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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?

Thanks for any answer in advance.

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    $\begingroup$ Because $z \leq (1-\sqrt{x} - \sqrt{x})^2$. Why is the assistant professor correct? $\endgroup$
    – Huy
    Apr 30, 2011 at 16:45
  • $\begingroup$ I answered too soon. It looks like you are right. :) $\endgroup$
    – Cheerful Parsnip
    Apr 30, 2011 at 16:48
  • $\begingroup$ PS: I of course meant $z \leq (1-\sqrt{x}-\sqrt{y})^2$. Somehow I can't edit my first comment anymore. $\endgroup$
    – Huy
    Apr 30, 2011 at 16:52

1 Answer 1

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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.)

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  • $\begingroup$ @Huy: to summarise, you did it right. $\endgroup$
    – Willie Wong
    May 1, 2011 at 10:09
  • $\begingroup$ @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? $\endgroup$
    – Huy
    May 1, 2011 at 11:29
  • $\begingroup$ @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? $\endgroup$
    – Huy
    May 1, 2011 at 11:37
  • $\begingroup$ @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) $\endgroup$
    – Andy
    May 1, 2011 at 12:00
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    $\begingroup$ @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. $\endgroup$
    – Christian Blatter
    May 1, 2011 at 16:06

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