What is the volume of $\{ (x,y,z) \in \mathbb{R}^3_{\geq 0} |\; \sqrt{x} + \sqrt{y} + \sqrt{z} \leq 1 \}$?

Solution 1:

An integral $(*)\ \int_B f(x){\rm d}(x)$ over a three-dimensional domain $B$ depends on the exact expression for $f(x)$, $\ x\in{\mathbb R}^n$, and on the exact shape of the domain $B$. The latter is usually defined by a set of inequalities of the form $g_i(x)\leq c_i$. The information about $B$ has to be entered in the course of the reduction of the integral $(*)$ to a sequence of nested integrals. So, as a rule, there is a lot of work involved in the process of reducing everything to the computation and evaluation of primitives.

Now sometimes there is another way of handling such integrals: Maybe we can set up a parametric representation of $B$ with a parameter domain $\tilde B$ which is a standard geometric object like a simplex, a rectangular box or a half sphere. In the case at hand we can use the representation $$g: \quad S\to B,\quad (u,v,w)\mapsto (x,y,z):=(u^2,v^2,w^2)$$ which produces $B$ as an essentially 1-1 image of the standard simplex $$S:=\{(u,v,w)\ |\ u\geq0, v\geq0, w\geq 0, u+v+w\leq1\}\ .$$ In the process we have to compute the Jacobian $J_g(u,v,w)=8uvw$ and obtain the following formula: $${\rm vol}(B)=\int_B 1\ {\rm d}(x)= \int_S 1 \> J_g(u,v,w) \> {\rm d}(u,v,w)=\int_0^1\int_0^{1-u}\int_0^{1-u-v} 8uvw \> dw dv du ={1\over 90}\ .$$ (In this particular example the simplification is only marginal.)