4
$\begingroup$

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

$\endgroup$
2
  • $\begingroup$ 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. $\endgroup$ – Michael Hardy Mar 13 '14 at 16:54
  • $\begingroup$ See here. There is a recursion that is similar to yours on the page. It should be easy to show equivalence of the terms. $\endgroup$ – Joseph Zambrano Mar 13 '14 at 16:57
5
$\begingroup$

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 ${\Bbb R}^{n+1}$"

$\endgroup$
4
$\begingroup$

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

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

$\endgroup$
2
  • $\begingroup$ How is the first line only limited to 2 integrals because we are integrating over n dimensions. Also how to obtain line 2 $\endgroup$ – clarkson Mar 14 '14 at 0:20
  • $\begingroup$ 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 $\sqrt{r^2-x^2}$. $\endgroup$ – Martín-Blas Pérez Pinilla Mar 14 '14 at 8:52

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.