I'm going to get the area of the parallelogram, defined by these two vectors. So if I take the parallel to that one, it will look something like this, and then a parallel to the orange one it will look something like that. So if I take the cross product of that and that, I'm going to get the area of that parallelogram. Now you might say, this is a surface, you're taking a straight-up parallelogram, but remember, these are super small changes. So you can imagine, a surface can be broken up into super small changes in parallelograms, or infinitely many parallelograms. And the more parallelograms you have, the better approximation of the surface you're going to have. And this is no different than when we first took integrals. We approximated the area under the curve with a bunch of rectangles. The more rectangles we had, the better. So let's call this little change in our surface d sigma, for a little change, for a little amount of our surface. And we could even say that, you know, the surface area of the surface will be the infinite sum of all of these infinitely small d sigmas. And there's actually a little notation for that. So surface area is equal to, we could integrate over the surface, and the notation usually is a capital sigma for a surface as opposed to a region or-- so you're integrating over the surface, and you do a double integral, because you're going in two directions, right? A surface is a kind of a folded, two-dimensional structure. And you're going to take the infinite sum of all of the d sigmas. So this would be the surface area. So that's what a d sigma is. Now we just figured out, we just said, well, that d sigma can be represented, that value, that area, of that little part of the surface, of that parallelogram, can be represented as a cross product of those two vectors. So let me write here. And this is not rigorous mathematics. The whole point here is to give you the intuition of what a surface integral is all about. So we can write that d sigma is equal to the cross product of the orange vector and the white vector. The orange vector is this, but we could also write it like this. This was the result from the last video. I'll write it in orange. So the partial of r with respect to, I'm running out of space, with respect to s, ds, and it's going-- well, d sigma is going to be the magnitude of the cross product, not just the cross product. The cross product by itself will just give you a vector, and that's going to be useful when we start doing vector-valued surface integrals, but just think about it this way. So this orange vector is the same thing is that. And we're going to take the cross product of that with this white vector. This white vector is the same thing as that, which we saw, which was the same thing as this. The partial of r with respect to t, dt. And we saw, if we take the magnitude of that, that's going to be equal to our little small change in area, or the area of this little parallelogram over here Now, you may or may not remember that if you take these, so let's just be clear. This and this, these are vectors, right? When you take the partial derivative of a vector-valued function, you're still getting a vector. This ds, this is a number. That's a number and that's a number. And you might remember, when we, in the linear algebra or whenever you first saw taking cross products, taking the cross product of some scalar multiple. You can take the scalars out. So if we take this number and that number, we essentially factor them out of the cross product. This is going to be equal to the magnitude of the cross product of the partial of r, with respect to s, crossed with the partial of r with respect to t, and then all of that times these two guys, over here. Times ds and dt. So I wrote this here, hey, maybe our surface area, if we were to take the sum of all of these little d sigmas, but there's no obvious way to evaluate that. But we know that all of the d sigmas, they're the same thing as if you take all of the ds's and all of the dt's. So you take all of the ds's, all of the dt's. So this is a ds times a dt, right? A ds times a dt, ds times a dt is right there. If we multiply this times the cross product of the partial derivative, this times this is going to give us this area. So if we summed up all of this times this, or this times this, if we summed them up over this entire region, we will get all of the parallelograms in this region. We will get the surface area. So we can write-- I know this is all a little bit convoluted. And you need to kind of ponder a little bit. Surface intervals, at least in my head, are one of the hardest things to really visualize, but it all hopefully makes sense. So we can say that this thing right over here, the sum of all of the little parallelograms on our surface, or the surface area, is going to be equal to, instead of taking the sum over the surface, let's take the sum of all the ds times dt's over this region right here. And of course, we're also going to have to take this cross product here. And we know how to do that. That's a double integral. for going to take the double integral over this, we could call it this region, or this area, right here. That area is the same thing as that whole area, right over there, of this thing. I'll just write it in yellow. Of the cross product of the partial of r with respect to s, and the partial of r with respect to t. ds and dt. And so you literally just take, and it seems very convoluted in how you're going to actually evaluate it, but we were able to express this thing called a surface-- well, this is a very simple surface integral-- in something that we can actually calculate. And in the next few videos, I'm going to show you examples of actually calculating it. Now, this right here will only give you the surface area. But what if, at every point here-- so over here what we've done in both of these expressions, is we're just figuring out the surface area of each of these parallelograms and then adding them all up. That's what we're doing. But what if, associated with each of those little parallelograms, we had some value, where that value is defined by some third function f of x, y, z? So every parallelogram, it's super small, it's around a point, you can say it's maybe the center of it, doesn't have to be the center. But maybe the center of it is at some point in three-dimensional space, and if you use some other function, f of x, y, and z, you'll get the value of that point. And what we want to do is figure out what happens if for every one of those parallelograms, we were to multiply it times the value of the function at that point? So we could write it this way. So this is where, you can imagine, the function is just one. We're just multiplying each of the parallelograms by one. But we could imagine we're multiplying each of the little parallelograms by f of x, y, and z, d sigma, and it's going to be the exact same thing, where this is each of the little parallelograms, we're just going to multiply it by f of x, y, and z there. So we're going to integrate it over the area, over that region, of f of x, y, and z, and then times the magnitude of the partial of r with respect to f, crossed with the partial of r with respect to t, ds, dt. And of course, we're integrating with respect to s and t. Hopefully we can express this function in terms of s and t, and we should be able to, because we have a parameterization there. Wherever we see an x there, it's really x is a function of s and t. y is a function of s and t. z is a function of s and t. And this might look super convoluted and hard. And the visualizations for this, of why you'd want to do this, it has applications in physics. It's a little hard to visualize. It's easier just to visualize the straight-up surface area. But we're going to see the next few videos that it's a little hairy to calculate these problems, but they're not too hard to do. That you just kind of have to stick with them.