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Let $v_n$ be the volume of the unit ball in $\mathbb{R}^n$. Show by using Fubini's theorem that $$v_n = 2\,v_{n-1}\,\int_0^1{(1-t^2)^{(n-1)/2}\,dt}\,.$$

I've been trying to do this inductively, and I've been thinking about the integral as kind of a generalization of the integral formula for the area of (part of) the unit disc in $\mathbb{R}^2$, but when I go into more dimensions, the exponent is tripping me up.

One of my thoughts was to just write down an integral for the volume of the n-ball as a big iterated integral, $$4\,\int_0^1{\cdots \int_0^1{(1-x_1^2)^{1/2}\cdots (1-x_n^2)^{1/2}\,dx_n\cdots dx_1}}\,,$$ but the more I look at this, the less sense it makes to me. At that point, I was thinking of, like, isolating one coordinate at a time and integrating, where the other coordinates are "smooshed" together in a "y-axis." Or something.

Please help, hahaha

Edit: Here's a solution (I believe).

In general, we have $$v_n = \idotsint\limits_{\sum\limits_{i=1}^n{x_i^2}\le 1}{dx_1\cdots dx_n}\,.$$ So by Fubini's theorem, we have $$\int_{-1}^1{\left[\idotsint\limits_{\sum\limits_{i=1}^{n-1}{x_i^2}\le 1-x_n^2}{dx_1\cdots dx_{n-1}}\right]\,dx_n}\,.$$

Now set $y_i=\dfrac{x_i}{\sqrt{1-x_n^2}}$. So $dy_i=\dfrac{dx_i}{\sqrt{1-x_n^2}}$. Thus, we have $\sum\limits_{i=1}^{n-1}{y_i^2} = \sum\limits_{i=1}^{n-1}{\dfrac{x_i^2}{1-x_n^2}}\le 1$.

So we make our change of variables: $$\int_{-1}^1{\left[\idotsint\limits_{\sum\limits_{i=1}^{n-1}{y_i^2}\le 1}{\left(\sqrt{1-x_n^2}\right)^{n-1}\,dy_1\cdots dy_{n-1}}\right]\,dx_n} = 2\,\int_0^1{\left(\sqrt{1-x_n^2}\right)^{n-1}v_{n-1}\,dx_n}\,.$$ Now we can just call $x_n$ whatever we like -- in this case, $t$.

How does this look?

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You are almost there. You should note that the leading constant factor in your expression for the volume of the $n$-ball should not be 4, but $2^n$. –  Willie Wong Jan 25 '11 at 0:15
    
@Willie: This clears up some of my trouble, but I still can't get the exponent to work out correctly... –  Bey Jan 25 '11 at 1:46
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Your integrand is also a little faulty. (This has to do with what Theo wrote below.) The ball is represented by $\sum x_i^2 \leq 1$, but your integrand admits points with coordinate $(1,1,1,\ldots,1)$ which is not in the sphere. –  Willie Wong Jan 25 '11 at 12:32
    
@Willie: I figured it out, I believe. I'll be editing my post to include the solution. –  Bey Jan 25 '11 at 16:38
    
looks good. The change of variable trick you used is similar to the scaling argument Theo described. Good work. –  Willie Wong Jan 25 '11 at 17:04
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1 Answer 1

up vote 2 down vote accepted

Hint: How does the $(n-1)$-dimensional volume of a ball scale with its radius $r$?

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I'm sorry, but I don't understand your question. What do you mean by scaling with its radius? –  Bey Jan 25 '11 at 1:10
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@Bey: Slice the ball with horizontal planes orthogonal to the $t$-axis at height $t$. Then $r = (1-t^{2})^{1/2}$ is the radius of the slice of the ball at height $t$. Now observe that $v_{n-1}(r) = r^{n-1}v_{n-1}(1)$, where $v_{n-1}(r)$ is the volume of the $(n-1)$-dimensional ball of radius $r$. –  t.b. Jan 25 '11 at 1:19
    
Thanks for your responses! I found another way to approach the problem. I'll be editing my post to include the solution I found. Since your answer was the only answer submitted, you get the check =) –  Bey Jan 25 '11 at 16:37
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