<s>[INST] The constraints are not all considered 
The solution says that domain of integration is delimited above by the sphere of equation $x ^ 2 + y ^ 2 + z ^ 2 = 2$ and below by the cone $z = \sqrt{x ^ 2 + y ^ 2}$.  I have the impression that we do not consider the fact that $x$ is between $0$ and $1$ and that $y$ is between $0$ and $\sqrt{1-x ^ 2}$. Am I wrong? I had the impression we could find an $x$ great than $1$ or less than $0$ respecting the fact that the $z$ could be selected between the cone and the sphere.
EDIT:
As we have to select a $y$ such that $0 \leq y \leq \sqrt{1-x^2}$, then $x$ is necessarily between $0$ and $1$, because otherwise $\sqrt{1-x^2}$ would be a complex number. However, we work in the real numbers. So it seems obvious that  the condition $0 \leq y \leq \sqrt{1-x^2}$ is not necessary to delimit the domain of integration. How about the condition that $0 \leq y \leq \sqrt{1-x^2}$?
 [/INST] You are right and the solution is incorrect. If we wished to consider the region bounded by the sphere $x^2 + y^2 + z^2 = 2$ and the cone $z = \sqrt{x^2 + y^2}$, then the correct bounds would be
$$\int_{-1}^1 \int_{-\sqrt{1 - x^2}}^{\sqrt{1 - x^2}}\int_{\sqrt{x^2 + y^2}}^{\sqrt{2 - x^2 - y^2}} \frac{1}{\sqrt{x^2 + y^2}} \, \mathrm{d}z \, \mathrm{d}y \, \mathrm{d}x.$$
Plotting the cone and sphere, you'll find that the intersection is a circle lying on the $z = 1$ plane, and the region is a conical segment of the sphere. If $(x, y, z)$ lies on the intersection, then
$$\sqrt{x^2 + y^2} = z = \sqrt{2 - x^2 - y^2} \implies 2(x^2 + y^2) = 2 \implies x^2 + y^2 = 1.$$
The entire intersection of the two regions lies in the cylinder defined by $x^2 + y^2 \le 1$, so the bounds of our integrals with respect to $x$ and $y$ must precisely capture the disk $x^2 + y^2 \le 1$ in the $x$-$y$ plane$^1$. That's what the integral bounds above do for us: the $y$ variable must range between the bottom semicircle $y = -\sqrt{1 - x^2}$ and the top semicircle $y = \sqrt{1 - x^2}$. The $x$ variable, on the disk $x^2 + y^2 \le 1$, ranges between $-1$ and $1$.
The proposed solution only considers the region in the positive orthant; the region is further bounded by $x \ge 0$ and $y \ge 0$. Given the symmetry of the region and the function, the proposed solution will be one quarter of the true value.

$^1$ Suppose that $(x, y, z) \in \Bbb{R}^3$ lay in our intersection. That is, $x^2 + y^2 + z^2 \le 2$ and $z \ge \sqrt{x^2 + y^2}$. Then $$x^2 + y^2 \le z^2 \le 2 - x^2 + y^2 \implies 2(x^2 + y^2) \le 2 \implies x^2 + y^2 \le 1,$$
i.e. basically the same argument as above, with inequalities instead.
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