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# Surface integral ex3 part 1

## Video transcript

Let's try another surface integral. And the surface that we're going to care about, s, is this shape, the outside of this shape right over here. And you can see we can kind of decompose it into three separate surfaces. The first surface is its base, which is really the filled-in unit circle down here. The second surface, which we have in blue, is you can kind of view it as the side of it. You can view it as kind of like the side of a cylinder, but the cylinder has been cut by a plane up here. And the plane that cuts it is the plane z is equal to 1 minus x. And obviously, the plane itself goes well beyond this shape. But where that plane cuts the cylinder kind of defines this shape. So the blue surface is above the boundary of the unit circle and below the plane. And then the third surface is the subset of the plane, of that purple plane, of z equals 1 minus x, that overlaps, that kind of forms the top of this cylinder. And so we can rewrite this surface integral as the sum of three surface integrals. It's the surface integral of z over s1 plus the surface integral of z over s2 plus the surface integral of z over s3. And we can just tackle each of these independently. So let's start with surface 1. And you might immediately want to start parameterizing things and all the rest, but there's actually a very fast way to handle this surface integral, especially because we're taking the surface integral of z. What value does z take on throughout this surface, throughout this filled-in unit circle right over here? Well, that surface is on the xy plane. When we're on the xy plane, z is equal to 0. So for this entire surface, z is going to be equal to 0. You're essentially integrating 0. 0 times dS is just going to be 0. So this whole thing is going to evaluate to 0. So that's about as simple as evaluating a surface integral can get. But it's always important to least keep a look out for things like that. And it'll keep you from kind of going down this wild goose chase-- that wouldn't be a wild goose chase. You would eventually get the answer anyway. But you don't have to waste so much time in parameterizing things and all the rest. Now let's tackle the other two surfaces, and we'll focus on surface 2. So the x and y-values, the valid x and y-values, are the x and y-values along the unit circle right over here. So we can really parameterize it the way that we would parameterize a traditional unit circle. So we could set x is equal to-- let's set it-- and our radius is 1, so x is equal to cosine of-- I'll use the parameter u. x is equal to cosine of u. And then let's say y is equal to sine of u. And then u is the angle between the positive x-axis and wherever we are essentially on that unit circle. So that right over there, that angle right over there is u. And so this would give us for u-- so as long as u is between 0 and 2 pi, that we're essentially going around this unit circle. So those are all the possible x and y-values that we can take on. And then the z-value is what takes us up above that boundary and gets it someplace along the surface. But this is interesting, because the z-value, it can take on obviously a bunch of different values, but it always has to be below this plane right over here. So for this z-value, I'm going to introduce a new parameter. Let's call it z v. That's the second parameter. And v is definitely going to be greater than 0. z and v, they're the same thing. They're always going to be positive, so they're definitely going to be greater than or equal to 0. But it's not less than or equal to some constant thing. This has kind of a variable roof on it. And so it's always going to be less than or equal to this plane right over here. And so we could say is less than or equal to 1 minus x. We could have said-- so we know that z is less than or equal to 1 minus x, but if we use the new parameters, v is z, and 1 minus x is the same thing as 1 minus cosine u. And so now we have our parameterization. We are ready to actually evaluate the surface integral. And to do that, first let's do the cross product. We want to figure out what dS is, and we have to take the magnitude of the cross product of the partial of our parameterization with respect to u, crossed with the partial with respect to v. Actually, let me just write all this stuff down instead of trying to do shortcuts. So dS, and this is all-- let me do it in blue still because we're talking about surface 2. dS is going to be equal to the magnitude of the cross product of the partial of our parameterization with respect to u, crossed with our partial of our parameterization with respect to v, du dv. So let's write the partial of our parameterization with respect to u. And you might say, wait, where's our parameterization? Well, it's right over here. I just didn't write it in the traditional ijk form, but I can. I could write r is equal to-- maybe I call it r2 because we're talking about surface 2. I shouldn't use that orange color because I used that for surface 1. I'll use that in a still bluish color. So r for surface 2 is equal to cosine of ui plus the sine of uj plus vk. And this is the range that the u's and the v's can take on. If I want to take the partial of r with respect to u, I would get-- let's see, the derivative of cosine u with respect to u is negative sine ui. The derivative of sine of u with respect to u is plus cosine uj. And the derivative of v with respect to u is just 0. So that's our partial with respect to u. Our partial-- let' me do this in a different color just for the sake of it. Our partial with respect to v is equal to-- well, this is going to be 0, this is going to be 0, and we're just going to have one k. So it's just going to be equal to k. And so now we're ready to at least evaluate this cross product. So we'll evaluate the cross product, and then we'll take the magnitude of that, so just this part, so just the cross product, the inside of that, before we take the magnitude. So copy and paste. So just that is going to be equal to the determinant, the 3 by 3 determinant. Let me write my components, i, j, k. And then for r sub u, we have this. Our i component is negative sine of u. Our j component is cosine of u. And it has no k component, so we'll put a 0 right there. And then r sub v, no i component, no j component, and it has 1 for its k coefficient. And so we can just evaluate this. And so first, we think about the i component. Well, we ignore that column, that row. It's going to be cosine of u minus 0. So it's going to be cosine of ui. And then we'll have minus j times-- ignore column, ignore row-- negative sine of u times 1 minus 0. So it's negative j times negative sine of u. So it's just going to be plus sine of uj. And then for the k, we have this times 0 minus this times 0. So we're just going to have 0. So this is the cross product. And if we want to take the magnitude of this, if we just take the magnitude of this-- so now we're ready to take the magnitude of this. It's going to be equal to-- the magnitude of this is going to be equal to the square root of cosine of u squared plus sine of u squared. There is no k component. And from the unit circle, this is the most basic trig identity. This is just going to be equal to 1. And so we have this part right over here just simplifies to 1, which is nice. And dS simplifies to du dv for at least this surface. And so now we are ready to attempt to evaluate the integral. But I'm almost running out of time, so I will do that in the next video.