Answer
The integral of $f\left( {x,y,z} \right) = {x^2} + {y^2}$ over ${\cal W}$ in both spherical and cylindrical coordinates yield:
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal W}^{} {x^2} + {y^2}{\rm{d}}V = \frac{{8\sqrt 2 \pi }}{5}$
Work Step by Step
We have $f\left( {x,y,z} \right) = {x^2} + {y^2}$ over ${\cal W}$, the region within the cylinder ${x^2} + {y^2} = 2$ between $z=0$ and the cone $z = \sqrt {{x^2} + {y^2}} $.
Case 1. Spherical coordinates
In spherical coordinates:
$f\left( {\rho \sin \varphi \cos \theta ,\rho \sin \varphi \sin \theta ,\rho \cos \varphi } \right) = {\rho ^2}{\sin ^2}\varphi $
The cone $z = \sqrt {{x^2} + {y^2}} $ in spherical coordinates is expressed by
$z = \rho \cos \varphi = \sqrt {{\rho ^2}{{\sin }^2}\varphi } $
$\cos \varphi = \sin \varphi $
$\tan \varphi = 1$, ${\ \ \ \ }$ $\varphi = \frac{\pi }{4}$
From the figure attached, we see that each ray starts from the cone $\varphi = \frac{\pi }{4}$ and ends at $\varphi = \frac{\pi }{2}$. Therefore, the ranges of $\varphi $ is $\frac{\pi }{4} \le \varphi \le \frac{\pi }{2}$.
Since $\rho$ starts from $0$ and intersects the cylinder, we need to express the cylinder ${x^2} + {y^2} = 2$ in spherical coordinates:
${\rho ^2}{\sin ^2}\varphi = 2$
$\rho \sin \varphi = \sqrt 2 $
$\rho = \frac{{\sqrt 2 }}{{\sin \varphi }}$
So, the range of $\rho$ is $0 \le \rho \le \frac{{\sqrt 2 }}{{\sin \varphi }}$.
Thus, the description of ${\cal W}$ in spherical coordinates:
${\cal W} = \left\{ {\left( {\rho ,\varphi ,\theta } \right)|0 \le \rho \le \frac{{\sqrt 2 }}{{\sin \varphi }},\frac{\pi }{4} \le \varphi \le \frac{\pi }{2},0 \le \theta \le 2\pi } \right\}$
Using Eq. (8) of Theorem 3, we evaluate the integral of $f\left( {x,y,z} \right)$ over ${\cal W}$ in spherical coordinates:
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal W}^{} {x^2} + {y^2}{\rm{d}}V = \mathop \smallint \limits_{\theta = 0}^{2\pi } \mathop \smallint \limits_{\varphi = \pi /4}^{\pi /2} \mathop \smallint \limits_{\rho = 0}^{\sqrt 2 /\sin \varphi } {\rho ^4}{\sin ^3}\varphi {\rm{d}}\rho {\rm{d}}\varphi {\rm{d}}\theta $
$ = \frac{1}{5}\mathop \smallint \limits_{\theta = 0}^{2\pi } \mathop \smallint \limits_{\varphi = \pi /4}^{\pi /2} \left( {{\rho ^5}|_0^{\sqrt 2 /\sin \varphi }} \right){\sin ^3}\varphi {\rm{d}}\varphi {\rm{d}}\theta $
$ = \frac{1}{5}\mathop \smallint \limits_{\theta = 0}^{2\pi } \mathop \smallint \limits_{\varphi = \pi /4}^{\pi /2} \left( {\frac{{4\sqrt 2 }}{{{{\sin }^5}\varphi }}} \right){\sin ^3}\varphi {\rm{d}}\varphi {\rm{d}}\theta $
$ = \frac{{4\sqrt 2 }}{5}\mathop \smallint \limits_{\theta = 0}^{2\pi } \mathop \smallint \limits_{\varphi = \pi /4}^{\pi /2} {\csc ^2}\varphi {\rm{d}}\varphi {\rm{d}}\theta $
