Integrating a jacobian to find the volume. I want to solve the following:
Prove that
$$\displaystyle \int_R \sin^{n-2}\phi_1 \sin^{n-3}\phi_2\cdots\sin \phi_{n-2} d\theta d\phi_1\cdots d\phi_{n-2} = \frac{2\pi^{n/2}}{\Gamma(n/2)}$$
where $ R=[0,2\pi] \times [0,\pi]^{n-2}$. Hint: Compute $ \int_{\mathbb R^n}e^{-|x|^2}dx$ in spherical coordinates.
So I am having problems to calculate the latter integral in spherical coordinates because I dont know how to integrate (in finite steps) $sin^{n}(x)$ and I dont know how this results to be a division of integrals.Can you help me to solve this please?, Thanks a lot in advance :)
 A: You don't want to do the integral directly: hence the hint. One way to integrate is to split
$$ e^{-|x|^2} = e^{-(x_1^2+x_2^2+\dotsb+x_n^2)}  = e^{-x_1^2} e^{-x_2^2} \dotsm e^{-x_n^2} , $$
which is a product of $n$ one-dimensional integrals, all of which we know evaluate to
$$ \int_{-\infty}^{\infty} e^{-x^2} \, dx = \sqrt{\pi}. $$
On the other hand, if you substitute in spherical coordinates, you have $e^{-|x|^2}=e^{-r^2}$, and the volume element is
$$ r^{n-1}\sin^{n-2}\phi_1 \sin^{n-3}\phi_2\cdots\sin \phi_{n-2} \, dr \, d\theta \, d\phi_1\cdots d\phi_{n-2}, $$
which means that the integral splits this time as:
$$ \left( \int_0^{\infty} r^{n-1} e^{-r^2} \, dr \right) \left( \int_R \sin^{n-2}\phi_1 \sin^{n-3}\phi_2\cdots\sin \phi_{n-2}  \, d\theta \, d\phi_1\cdots d\phi_{n-2} \right). $$
The second integral is the one you want, and you can do the first using the Gamma function, which will give you the answer you want when you divide: that is, you have two different evaluations of the same quantity, so they must be equal,
$$ \pi^{n/2} = \int_{\mathbb{R}^n} e^{-|x|^2} \, dx = \left( \int_0^{\infty} r^{n-1} e^{-r^2} \, dr \right) \left( \int_R \sin^{n-2}\phi_1 \sin^{n-3}\phi_2\cdots\sin \phi_{n-2}  \, d\theta \, d\phi_1\cdots d\phi_{n-2} \right), $$
so
$$ \int_R \sin^{n-2}\phi_1 \sin^{n-3}\phi_2\cdots\sin \phi_{n-2}  \, d\theta \, d\phi_1\cdots d\phi_{n-2} = \frac{\pi^{n/2}}{\int_0^{\infty} r^{n-1} e^{-r^2} \, dr}. $$
A: Hint all the variables $\phi_i$ can be separted in the integral, so you can transform the integral into:
$$\displaystyle \int_R \sin^{n-2}\phi_1 \sin^{n-3}\phi_2\cdots\sin \phi_{n-2} d\theta d\phi_1\cdots d\phi_{n-2}=\int_{0}^{2\pi}d\theta \prod_{i=1}^{n-2}\int_{0}^{\pi}\sin^i\phi_i d\phi_i$$
