Integral of $\sqrt{1-\|x\|^2}$ I am trying to calculate the next integral:
$$\int_{Q}\sqrt{1-\|x\|^2}dx$$
where $Q =\{x\in\mathbb{R}^n: \|x\|\leq 1\}$ and $\|x\|$ is the usual norm of $\mathbb{R}^n.$
For the cases $n = 2$ and $n = 3$ polar and spheric coordinates are useful, however, is there an easier form to compute this? I am trying to find a nice variable change but I have not gotten any useful.
Any kind of help is thanked in advanced.
 A: Notice that the integral can be rewritten as
$$\int_Q \:dx \int_0^{\sqrt{1-||x||^2}}\:dy$$
by introducing a new variable in $\Bbb{R}^{n+1}$. Thus the integral is equal to
$$\frac{\pi^{\frac{n+1}{2}}}{2\Gamma\left(\frac{n+3}{2}\right)}$$
or half the volume of the unit $(n+1)$-ball
A: Just use $n$-dimensional spherical coordinates. You have $dV = r^{n-1}dr\,d\sigma$, where $d\sigma$ is the "element of surface area" of the unit sphere in $\Bbb R^n$. Since the function depends only on $r$, you're going to get
$$\int_0^1 r^{n-1}\sqrt{1-r^2}\,dr$$
times the surface area of the unit sphere in $\Bbb R^n$. (The volume of the unit ball is a standard inductive computation, and then you get the surface area of the sphere by differentiating the volume $V(r)=r^n V(1)$ of the ball of radius $r$.)
A: If you decompose the domain in elementary spherical slices, the integral is the sum of $\sqrt{1-r^2}$ times the volume of a slice.
$$I=\int_0^1\sqrt{1-r^2}dV=\frac{\pi^{n/2}}{\Gamma\left(\dfrac n2+1\right)}\int_0^1\sqrt{1-r^2}nr^{n-1}dr
\\=\frac{n\pi^{n/2}}{2\Gamma\left(\dfrac n2+1\right)}\int_0^1(1-t)^{1/2}t^{n/2-1}dt
\\=\frac{n\pi^{n/2}}{2\Gamma\left(\dfrac n2+1\right)}\frac{\Gamma\left(\dfrac 32\right)\Gamma\left(\dfrac n2\right)}{\Gamma\left(\dfrac{n+3}2\right)}
\\=\frac{n\pi^{(n+1)/2}}{4\left(\dfrac n2+1\right)\Gamma\left(\dfrac{n+3}2\right)}.$$
