Volume of n-dimensional sphere equation

[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

 

[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

<br /> \begin{equation}<br /> V_n=\int_0^R S_n r^{n-1} dr=\frac{S_n R^n}{n}<br /> \end{equation}<br />.

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.

<br /> \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}<br />

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

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

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

Now solve for S_n and plug it into equation one.

 

[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?