Let $V_n(R)$ be the volume of the sphere of center $0$ and radius $R>0$ in $\mathbb{R}$. Prove that for $n \geq 3$ holds:
$$V_n(1)= \frac{2\pi}{n} V_{n-2}(1)$$
Solution
By definition holds:
$$V_n(1)=\idotsint\limits_{x_1^2+x_2^2+\cdots+x_n^2=1} 1\, {\rm d}(x_1, \; x_2, \dots\, x_n)$$
Converting into spherical coordinates and through the Jacobian we have that:
$$V_n(1)= \int_{0}^{2\pi}\int_{0}^{\pi}\cdots \int_{0}^{\pi}\int_{0}^{1}r^{n-1}\sin^{n-2}\theta_1 \sin^{n-3}\theta_2\cdots \sin \theta_{n-2}\, {\rm d}(r, \theta_1, \dots, \theta_{n-1})$$
The final expression is an expression containing single integrals which are reduced down, due to Walli's formula, in a Gamma form. Indeed:
$$V_n(1)= \frac{\pi^{n/2}}{\Gamma \left ( 1+ \frac{n}{2} \right )}$$
Now the result follows immediately. In more general it holds that:
$$V_n(R)= \frac{2\pi R^n}{n}V_{n-2}(R)$$
(due to Walli's formula)
More informations one can find here.
$$V_n(1)= \frac{2\pi}{n} V_{n-2}(1)$$
Solution
By definition holds:
$$V_n(1)=\idotsint\limits_{x_1^2+x_2^2+\cdots+x_n^2=1} 1\, {\rm d}(x_1, \; x_2, \dots\, x_n)$$
Converting into spherical coordinates and through the Jacobian we have that:
$$V_n(1)= \int_{0}^{2\pi}\int_{0}^{\pi}\cdots \int_{0}^{\pi}\int_{0}^{1}r^{n-1}\sin^{n-2}\theta_1 \sin^{n-3}\theta_2\cdots \sin \theta_{n-2}\, {\rm d}(r, \theta_1, \dots, \theta_{n-1})$$
The final expression is an expression containing single integrals which are reduced down, due to Walli's formula, in a Gamma form. Indeed:
$$V_n(1)= \frac{\pi^{n/2}}{\Gamma \left ( 1+ \frac{n}{2} \right )}$$
Now the result follows immediately. In more general it holds that:
$$V_n(R)= \frac{2\pi R^n}{n}V_{n-2}(R)$$
(due to Walli's formula)
More informations one can find here.
The last relation holds since : \(\displaystyle{V_{n}(R)=R^{n}\,V_{n}(1)\,,n\in\mathbb{N}}\) . Indeed,
ReplyDelete[b]PROOF[/b]
Let \(\displaystyle{n\in\mathbb{N}}\) . We have that :
\(\displaystyle{V_{n}(1)=\int_{y_1^2+...+y_n^2=1}1\,\mathrm{d}y_1...\mathrm{d}y_n}\) .
Consider the \(\displaystyle{C^{\infty}}\) function \(\displaystyle{f:\mathbb{R}^{n}\longrightarrow \mathbb{R}^{n}}\) defined by
\(\displaystyle{f\left(x_1,...,x_n\right)=\left(y_1,...,y_n\right)=\left(\dfrac{x_1}{R},...,\dfrac{x_n}{R}\right)}\) and
\(\displaystyle{\rm{grad}f(x_1,...,x_n)=\begin{pmatrix}
1/R& 0 & ... & 0\\
0& 1/R &... &0 \\
...&... & ... &... \\
0& 0 &... & 1/R
\end{pmatrix}\,\,,\forall\,\left(x_1,...,x_n\right)\in\mathbb{R}^{n}}\)
with \(\displaystyle{\rm{det}\left(\rm{grad}f(x_1,...,x_n)\right)=\dfrac{1}{R^{n}}\,,\forall\,\left(x_1,...,x_n\right)\in\mathbb{R}^{n}}\) .
Also, \(\displaystyle{f\,\left(S^{n+1}(R)\right)=S^{n+1}(1)}\) .
Check here : http://www.mathimatikoi.org/index.php/change-of-variables
Therefore,
\(\displaystyle{\begin{aligned} V_{n}(1)&=\int_{S^{n+1}(1)}1\,\mathrm{d}y_1...\mathrm{d}y_n\\&=\int_{S^{n+1}(R)}\rm{det}\left(\rm{grad}f(x_1,...,x_n)\right)\,\mathrm{d}x_1...\mathrm{d}x_n\\&=\dfrac{1}{R^{n}}\,\int_{S^{n+1}(R)}1\,\mathrm{d}x_1...\mathrm{d}x_n\\&=\dfrac{1}{R^{n}}\,V_{n}(R)\end{aligned}}\)
so: \(\displaystyle{V_{n}(R)=R^{n}\,V_{n}(1)}\) .
Also:
ReplyDelete\(\displaystyle{V_{n}(1)=\dfrac{\pi^{n/2}}{\Gamma\,\left(1+\dfrac{n}{2}\right)}\,\,,n\in\mathbb{N}}\) .
\(\displaystyle{V_{1}(1)=\dfrac{\sqrt{\pi}}{\Gamma\,\left(1+\dfrac{1}{2}\right))}=\dfrac{2\,\sqrt{\pi}}{\Gamma\,(\dfrac{1}{2})}=2}\)
since \(\displaystyle{V_{1}(1)=\mu\,\left(\left\{-1,1\right\}\right)=0}\) .
\(\displaystyle{V_{2}(1)=\dfrac{\pi}{\Gamma(1)}=\pi}\) ( circle or radius \(\displaystyle{1}\)) .
If \(\displaystyle{n\geq 3}\), then :
\(\displaystyle{\begin{aligned} V_{n-2}(1)&=\dfrac{\pi^{(n-2)/2}}{\Gamma\,\left(1+\dfrac{n-2}{2}\right)}\\&=\dfrac{\pi^{n/2}\cdot \pi^{-1}}{\Gamma\,\left(\dfrac{n}{2}\right)}\\&=\dfrac{n}{2}\,\dfrac{1}{\pi}\,\dfrac{\pi^{n/2}}{\displaystyle{\dfrac{n}{2}\,\Gamma\,(\dfrac{n}{2})}}\\&=\dfrac{n}{\,2\,\pi}\,\dfrac{\pi^{n/2}}{\Gamma\,\left(1+\dfrac{n}{2}\right)}\\&=\dfrac{n}{2\,\pi}\,V_{n}(1)\end{aligned}}\)
so: \(\displaystyle{V_{n}(1)=\dfrac{2\,\pi}{n}\,V_{n-2}(1)\,\,,n\geq 3}\) .