Volume of n dimensional ball

Question

• The open ball of radius $r$ in $\mathbb{R}^{N}$ is the set $\left\{\left(x_{1}, x_{2},\ldots,x_{n}\right)\in \mathbb{R}^{N} \mid \sum_{i = 1}^{N} x_{i}^{2} < r^{2}\right\}$.
• By definition its volume $V_N\left(r\right) = \int\int\cdots\int 1\,\,{\rm d}x_{1}\,{\rm d}x_{2}\ldots {\rm d}x_{N}$

$$ \mbox{How to prove that}\quad V_{N} = V_{N - 1}\int_{0}^{\pi/2}\cos^{n}\left(\theta\right) \,{\rm d}\theta\,\,\, ? $$

$V_{n}$ is the volume of the $n$-dimensional unit ball.

Any idea how I can show this please. I have no idea what sort of approach I should take.

Answer 1

Search:

"Finding Volume and Surface Area of Hyperspheres in ${\Bbb R}^n$" (Mario Sracic)

"The volume of a n-dimensional hypersphere" (A. E. Lawrence)

"The volume of a n-dimensional sphere in ${\Bbb R}^{n+1}$"

Answer 2

$$ V_n(r) = \int_0^r \int_{x_1^2+\cdots+x_{n-1}^2\le r^2-x^2} dx_1\ldots dx_{n-1} dx \\ = \int_0^r V_{n-1}(\sqrt{r^2-x^2}) dx\\ = \int_0^{\frac\pi 2} V_{n-1}(r\cos\theta) r\cos\theta \ \ d\theta $$

Now use the fact that $V_n(r) = r^n V_n(1) =: r^n V_n$: $$ r^n V_n = \int_0^{\frac\pi 2} V_{n-1}(r\cos\theta)^{n-1} r\cos\theta \ \ d\theta\\ V_n = V_{n-1}\int_0^{\frac\pi 2} (\cos\theta)^{n}\ d\theta $$

Answer 3

As a matter of fact, you cannot prove that

$$ V_n \stackrel?= V_{n-1} \int _0 ^{\pi/2} \cos^n \theta \, \mathrm d\theta.$$

The two sides of this alleged "equation" are not equal.

For example, let $n = 3.$ Then $V_{n-1} = V_2 = \pi$ and

$$ \int_0^{\pi/2} \cos^n \theta \, \mathrm d\theta = \int_0^{\pi/2} \cos^3 \theta \, \mathrm d\theta = \frac23. $$

Hence $$ V_{n-1} \int _0 ^{\pi/2} \cos^n \theta \, \mathrm d\theta = \frac23\pi. $$

But $$ V_n = V_3 = \frac43\pi. $$

To correct the equation, we can either integrate from $-\frac\pi2$ to $\frac\pi2$ instead of $0$ to $\frac\pi2$ (equivalent to integrating in Cartesian coordinates from $-r$ to $r$ instead of from $0$ to $r$, as suggested in a comment under another answer) or we can recognize the symmetry of that integral and simply double the original integral. That is,

$$ V_n = V_{n-1} \int _{-\pi/2} ^{\pi/2} \cos^n \theta \, \mathrm d\theta = 2 V_{n-1} \int _0 ^{\pi/2} \cos^n \theta \, \mathrm d\theta. $$

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A correct derivation in the style of the answer mentioned above could be \begin{align} V_n &= \int_{x_1^2+\cdots+x_{n-1}^2+x_n^2\leq 1} \mathrm dx_1 \cdots \mathrm dx_{n-1}\, \mathrm dx_n \\ &= \int_{x_n^2\leq 1} \int_{x_1^2+\cdots+x_{n-1}^2\le 1-x^2} \mathrm dx_1\cdots \mathrm dx_{n-1}\, \mathrm dx_n \\ &= \int_{-1}^1 \left(\sqrt{1-x^2}\right)^{n-1} V_{n-1}\, \mathrm dx \\ &= \int_{-\frac\pi2}^{\frac\pi2} (\cos^{n-1}\theta) V_{n-1}\cdot \cos\theta \,\mathrm d\theta \\ &= V_{n-1} \int_{-\frac\pi2}^{\frac\pi2} \cos^n\theta \,\mathrm d\theta \end{align}

using the notational convention that the volume of an $n$-ball of radius $r$ is $r^n V_n.$