Multidimensional Integration 8 | Example of Surface Integral [dark version]

Añadido: 2026-05-15

[00:00:00]Hello and welcome back to multi-dimensional integration, the video series where we learn about a lot of different concepts for integrals.

[00:00:09]Indeed, in today's part 8, we will continue our discussion about the surface integral on hypersurfaces.

[00:00:16]In particular, I will show you how this integration works on the two-dimensional sphere. But as always before we start with that I first want to thank all the nice people who support this channel on steady here on YouTube or via other means and please use the link in the description to download the additional material for the videos. Okay. And then I would say let's immediately recall the definition of the surface integral and the concept works in any dimension. So let's say that we have a generalized surface which we call m and now the surface is n dimensional which means it's parameterized with a map fi from rn to rm. Furthermore we can look at a function that is defined on m which means it maps into r and it's called the function f and this one is the one we want to integrate over the whole surface. So in the formula you would write integral of fd sigma over m. And now as we have discussed in the last video this integral can be defined by using the parameterization phi which means we just integrate over u and then we look at the composition f after phi.

[00:01:27]So this is what we integrate in rn which means we just have an ordinary volume integral in n dimensions. However, here it's quite important that we incorporate the stretching of area that the parameterization phi makes and this can be done by using the so-called crian determinant. The name already suggests that it comes from a determinant and this is what we have done in the last video. So just as a quick reminder what you calculate is a determinant of a square matrix which is of size n +1 * n +1. Therefore, for this formula, it's important that we have a hypers surface, which means we map Rn to Rn + one. And here I will not talk about generalizations of that because for that I have my manifold series. But I will definitely tell you more about this definition here because it's quite important. And there you see the standard partial derivatives of phi give us n tangent vectors. And by assumption they should span an n dimensional space which means we can also find a non-vanishing normal vector n. And for that reason we can also normalize it. So we have a unit normal vector. So this is what we need in general for calculating this integral. But now it turns out that for a two-dimensional surface the whole thing gets simpler. In other words in the classical case of a two-dimensional surface we can reduce the whole integral. And that's simply because in R3 we have the crossroduct the vector product to calculate a normal vector. In fact you can just take the first tangent vector here and form the cross productduct with the second tangent vector. So this gives us the orthogonal direction to both vectors which should be our normal vector. Hence the only thing we need to do now is to divide by the norm to get a unit vector. This means here we just have the standard ukidian norm in R3. So finding our capital n here is quite simple and also calculating this determinant gets much simpler in this case. In order to see that let's simply do it. Let's write down the calculation.

[00:03:38]So here we have a determinant of a 3 * 3 matrix which can be calculated with the inner product together with the cross productduct. So you found the crossroduct of two vectors and put that into the inner product with the third vector. So for example this formula here is correct. Of course you have to use the standard inner product in R3 and then this formula is really easy to show. However, now we can just put in what we already know of NU and then you see that we actually have the same thing left and right in the inner product. In both entries we have the crossroduct of the two tangent vectors and we also divide by the length of it. So what we actually have is the norm squared divided by the norm itself. Therefore what remains here is just the norm to the power 1. This means the determinant of kam that we need in our calculation of the integral is just given by the standard ukidian norm of the crossroduct of the tangent vectors. So actually quite easy to calculate and that's what I want to show you now with an example.

[00:04:44]It's a standard one but a really important one and that's why I have to show it. And as already mentioned before this will be the two-dimensional sphere in R3. And the common name for this two-dimensional sphere is just S2. And indeed here we have a little problem namely just one parameterization fi cannot cover the whole sphere. However, this is not really a problem here because as we already know for the integration sets of measure zero don't matter at all. So the idea is simply that we cover almost everything on the sphere. And you might already know that this can be done by considering two different angles and one goes from 0 to 2 pi and the other one just from 0 to pi. The idea is quite simple because the first angle you can visualize here on the equator plane. So it just goes around in a full circle here. And then the other angle just starts here at the north pole and can go around in a half circle around the whole globe. And then together you should see that we almost hit all the points on the sphere. We just miss some line here on the surface because we have open intervals. But as already mentioned, this does not matter at all for our integration. So what we have here are two parameters which are angles on the sphere but we still call them u1 and u2 and now the two parameters are mapped to a three-dimensional vector. So we have three components and you already know that the s and cosine functions have to come in. Indeed for the second one the polar angle we already know that the third component the z coordinate is given by the cosine of u2. But now this also implies that we have the sign of U2 in the two other components. And there I should tell you that this polar angle is often called theta. On the other hand, our first angle is often denoted by phi and called the aimethal angle. And this one does not care about the z coordinate. So we just have to rotate x and y. So we have the cosine of u1 and the s of u1 and y. And that's already it. This is how we can parameterize our unit sphere in R3. So definitely something you can remember here. But you should also know that there are different possibilities for the angles as well. So if you see some other combination of s and cosine functions, then this might be a different parameterization of the sphere. So very well now we have everything to calculate the tangent vectors which means we have to form the partial derivatives of phi.

