Derive the equation for **the volume of an n-dimensional sphere**

\[\displaystyle \text{the volume of an n-dimensional sphere } = \frac{\sqrt{{\pi}^{n}}}{\Gamma\left(\frac{n}{2} + 1\right)} {R}^{n}\]

\(R\) is the radius of the sphere.

This one is something special because I myself am not able to prove it until now.

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TopNewestLet \(V_{n}\) be the volume of the \(n\) dimension sphere. Note that,\(V_{n} \propto R^n\) where \(R\) is the radius of the sphere.

\(\implies V_{n} = k_{n} R^n\)

From the definition of an \(n+1\) dimension sphere,

\(\displaystyle {x_{1}}^2 + {x_{2}}^2 + \ldots + {x_{n+1}}^2 = R^2\)

\(\displaystyle \implies {x_{1}}^2 + {x_{2}}^2 + \ldots + {x_{n}}^2 = R^2 - {x_{n+1}}^2\)

Divide the \(n+1\) sphere into small "pieces" of \(n\) sphere, each of length \(d x_{n+1}\) with radius as \(R_{*} = \sqrt{R^2 - {x_{n+1}}^2}\)

\(\displaystyle \implies V_{n+1} = 2 \int_{0}^{R} k_{n} \left(\sqrt{R^2 - {x_{n+1}}^2}\right)^n d x_{n+1}\)

\(\displaystyle \implies k_{n+1} R^{n+1} = 2 \int_{0}^{R} k_{n} \left(\sqrt{R^2 - y^2}\right)^n d y\)

\(\displaystyle \implies k_{n+1} R^{n+1} = 2 R^n \int_{0}^{R} k_{n} \left(1 - \dfrac{y^2}{R^2} \right)^{\dfrac{n}{2}} d y\)

Putting \(\displaystyle t = \left(1 - \frac{y^2}{R^2} \right)\), we have,

\(\displaystyle k_{n+1} = k_{n} \int_{0}^{1} t^{\frac{n}{2}} (1-t)^\frac{-1}{2}\)

\(\displaystyle \implies \dfrac{k_{n+1}}{k_{n}} = \operatorname{B} \left(\frac{n}{2} + 1 , \frac{1}{2} \right)\)

\(\displaystyle \implies \dfrac{k_{n+1}}{k_{n}} = \dfrac{\Gamma \left( \frac{n}{2} +1 \right) \Gamma \left(\frac{1}{2}\right)}{\Gamma \left(\frac{n}{2} + \frac{3}{2} \right)}\)

\(\displaystyle \implies \prod_{n=1}^{m} \dfrac{k_{n+1}}{k_{n}} = \prod_{n=1}^{m} \dfrac{\Gamma \left( \frac{n}{2} +1 \right) \Gamma \left(\frac{1}{2}\right)}{\Gamma \left(\frac{n}{2} + \frac{3}{2} \right)} \ (*)\)

Clearly, \((*)\) telescopes, thus,

\[V_{m} = \dfrac{\sqrt{{\pi}^m}}{\Gamma\left(m + \dfrac{1}{2}\right)} R^m\] – Ishan Singh · 2 years ago

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I'm sorry I didn't see it before. It had been a week? How I missed it?

But I must tell you that you should also see other approaches available on the net using multiple integrals. That's because there are many questions relating to this one which is easiest to get done by that approach. – Kartik Sharma · 1 year, 12 months ago

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