Proof for volume of n-ball with radius 1

Question

I'm trying to prove this formula from here, that the volume of a n-ball with radius 1 (let's call it $B_n$) is: $$\frac{\pi^{n/2}}{\Gamma(\frac{n}{2}+1)}$$

However, I come to the wrong result and I cannot find the mistake.

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The intersection of an n-ball with a hyperplane is an $(n − 1)$-ball. Therefore:

$$\text{vol}(B_{n+1}) = \int_{-1}^1 \text{vol}((B_{n} \,\,| \,\,\text{Radius = }\sqrt{1-x^2})) \,\,dx =$$

$$2 \int_{0}^1 \text{vol}((B_{n} \,\,| \,\,\text{Radius = }\sqrt{1-x^2})) \,\,dx$$

Now I want to find $\text{vol}((B_{n} \,\,| \,\,\text{Radius = }\sqrt{1-x^2}))$. That is per definition $\int_{-1}^1 \,\,1_K$ where $K = \{(x_1, ..., x_{n}) \,\,|\,\, x_1^2 + ...+ x_n^2 \leq \sqrt{1-x^2} \}$.

It holds: $ x_1^2 + ...+ x_n^2 \leq \sqrt{1-x^2} \Leftrightarrow \sum \frac{x_i^2}{\sqrt{1-x^2}} \leq 1 $

Now I do the transformation $x_i \mapsto x_i \cdot (\sqrt{1-x^2})^{1/2}$. The determinant of the Jacobian is $(\sqrt{1-x^2})^{n/2}$.

So I get:

$$2 \int_{0}^1 \text{vol}((B_{n} \,\,| \,\,\text{Radius = }\sqrt{1-x^2})) \,\,dx = $$

$$2 \int_{0}^1 \text{vol}(B_{n}) \cdot (\sqrt{1-x^2})^{n/2} \,\,dx = $$

$$2 \int_{0}^1 \text{vol}(B_{n}) \cdot (1-x^2)^{n/4} \,\,dx$$

Here I would use induction now.

But above formula isn't correct. For $n=2$ it's correct, but for $n=3$ it would state that $$\text{vol}(B_{3}) = 2 \int_{0}^1 \text{vol}(B_{2}) \cdot (1-x^2)^{3/4} \,\,dx = 2 \pi \cdot \int_{0}^1 (1-x^2)^{3/4} \,\,dx$$

which according to Wolfram Alpha doesn't equal $\frac{3}{4}\pi$, which would be the correct answer.

Answer 1

You have two errors. The formula for volume is wrong by a factor of 2. The volume is proportional to $r^n$, not $r^{n/2}$ $$\text{vol}((B_{n} \,\,| \,\,\text{Radius = }\sqrt{1-x^2}))$$ The second error is that if you want to calculate the volume of the 3D sphere, the $n$ on the right hand side is $2$, not $3$. So $$\text{vol}(B_{3}) = 2 \int_{0}^1 \text{vol}(B_{2}) \cdot (1-x^2)^{2/2} \,\,dx=2\pi\frac23$$

Answer 2

The correct recurrence is $\text{Vol}(B_1)=\text{length}((-1,1))=2$, and for any $n\geq 1$, \begin{align} \text{Vol}(B_{n+1})&=\int_{-1}^1\text{Vol}(B_n\,; \text{radius} \sqrt{1-x^2})\,dx\\ &=2\int_0^1\text{Vol}(B_n) \left(\sqrt{1-x^2}\right)^n\,dx\\ &=2\text{Vol}(B_n)\int_0^1(1-x^2)^{n/2}\,dx \end{align}

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If you wish to evaluate this more explicitly, make the substitution $t=x^2$, then \begin{align} 2\int_0^1(1-x^2)^{n/2}\,dx&=2\int_0^1(1-t)^{n/2}\frac{dt}{2\sqrt{t}}\\ &=\int_0^1t^{-1/2}(1-t)^{n/2}\,dt\\ &:=B\left(\frac{1}{2},\frac{n}{2}+1\right), \end{align} where $B(x,y)=\int_0^1t^{x-1}(1-t)^{y-1}\,dt$ is the Beta function. Using the functional equation, we have that $B(x,y)=\frac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)}$, and thus $B\left(\frac{1}{2},\frac{n}{2}+1\right)=\frac{\sqrt{\pi}\,\Gamma\left(\frac{n}{2}+1\right)}{\Gamma\left(\frac{n+1}{2}+1\right)}$, where we used the fact that $\Gamma(\frac{1}{2})=\sqrt{\pi}$. Therefore, \begin{align} \text{Vol}(B_{n+1})&=\text{Vol}(B_n)\cdot \frac{\sqrt{\pi}\,\Gamma\left(\frac{n}{2}+1\right)}{\Gamma\left(\frac{n+1}{2}+1\right)}. \end{align} Now, you can solve this recurrence.

Answer 3

Let $p\geq 3$ be an integer, I note $V_p(r)$ the volume of a $p$-dimensional ball of radius $r$. Here is the integral definition of the volume: $$ V_p(r) = \int_{\mathbb{R}^p}\mathbf{1}_{||x||\leq r}\text{Leb}_p(dx), $$ with $\text{Leb}_p$ the Lebesgue measure on $\mathbb{R}^p$. The change of variables $y = r^{-1}x$ gives $$ V_p(r) = \int_{\mathbb{R}^p}\mathbf{1}_{||y||\leq 1}\det(r^{-1}\mathbf{I_p})^{-1}\text{Leb}_p(dy) $$ so we have $V_p(r)= r^p V_p(1).$

An application of Fubini theorem gives: \begin{align} V_p(1) &= \int_{\mathbb{R}^p}\mathbf{1}_{\sum x_i^2\leq 1}dx_1\ dx_2 \dots dx_p,\\ &= \int_{\mathbb{R}^2}\mathbf{1}_{ x_1^2+x_2^2\leq 1}\int_{\mathbb{R}^{p-2}}\mathbf{1}_{x_3^2+\dots+x_p^2\leq 1 - x_1^2-x_2^2}dx_3 \dots dx_p\ dx_1\ dx_2,\\ &= \int_{x_1^2+x_2^2\leq 1}V_{p-2}(\sqrt{1-x_1^2-x_2^2})dx_1\ dx_2. \end{align} Now we can compute a polar change of variables: \begin{align} V_p(1) &= \int_0^{+\infty}\int_0^{2\pi}\mathbf{1}_{r^2\leq 1}V_{p-2}(\sqrt{1-r^2})rd\theta\ dr,\\ &= 2\pi\int_0^1 r V_{p-2}(\sqrt{1-r^2}) dr,\\ &= 2\pi V_{p-2}(1)\int_0^1 r(1-r^2)^{p/2-1} dr. \end{align} The term under the integral is the derivative of $r\mapsto -p^{-1}(1-r^2)^{p/2}$. We have $V_p(1) = \frac{2\pi}{p}V_{p-2}(1).$

Now we proceed by an immediate recurrence, for all $p\in\mathbb{N}^*$, $$ V_{2p}(1) =\frac{\pi^{p-1}}{p!}V_2(1)\ \text{and}\ V_{2p-1} = \frac{(2\pi)^{p-1}}{\Pi_{k=1}^{p-1}(2p-1-2k)}V_1(1). $$ So finally $V_{2p}(1) = \frac{\pi^p}{p!}$ and $V_{2p+1}(1) = \frac{2^{p+1}\pi^p}{(2p+1)(2p-1)\dots 3}.$