Calculus Challenge 6!

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

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

RR 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
5 years, 9 months ago

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Comments

Let VnV_{n} be the volume of the nn dimension sphere. Note that,Vn∝RnV_{n} \propto R^n where RR is the radius of the sphere.

  ⟹  Vn=knRn\implies V_{n} = k_{n} R^n

From the definition of an n+1n+1 dimension sphere,

x12+x22+…+xn+12=R2\displaystyle {x_{1}}^2 + {x_{2}}^2 + \ldots + {x_{n+1}}^2 = R^2

  ⟹  x12+x22+…+xn2=R2−xn+12\displaystyle \implies {x_{1}}^2 + {x_{2}}^2 + \ldots + {x_{n}}^2 = R^2 - {x_{n+1}}^2

Divide the n+1n+1 sphere into small "pieces" of nn sphere, each of length dxn+1d x_{n+1} with radius as R∗=R2−xn+12R_{*} = \sqrt{R^2 - {x_{n+1}}^2}

  ⟹  Vn+1=2∫0Rkn(R2−xn+12)ndxn+1\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}

  ⟹  kn+1Rn+1=2∫0Rkn(R2−y2)ndy\displaystyle \implies k_{n+1} R^{n+1} = 2 \int_{0}^{R} k_{n} \left(\sqrt{R^2 - y^2}\right)^n d y

  ⟹  kn+1Rn+1=2Rn∫0Rkn(1−y2R2)n2dy\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 t=(1−y2R2)\displaystyle t = \left(1 - \frac{y^2}{R^2} \right), we have,

kn+1=kn∫01tn2(1−t)−12\displaystyle k_{n+1} = k_{n} \int_{0}^{1} t^{\frac{n}{2}} (1-t)^\frac{-1}{2}

  ⟹  kn+1kn=B⁡(n2+1,12)\displaystyle \implies \dfrac{k_{n+1}}{k_{n}} = \operatorname{B} \left(\frac{n}{2} + 1 , \frac{1}{2} \right)

  ⟹  kn+1kn=Γ(n2+1)Γ(12)Γ(n2+32)\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)}

  ⟹  ∏n=1mkn+1kn=∏n=1mΓ(n2+1)Γ(12)Γ(n2+32) (∗)\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,

Vm=πmΓ(m+12)RmV_{m} = \dfrac{\sqrt{{\pi}^m}}{\Gamma\left(m + \dfrac{1}{2}\right)} R^m

Ishan Singh - 5 years, 9 months ago
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