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

I have to calculate the volume of the set

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

and I did this by evaluating the integral

$$\int_0^1 \int_0^{(1-\sqrt{x})^2} \int_0^{(1-\sqrt{x}-\sqrt{y})^2} \mathrm dz \; \mathrm dy \; \mathrm dx = \frac{1}{90}.$$

However, a friend of mine told me that his assistant professor gave him the numerical solutions and it turns out the solution should be $\frac{1}{70}$. Also, I found out that this would be the result of the integral

$$\int_0^1 \int_0^{1-\sqrt{x}} \int_0^{1-\sqrt{x}-\sqrt{y}} \mathrm dz \; \mathrm dy \; \mathrm dx,$$

which is pretty much the same as mine just without squares in the upper bounds. My question is: Is the solution provided by the assistant professor wrong or why do I have to calculate the integral without squared upper bounds?

Also, is there any tool to compute the volume of such sets without knowing how one has to integrate?

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Because $z \leq (1-\sqrt{x} - \sqrt{x})^2$. Why is the assistant professor correct? – Huy Apr 30 '11 at 16:45
I answered too soon. It looks like you are right. :) – Grumpy Parsnip Apr 30 '11 at 16:48
PS: I of course meant $z \leq (1-\sqrt{x}-\sqrt{y})^2$. Somehow I can't edit my first comment anymore. – Huy Apr 30 '11 at 16:52

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.)
@Willie: I did understand that. :) @Christian: In order to apply this kind of substitution, we needed $g$ to be a diffeomorphism. I'm pretty sure in general such a mapping wouldn't be a diffeomorphism, but is this here the case due to the fact that both $B$ and the standard simplex only contain non-negative components? – Huy May 1 '11 at 11:29
@Christian: As far as I see, even with only non-negative components, this is no diffeomorphism, as the inverse function is not differentiable in $0$. So why is one allowed to do this transformation here? – Huy May 1 '11 at 11:37
@Huy: it's the same this as with polar coordinates: they are not a diffeomorphism on the whole plane, but it is in the plane withouth $x \geq 0$. Here it is a diffeomorphism on the whole plane without the origin; does this fact change your integral? (think about polar coordinates again) – Andy May 1 '11 at 12:00
@Huy: The map $g$ has to be a diffeomorphism $S'\to B'$ where $S\triangle S'$ and $B\triangle B'$ are sets of measure $0$ in their respective ${\mathbb R^d}$'s. – Christian Blatter May 1 '11 at 16:06