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Calculus Challenge 6!

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.

Note by Kartik Sharma
2 years ago

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Let \(V_{n}\) 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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Comment deleted Dec 05, 2015

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@Ishan Singh Brilliant! I have seen other proofs of the same which done using Gaussian integral but i must tell you how much I admired your solution.

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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