In cylindrical coordinates, we have [tex]r^2=x^2+y^2[/tex], so that
[tex]z = \pm \sqrt{2-r^2} = \pm \sqrt{2-x^2-y^2}[/tex]
correspond to the upper and lower halves of a sphere with radius [tex]\sqrt2[/tex]. In spherical coordinates, this sphere is [tex]\rho=\sqrt2[/tex].
[tex]1 \le r \le \sqrt2[/tex] means our region is between two cylinders with radius 1 and [tex]\sqrt2[/tex]. In spherical coordinates, the inner cylinder has equation
[tex]x^2+y^2 = 1 \implies \rho^2\cos^2(\theta) \sin^2(\phi) + \rho^2\sin^2(\theta) \sin^2(\phi) = \rho^2 \sin^2(\phi) = 1 \\\\ \implies \rho^2 = \csc^2(\phi) \\\\ \implies \rho = \csc(\phi)[/tex]
This cylinder meets the sphere when
[tex]x^2 + y^2 + z^2 = 1 + z^2 = 2 \implies z^2 = 1 \\\\ \implies \rho^2 \cos^2(\phi) = 1 \\\\ \implies \rho^2 = \sec^2(\phi) \\\\ \implies \rho = \sec(\phi)[/tex]
which occurs at
[tex]\csc(\phi) = \sec(\phi) \implies \tan(\phi) = 1 \implies \phi = \dfrac\pi4+n\pi[/tex]
where [tex]n\in\Bbb Z[/tex]. Then [tex]\frac\pi4\le\phi\le\frac{3\pi}4[/tex].
The volume element transforms to
[tex]dx\,dy\,dz = r\,dr\,d\theta\,dz = \rho^2 \sin(\phi) \, d\rho \, d\theta \, d\phi[/tex]
Putting everything together, we have
[tex]\displaystyle \int_0^{2\pi} \int_1^{\sqrt2} \int_{-\sqrt{2-r^2}}^{\sqrt{2-r^2}} r \, dz \, dr \, d\theta = \boxed{\int_0^{2\pi} \int_{\pi/4}^{3\pi/4} \int_{\csc(\phi)}^{\sqrt2} \rho^2 \sin(\phi) \, d\rho \, d\phi \, d\theta} = \frac{4\pi}3[/tex]