[00:07:24]And obviously it makes sense to write the first partial derivative as deli delu1.

[00:07:30]Hence our u2 is a constant. So we definitely get zero in the third component. And we can immediately do the other calculations which means in the first component we have minus s of u1 and in the second the cosine of u1. So that's already it. And then let's go to the second tangent vector. And there in the first two components we just need the derivative of sine of u2. So we just get the cosine of u2 and in the last component we get minus the s of u2. So I think we don't have any issues here. And then in the next step we can go to the crossroduct of both vectors. So this is definitely something you should know how to do very quickly. So just calculate the combinations. In the first component, we just have minus sin u2 ^ 2 time the cosine of u1. And in order to make it more compact, I will write the power directly to the function here.

[00:08:24]This is helpful because in the second component, we also have sine of u2 squared. The only difference there is that it is combined with the sine function of u1. And then finally we can go to the last component which is a combination of two terms. And in the first part we have the s of u1^ squar and then in the second part we have the cosine of u1^ 2. So it's a little bit longer but we can actually combine the sin^ squar with the cosine squar. Indeed we see that the other functions are the same in both terms. So we can pull them out. So we have cossine^ squar + sin^ squar and there should always equal to 1. Hence the only thing that remains here are the two functions in the front.

[00:09:12]And that's it. Now we can use this normal vector to calculate the common determinant that we need for the integration. This means we just need to calculate the standard ukidian norm of this vector. In other words, you square up the components, sum them up, and then you take the square root of that. So I would say let's simply and quickly do that. And you immediately see that squaring these components is not complicated at all. In fact, we also recognize immediately that we can factor out some things as before. Hence, in the same way as before, you can see that here in the first term, we have the cosine squar + sin^ squ again. So only the factor sine ^ 4 of u2 remains here which also tells us that we can factor out sin ^ 2. Hence in the first part we have sin squar remaining and in the second part we have the cosine squar remaining which makes everything much easier again because we already know this is just the constant 1. So in the end what we get here is the square root of sin^ squar which we can simply write as the absolute value of u2. However you might remember that u2 only goes from 0 to pi. So sine of u2 is always positive anyway. So there is no need to write the absolute value here. Our karmian determinant is simply the s of u2. And now you know with this nice result we can actually write down all possible integrals over the sphere. And maybe the first example of such a surface integral should be a constant function f. So we could integrate the constant one which means we should get out the area of s_ub_2. So this will be the two-dimensional area of the two-dimensional sphere in R3. And now by the definition of the surface integral we know we can calculate it by using our parameterization phi. And please note here we use the fact that we ignore a set of measure zero on the sphere. Okay.

[00:11:22]But now the constant one here stays the constant one in the parameterization and then we just have to multiply with our crian here and then we have the integral over u1 and u2. And at this point you already know how to solve such ordinary two-dimensional integrals because we can use Fubini's theorem.

[00:11:43]This means we have the integral over u2 and the 1 / u1. And obviously the integral for u1 is quite simple because there is no u1 in the function that we want to integrate. So this means the integration with respect to u1 just gives us 2 pi. And on the other hand, the integral with respect to u2 is also not complicated at all. In fact, we can just use the anti-derivative which is given by minus cosine of u2. And by putting in the limits, you see we have 1 + 1. So in the end, we actually get 4 pi out which is the area of the sphere. And with that, I think this is good enough for this example. Now I hope you know how such a surface integral works. And in fact, I can already tell you these theorical coordinates we have here can be generalized to n dimensions as well.

[00:12:39]And indeed, these coordinates can be really helpful in integrating some functions that have some symmetries.

[00:12:46]But that's definitely something I want to show you in the next video. So, I really hope I meet you there again and have a nice day. Bye-bye.

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