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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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    looks good. The change of variable trick you used is similar to the scaling argument Theo described. Good work.2011-01-25

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

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    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 =)2011-01-25