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How can I prove that: $$v_d= 2 v_{d-1} \int_{0} ^{1}(1-x^2)^{\frac{d-1}{2}}dx $$

without using polar coordinates.

$v_d$ denotes the volume of the unit ball in $\mathbb{R}^d$

Burrrrb
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    This has nothing to do with polar coordinates - it's just Fubini. (Polar coordinates might come in in the formula for the volume of the ball in terms of the area of the unit sphere - this is different.) – David C. Ullrich Nov 21 '17 at 18:45
  • Closely related to Volume of a hypersphere. – robjohn Nov 21 '17 at 18:54
  • @robjohn I know the proof in that answer, it's not what I'm looking for. – Burrrrb Nov 21 '17 at 18:55
  • The integral in your question can be evaluated as a Beta function integral. – robjohn Nov 21 '17 at 18:56
  • @PotatoHead47: In a comment to my now deleted answer, you stated: "I'm trying to establish the reduction formula without knowledge of $v_d$, I'm also not trying to evaluate the integral!" What is the reduction formula you are trying to establish? If it is the formula in the question, that formula is derived by integrating slices in $\mathbb{R}^{d-1}$ in the $d^\text{th}$ direction; it doesn't use polar coordinates. – robjohn Nov 21 '17 at 19:27
  • Could you please elaborate? – Burrrrb Nov 21 '17 at 19:35
  • Oh, you are simply asking how to prove the formula in the question. The part about polar coordinates turns out to be a red herring. I will update my answer. – robjohn Nov 21 '17 at 19:43

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The slice of the unit sphere in $\mathbb{R}^d$ where $t\le x_d\le t+dt$ is essentially a sphere in $\mathbb{R}^{d-1}$ with radius $\left(1-t^2\right)^{1/2}$; that is, since $x_d$ is essentially $t$, we must have $$ \sum\limits_{k-1}^{d-1}x_k^2\le1-t^2 $$ This sphere has volume $v_{d-1}\left(1-t^2\right)^{\frac{d-1}2}$ since volumes in $\mathbb{R}^{d-1}$ scale as the $d-1$ power of the scale factor.

Integrating the volumes of these slices from $-1$ to $1$ gives the volume of the unit sphere in $\mathbb{R}^d$. Since the integrand is even, we can integrate from $0$ to $1$ and double.

robjohn
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