Volume of a $d$-dimensional sphere with radius $R$

Question

I am trying to calculate the integral, but I've been stuck at an integral, which I don't know how to solve:

Determinant of the Jacobian for variable transformation from $x_1,x_2,...,x_d$ to $r,\phi_1,\phi_2,...\phi_{d-1}$ is (where $\phi_{d-1}$ goes from $0$ to $2\pi$ while the rest of $\phi$'s go from $0$ to $\pi)$ :

$$|J|= r^{d-1}\Pi_i^{d-2} [\sin \phi_i]^{d-1-i}$$

I then try to find the volume by integrating and I have (after some calculations) :

$$V_d= \frac {R^d} d \cdot 2\pi \cdot \Pi_i^{d-2} \int_{\phi_i=0}^{\pi} [\sin \phi_i]^{d-1-i}d\phi_i $$.

Now I know that that integral has something to do with the Beta function, which is somehow connected to the Gamma function. The difference is though, that in Wikipedia for the Beta function, the boundaries of the integral are from $0$ to $\frac \pi 2$ , while here as it can be seen from $0$ to $\pi $. How can I find that integral?

Answer 1

Take the integral $$ \int_0^\pi \sin(x)^k \, dx$$ and substitute $x=\arcsin(u)$. Then this transforms to $$ 2\int_0^1 u^k/\sqrt{1-u^2}\, du$$ (where the $2$ comes from $\sin(x)=\sin(\pi-x)$).

Next note $$ B(x,y) = \int_0^1t^{x-1}(1-t)^{y-1}$$ So we further substitute $u=v^{1/2}$. Then this turns into $$ 2\cdot (1/2) \int_0^1 v^{k/2} (1-v)^{-1/2} v^{-1/2}\,dv = \int_0^1 v^{k/2-1/2}(1-v)^{-1/2}=B(k/2+1/2,\,1/2)$$

Answer 2

Note that $\sin(\pi-x)=\sin x$. Then $$\int_0^\pi\sin^p x dx=\int_0^{\pi/2}\sin^pxdx+\int_{\pi/2}^{\pi}\sin^pxdx$$ Let's change variable in the second integral from $x$ to $y=\pi-x$. At $x=\pi/2$ you have $y=\pi/2$, and at $x=\pi$ you get $y=0$. In addition $dy=-dx$. Then $$\int_{\pi/2}^{\pi}\sin^pxdx=\int_{\pi/2}^0\sin^p(\pi-y)(-dy)=\int_0^{\pi/2}\sin^py dy$$ So $$\int_0^\pi\sin^p x dx=2\int_0^{\pi/2}\sin^p x dx$$