Volume of Ball on Spherical Manifold

Question

Let $S^n$ denote the $n$-dimensional spherical manifold, represented by the sphere model.

Let $B_x(r)$ be the ball centered at $x\in S^n$ of radius $r$.

What I'd like to know is how the volume of $B_x(r)$ behaves asymptotically in terms of $r$.

Is it $Vol(B_x(r))\in\Theta(r)$?

How does one compute the volume of $B_x(r)$?

Answer 1

Take the case of a 2-sphere. Without loss of generality, let's center the 2-ball at the north pole. Then the area of the 2-ball of radius $r$ lying on a 2-sphere of radius $R$ is given by the integral $$R^2\int_0^{2\pi}\int_0^{r/R} \mathrm{\sin\theta d\theta d\phi}= 2\pi R^2\left(1-\cos\left(\frac{r}{R}\right)\right)$$ Notice that when $r=\pi R$, we retrieve the surface area of the whole sphere.

Now let's imagine if we were to do the same integral but on a sphere of $n$ dimensions. What coordinate system would we pick? We would do another version of "spherical" coordinates and we would apply the same trick - set the center of the circle at the "north pole".

We can also always pick our coordinate system so that we only ever have to do one angular integral. The reason for this is best illustrated in the jumps from 2D to 3D, and again 3D to 4D. In 2D, "spherical" coordinates is just polar coordinates. When we move to 3D, we add a third axis. $$\begin{align} & x^1 = \rho\cos\phi \\ & x^2 = \rho\sin\phi \\ & x^3 = x^3 \\ \end{align}$$ We then apply $$\begin{align} & \rho = r\sin\theta \\ & x^3 = r\cos\theta \\ \end{align}$$ to retrieve our familiar spherical coordinates $$\begin{align} & x^1 = r\sin\theta\cos\phi \\ & x^2 = r\sin\theta\sin\phi \\ & x^3 = r\cos\theta \\ \end{align}$$ We can move to 4D by adding another axis $$\begin{align} & x^1 = r\sin\theta\cos\phi \\ & x^2 = r\sin\theta\sin\phi \\ & x^3 = r\cos\theta \\ & x^4 = x^4 \\ \end{align}$$ and then applying the following coordinate transformation $$\begin{align} & r = \chi\sin\gamma \\ & x^4 = \chi\cos\gamma \\ \end{align}$$ leaving us with the following coordinate system $$\begin{align} & x^1 = \chi\sin\gamma\sin\theta\cos\phi \\ & x^2 = \chi\sin\gamma\sin\theta\sin\phi \\ & x^3 = \chi\sin\gamma\cos\theta \\ & x^4 = \chi\cos\gamma \\ \end{align}$$ And we can repeat this process for as long as we need to. Notice that the last coordinate is always only in terms of cosine of the newest angular coordinate, this is why we'll only ever have to do that one integral.

To figure out how the Jacobian changes each time, let's look at the 2D to 3D example. The Jacobian for 2D polar is always the radial coordinate. Notice that the 2D radial coordinate and the third Cartesian coordinate also form a 2D polar pair. This means that $d\rho dx^3 = rd\theta dr$. So we can "derive" the Jacobian for 3D spherical from 2D spherical by plugging in: $$\mathrm{d^3x=\rho d\phi(d\rho dx^3) = (r\sin\theta)d\phi(rd\theta dr) = r^2\sin\theta d\phi d\theta dr}$$ Repeating the process for 3D to 4D the Jacobian is $\chi^3\sin^2\gamma\sin\theta$

In other words, we can start to see a pattern for the Jacobian of higher dimensional "spherical" coordinates. We can guess that the volume of an $n$-ball of radius $r$ lying on an $n$-sphere of radius $R$ is given by the integral $$R^n\int_{S^{n-1}} d\Omega\int_0^{\frac{r}{R}} \sin^{n-1}(\omega)d\omega$$ where $\omega$ is the last angular coordinate and $d\Omega$ is the surface measure of $S^{n-1}$ The integral, at least partially, evaluates to $$\frac{2\pi^{\frac{n}{2}}}{\Gamma\left(\frac{n}{2}\right)}R^n \int_0^{\frac{r}{R}} \sin^{n-1}(\omega)d\omega$$ where the $d\Omega$ integral evaluated to the total surface area of a unit $(n-1)$-sphere, a famous formula.

Edit: Mathematica finally finished evaluating the sine integral and boy is it not pretty. I'll do two interesting cases for asymptotic expressions. First, suppose $r\ll R$, then we can use small angle approximation and integrate. $$\frac{2\pi^{\frac{n}{2}}}{\Gamma\left(\frac{n}{2}\right)}R^n \int_0^{\frac{r}{R}} \omega^{n-1}d\omega = \frac{2\pi^{\frac{n}{2}}}{n\Gamma\left(\frac{n}{2}\right)}r^n = \frac{\pi^{\frac{n}{2}}}{\Gamma\left(\frac{n}{2}+1\right)}r^n$$ which happens to be exactly the formula for the volume of an $n$-ball in flat space.

The other case is where $\pi R - r \ll R$ Rewriting the integral $$\frac{2\pi^{\frac{n}{2}}}{\Gamma\left(\frac{n}{2}\right)}R^n \int_0^{\frac{r}{R}} \sin^{n-1}(\omega)d\omega = \frac{2\pi^{\frac{n}{2}}}{\Gamma\left(\frac{n}{2}\right)}R^n\left( \int_0^{\pi} \sin^{n-1}(\omega)d\omega - \int_{\frac{r}{R}}^\pi \sin^{n-1}(\omega)d\omega\right)$$ $$\frac{2\pi^{\frac{n+1}{2}}}{\Gamma\left(\frac{n+1}{2}\right)}R^n - \frac{2\pi^{\frac{n}{2}}}{\Gamma\left(\frac{n}{2}\right)}R^n \int_0^{\pi-\frac{r}{R}} \sin^{n-1}(\omega)d\omega$$ where the first integral retrieves the full surface area of the $n$-sphere and the second used a substitution making use of the fact that $\sin(\pi-\theta) = \sin(\theta)$. Now we are in a position to evaluate the integral with small angle approximation $$=\frac{2\pi^{\frac{n+1}{2}}}{\Gamma\left(\frac{n+1}{2}\right)}R^n - \frac{\pi^{\frac{n}{2}}}{\Gamma\left(\frac{n}{2}+1\right)}(\pi R - r)^n$$ which is, as a sanity check, equal to the total surface area of the sphere minus the volume of a ball in flat space with that smaller radius.