Tag: Higher-dimensional geometry

Consider the integral

$$I(u)=\underset{S^{n-1}}{\overset{}{\int }}|u·x|dS(x)$$

where $S^{n-1}$ is the unit sphere in $R^n$ and $u$ is a fixed vector. At first glance, this looks like a difficult high-dimensional integral. The absolute value creates a nonsmooth integrand, but the symmetry of the sphere makes the calculation surprisingly simple.

The key observation

Write

$$u=\parallel u\parallel e$$

where $e$ is a unit vector. Then

$$|u·x|=\parallel u\parallel |e·x|$$

Therefore,

$$I(u)=\parallel u\parallel \underset{S^{n-1}}{\overset{}{\int }}|e·x|dS(x)$$

The remaining integral does not depend on the direction of $e$. The sphere is rotationally symmetric, so we may rotate the coordinate system and assume that

$$e=(1,0,\dots ,0)$$

Then

$$e·x=x_1$$

and therefore

$$I(u)=\parallel u\parallel \underset{S^{n-1}}{\overset{}{\int }}\left|x_1\right|dS(x)$$

A geometric interpretation

For a point $x$ on the sphere, let $\theta$ be the angle between $x$ and the chosen direction $e$. Then the projection of $x$ onto this direction is

$$e·x=\cos (\theta )$$

so

$$|u·x|=\parallel u\parallel |\cos (\theta )|$$

The integral is therefore the total absolute projection of all points on the sphere onto a fixed direction.

Measuring spheres

The notation $\left|S^n\right|$ means the surface area of the unit sphere $S^n$.

For example,

• $S^0$ consists of two points, so $\left|S^0\right|=2$.
• $S^1$ is the unit circle, so $\left|S^1\right|=2\pi$.
• $S^2$ is the ordinary unit sphere, so $\left|S^2\right|=4\pi$.

Now consider the sphere $S^{n-1}$. Fix the angle $\theta$ between a point $x$ on this sphere and a fixed direction $e$.

All points with the same angle $\theta$ form a lower-dimensional sphere $S^{n-2}$. Therefore, $\left|S^{n-2}\right|$ is the surface area of this slice of the sphere.

This is the geometric reason that $\left|S^{n-2}\right|$ appears when we compute the integral using spherical coordinates.

The lower-dimensional sphere

To compute the integral, we slice the sphere by fixing the angle $\theta$. Each slice is itself a sphere of one lower dimension.

The notation

$$\left|S^{n-2}\right|$$

means the surface area of the unit sphere $S^{n-2}$ one dimension lower. For example,

• $\left|S^0\right|=2$, because it consists of two points;
• $\left|S^1\right|=2\pi$, because it is the unit circle;
• $\left|S^2\right|=4\pi$, because it is the usual sphere.

Using spherical coordinates, the surface element becomes

$$dS={\sin (\theta )}^{n-2}d\theta d{S}^{n-2}$$

Therefore,

$${\int }_{S^{n-1}}\left|x_1\right|dS=2\left|S^{n-2}\right|{\int }_0^{\pi /2}\cos (\theta ){\sin (\theta )}^{n-2}d\theta$$

Finishing the computation

The remaining one-dimensional integral is elementary. Let

$$y=\sin (\theta )$$

so that

$$dy=\cos (\theta )d\theta$$

Therefore,

$${\int }_0^{\pi /2}\cos (\theta ){\sin (\theta )}^{n-2}d\theta ={\int }_0^1{y}^{n-2}dy=\frac{1}{n-1}$$

Substituting this result gives

$${\int }_{S^{n-1}}\left|x_1\right|dS=\frac{2\left|S^{n-2}\right|}{n-1}$$

Finally,

$$\int S^{n-1}|u·x|dS=\frac{2\left|S^{n-2}\right|}{n-1}\Vert u\Vert$$

Examples

The formula becomes especially simple in low dimensions.

The circle $S^1$

For the unit circle we have $n=2$ . The lower-dimensional sphere is

$$S^0$$

which consists of two points, so

$$\left|S^0\right|=2$$

Therefore,

$$\int S^1|u·x|dS=\frac{2·2}{1}\Vert u\Vert =4\Vert u\Vert$$

The sphere $S^2$

For the ordinary unit sphere in $R^3$ we have

$$n=3$$

and the lower-dimensional sphere is the unit circle:

$$\left|S^1\right|=2\pi$$

Hence,

$$\int S^2|u·x|dS=\frac{2·2\pi }{2}\Vert u\Vert =2\pi \Vert u\Vert$$

In both examples, the direction of $u$ does not matter. Only its length remains. This is a direct consequence of the rotational symmetry of the sphere.

The main idea

The calculation is simple because the sphere has no preferred direction. A rotation can move any vector $u$ to a coordinate axis without changing the geometry of the sphere.

Therefore, the integral depends only on the length of the vector:

$${\int }_{S^{n-1}}^{}|u·x|dS=C(n)\Vert u\Vert$$

where $C(n)$ is a constant that depends only on the dimension.

The important lesson is not the integration itself, but the symmetry behind it: whenever a problem on a sphere involves a single fixed vector, the first question should be whether a rotation can remove the direction completely.