Recall from Eq. (16) of the Table of Trigonometric Integrals (Section 8.2):
$\smallint {\csc ^2}x{\rm{d}}x = - \cot x$
Therefore,
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal W}^{} {x^2} + {y^2}{\rm{d}}V = \frac{{4\sqrt 2 }}{5}\mathop \smallint \limits_{\theta = 0}^{2\pi } \mathop \smallint \limits_{\varphi = \pi /4}^{\pi /2} {\csc ^2}\varphi {\rm{d}}\varphi {\rm{d}}\theta $
$ = - \frac{{4\sqrt 2 }}{5}\mathop \smallint \limits_{\theta = 0}^{2\pi } \left( {\cot \varphi |_{\pi /4}^{\pi /2}} \right){\rm{d}}\theta $
$ = - \frac{{4\sqrt 2 }}{5}\mathop \smallint \limits_{\theta = 0}^{2\pi } \left( {0 - 1} \right){\rm{d}}\theta $
$ = \frac{{4\sqrt 2 }}{5}\mathop \smallint \limits_{\theta = 0}^{2\pi } {\rm{d}}\theta = \frac{{8\sqrt 2 \pi }}{5}$
Thus, $\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal W}^{} {x^2} + {y^2}{\rm{d}}V = \frac{{8\sqrt 2 \pi }}{5}$.
Case 2. Cylindrical coordinates
In cylindrical coordinates:
$f\left( {r\cos \theta ,r\sin \theta ,z} \right) = {r^2}$
The cone $z = \sqrt {{x^2} + {y^2}} $ in cylindrical coordinates is $z=r$. So, $z$ varies from $z=0$ to $z=r$.
The projection of ${\cal W}$ onto the $xy$-plane is ${\cal D}$, a disk of radius $\sqrt 2 $, that is, ${x^2} + {y^2} \le 2$. So, ${\cal D}$ is described by
${\cal D} = \left\{ {\left( {r,\theta } \right)|0 \le r \le \sqrt 2 ,0 \le \theta \le 2\pi } \right\}$
Thus, the description of ${\cal W}$ in cylindrical coordinates:
${\cal W} = \left\{ {\left( {r,\theta ,z} \right)|0 \le r \le \sqrt 2 ,0 \le \theta \le 2\pi ,0 \le z \le r} \right\}$
Using Eq. (5) of Theorem 2, we evaluate the integral of $f\left( {x,y,z} \right)$ over ${\cal W}$ in cylindrical coordinates:
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal W}^{} {x^2} + {y^2}{\rm{d}}V = \mathop \smallint \limits_{\theta = 0}^{2\pi } \mathop \smallint \limits_{r = 0}^{\sqrt 2 } \mathop \smallint \limits_{z = 0}^r {r^3}{\rm{d}}z{\rm{d}}r{\rm{d}}\theta $
$ = \mathop \smallint \limits_{\theta = 0}^{2\pi } \mathop \smallint \limits_{r = 0}^{\sqrt 2 } {r^3}\left( {z|_0^r} \right){\rm{d}}r{\rm{d}}\theta $
$ = \mathop \smallint \limits_{\theta = 0}^{2\pi } \mathop \smallint \limits_{r = 0}^{\sqrt 2 } {r^4}{\rm{d}}r{\rm{d}}\theta $
$ = \frac{1}{5}\mathop \smallint \limits_{\theta = 0}^{2\pi } \left( {{r^5}|_0^{\sqrt 2 }} \right){\rm{d}}\theta $
$ = \frac{{4\sqrt 2 }}{5}\mathop \smallint \limits_{\theta = 0}^{2\pi } {\rm{d}}\theta = \frac{{8\sqrt 2 \pi }}{5}$
Thus, $\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal W}^{} {x^2} + {y^2}{\rm{d}}V = \frac{{8\sqrt 2 \pi }}{5}$.
