An improper triple integral where the domain is a cube

I am trying to find the value of the following integral:

$$\iiint\limits_D {{x^2} + {y^2}dV}$$

where the domain $D$ is a cube $0 \leqslant x,y,z \leqslant 1$.

Solution (attempt 1) The first thing I did was to rewrite the domain, adding the upper/lower bounds (infinity in my case), so the domain becomes $$0 \leqslant x < \infty , \quad -\infty < y < \infty ,\quad -\infty < z \leqslant 1$$

This is good since it states clear upper/lower bounds for each iteration.

$$\iiint\limits_D {{x^2} + {y^2}dV} = \mathop {\lim }\limits_{R \to \infty } \int\limits_{ - R}^R {{y^2}dy\int\limits_0^R {{x^2}dx\int\limits_0^1 {dz = } } } \\ =\mathop {\lim }\limits_{R \to \infty } \left[ {\frac{{{y^3}}}{3}} \right]_{ - R}^R\left[ {\frac{{{x^3}}}{3}} \right]_0^R\left[ z \right]_0^1 = \mathop {\lim }\limits_{R \to \infty } \frac{{2{R^6}}}{9}(1 - R)$$

Solution (attempt 2) I observed at least one symmetry in the integral with respect to $y$. In other words, it can be rewritten as:

$$\iiint\limits_D {{x^2} + {y^2}dV} = \mathop {\lim }\limits_{R \to \infty } \int\limits_{ - R}^R {{y^2}dy\int\limits_0^R {{x^2}dx\int\limits_0^1 {dz = } } } \mathop {\lim }\limits_{R \to \infty } 2\int\limits_0^R {{y^2}dy\int\limits_0^R {{x^2}dx\int\limits_0^1 {dz =\\= \mathop {\lim }\limits_{R \to \infty } } } } 2\left[ {\frac{{{y^3}}}{3}} \right]_0^R\left[ {\frac{{{x^3}}}{3}} \right]_0^R\left[ z \right]_0^1 = \frac{{2{R^6}}}{9}$$

The strange observation

It's strange that both values support the idea that the hypervolume is infinitely big, but when I look in the key, it's simply $\frac{2}{3}$. Does anyone see the flaw in my reasoning?

• I think $0\leq x,y,z \leq 1$ simply means $0\leq x\leq 1, 0\leq y\leq 1,0\leq z\leq 1$. So it is not improper. – KittyL Mar 31 '15 at 14:19
• The domain doesn't look like a cube at all. Also, your integration in both attempts is wrong. $x^2+y^2$ doesn't integrate as you have mentioned. It is $x^2y^2$ that integrates into the one you wrote. – cgo Mar 31 '15 at 14:20
• Indeed, as you said, it's a cube, so $D=[0,1]^3$. – sranthrop Mar 31 '15 at 14:20
• @KittyL,@cgo,@sranthrop, thank you for your responses! I will use your suggestions to try to solve it again. – Artem Mar 31 '15 at 14:25

$$\begin{gathered} \iiint\limits_D {{x^2} + {y^2}dV = \int\limits_0^1 {dz\int\limits_0^1 {dx\int\limits_0^1 {{x^2} + {y^2}dy} } } } = \hfill \\ = \int\limits_0^1 {dz\int\limits_0^1 {dx} \left[ {{x^2}y + \frac{{{y^3}}}{3}} \right]_{y = 0}^{y = 1}} = \int\limits_0^1 {dz} \int\limits_0^1 {{x^2} + \frac{1}{3}dx} = \hfill \\ = \int\limits_0^1 {dz} \left[ {\frac{{{x^3}}}{3} + \frac{x}{3}} \right]_0^1 = \int\limits_0^1 {\frac{2}{3}dz} = \frac{2}{3}\left[ z \right]_0^1 = \frac{2}{3} \hfill \\ \end{gathered}$$
• As @cgo pointed out, we cannot just factor out the $x^2$ from $x^2+y^2$.