# Volume of n-dimensional sphere equation

1. Jul 31, 2009

### Petar Mali

I need help in understanding this problem.
Equation of sphere in n-dimensional space is:

$$x^2_1+x^2_2+...+x^2_n=R^2$$

We serch volume as $$V=C_nR^n$$. Why? Perhaps its analogy with $$CR^3$$.

Now we calculate this integral

$$I=\int^{\infty}_{-\infty}dx_1\int^{\infty}_{-\infty}dx_2\int^{\infty}_{-\infty}dx_3...\int^{\infty}_{-\infty}dx_ne^{-a(x^2_1+x^2_2+...+x^2_n)}$$
Why we do this?

And we get $$(\frac{\pi}{a})^{\frac{N}{2}}$$

And then

$$I=\int dV_n e^{-ar^2}$$

we get

$$V_n=\frac{(\pi)^{\frac{N}{2}}}{\Gamma(\frac{N}{2}+1)}R^n$$

Can someone tell me idea of all this. Thanks

2. Jul 31, 2009

### Cyosis

Some gaps need to be filled to arrive at the answer.

The reason we seek a volume of that form is because any volume is proportional to the dimension of the space. Intuitively think about cubic meters. A 'normal' volume would have m^3 as a unit, a 4-dimensional volume would have m^4 as unit and an n-dimensional volume would have m^n as unit. So we know that $V \approx R^n$. The only thing that is missing is the proportionality constant, which we will denote $C_n$. Therefore we are looking for an expression of the form $V_n=C_n R^n$.

Similarly we know that the area of the n-sphere is proportional to R^(n-1), introducing another proportionality constant, $S_n$, we get $A_n=S_n R^{n-1}$. Note that $S_n$ is the surface area of a unit n-sphere.
Now think about a solid sphere as a lot of spherical shells stacked on top of each other. Adding all these shells together will give us the volume of the sphere. Therefore

$$V_n=\int_0^R S_n r^{n-1} dr=\frac{S_n R^n}{n}$$.

We will need this result later on.

Now we use a trick by equating two integrals to each other and then solving for $S_n$

The integral over all space of the Gaussian function in Cartesian coordinates is.

$$\int_{\mathbb{R}^n} e^{-a ||\vec{x}||^2 }d\vec{x}=\left( \int_{-\infty}^\infty e^{-a x^2}dx \right)^n=\left(\frac{\pi}{a}\right)^{n/2}$$

Switching to spherical coordinates we should of course get the same answer.

\begin{align*} \int_{\mathbb{R}^n} e^{-a r^2}dV_n & =\int_0^\infty S_n r^{n-1} e^{-a r^2}dr \;\;\;\;\;\;\;(\mathb{substitute \;\; ar^2=x}) \\ & = \frac{S_n}{2 a^{\frac{n}{2}}} \int_0^\infty x^{\frac{n}{2}-1} e^{-x} dx \\ & =\frac{S_n}{2 a^{\frac{n}{2}}} \Gamma(\frac{n}{2}) \end{align*}

Therefore
$$\frac{S_n}{2 a^{\frac{n}{2}}} \Gamma(\frac{n}{2})=\left(\frac{\pi}{a}\right)^{n/2}$$

Now solve for S_n and plug it into equation one.

Last edited: Jul 31, 2009
3. Jul 31, 2009

### Petar Mali

Thanks! I have one more question. Why you solve this integral

$$\int_{\mathbb{R}^n} e^{-a ||\vec{x}||^2 }d\vec{x}$$?

Is that because is that easiest way?

Principial you can solve some other integral? Am I right?

Last edited: Jul 31, 2009