# Volume of n dimensional ball

The open ball of radius $r$ in $\mathbb R^N$ is the set $\{(x_1,x_2,\ldots,x_n)\in R^N \mid \sum_{i=1}^N x_i^2< r^2\}$
By definition its volume $V_N(r)={\int\int\cdots\int} 1 \, dx_1 \, dx_2\cdots dx_N$

How to prove that
$$V_N = V_{N-1} \int _0 ^{\pi/2} \cos^n \theta \, d\theta$$

V$_n$ is the volume of the n- dimensional unit ball.
Any idea how I can show this please.I have no idea what sort of approach I should take

• Your MathJax style is quite uninformed. You shouldn't keep alternating in and out of MathJax in the middle of one block of math notation. See my edits. One doesn't put "=" outside of MathJax while the things on either side of it are in MathJax, and lots of other things like that appeared. – Michael Hardy Mar 13 '14 at 16:54
• See here. There is a recursion that is similar to yours on the page. It should be easy to show equivalence of the terms. – Joseph Zambrano Mar 13 '14 at 16:57

Search:

"Finding Volume and Surface Area of Hyperspheres in ${\Bbb R}^n$" (Mario Sracic)

"The volume of a n-dimensional hypersphere" (A. E. Lawrence)

"The volume of a n-dimensional sphere in ${\Bbb R}^{n+1}$"

$$V_n(r) = \int_0^r \int_{x_1^2+\cdots+x_{n-1}^2\le r^2-x^2} dx_1\ldots dx_{n-1} dx \\ = \int_0^r V_{n-1}(\sqrt{r^2-x^2}) dx\\ = \int_0^{\frac\pi 2} V_{n-1}(r\cos\theta) r\cos\theta \ \ d\theta$$

Now use the fact that $V_n(r) = r^n V_n(1) =: r^n V_n$: $$r^n V_n = \int_0^{\frac\pi 2} V_{n-1}(r\cos\theta)^{n-1} r\cos\theta \ \ d\theta\\ V_n = V_{n-1}\int_0^{\frac\pi 2} (\cos\theta)^{n}\ d\theta$$

• How is the first line only limited to 2 integrals because we are integrating over n dimensions. Also how to obtain line 2 – clarkson Mar 14 '14 at 0:20
• The interior integral is $(n-1)$-dimensional and ${x_1^2+\cdots+x_{n-1}^2\le r^2-x^2}$ is a $n-1$ dimensional ball of radius $\sqrt{r^2-x^2}$. – Martín-Blas Pérez Pinilla Mar 14 '14 at 8:52