I am having issue to understand, why using two approaches I get two different results, i.e. where I am making a mistake:
Calculate integral: $$\int_{M}{\sqrt{x^2 + y^2 + z^2}}$$
where: $M := \{(x, y, z) \in \mathbb{R^3}: x^2 + y^2 + z^2 \leq 2, x^2 + y^2 + z^2 \leq 2z \}$.
Using substitution (spherical coordinates): $\rho(r,t,s) = (r\cos{t}\cos{s}, r\sin{t}\cos{s}, r\sin{s})$ on $(0, \infty)\times(0, 2\pi)\times(-\frac{\pi}{2}, \frac{\pi}{2})$. Jacobian of $\rho$ is $r^{2}\cos{s}$.
Which gives us (both conditions have to be true):
\begin{align} 0& < r < \sqrt{2} \\ 0& < r < 2\sin{s} \end{align}
$2\sin{s} < \sqrt{2}$ at $(0, \frac{\pi}{4})$, so integral we want to calculate is:
$$\int_{0}^{2\pi}{\int_{0}^{\frac{\pi}{4}}{ \int_{0}^{2\sin{s}} r^3 \cos{s} \mathrm dr\mathrm ds \mathrm dt}} + \int_{0}^{2\pi}{\int_{\frac{\pi}{4}}^{\frac{\pi}{2}}{ \int_{0}^{\sqrt{2}} r^3 \cos{s} \mathrm dr\mathrm ds \mathrm dt}} = 2\left(1 - \frac{2\sqrt{2}}{5}\right)\pi \approx 2.72$$
Another approach is merging two inequalities into single one, which gives us:
$0 < r < \sin{s} + \frac{1}{\sqrt{2}}$, then integral will be:
$$\int_{0}^{2\pi}{\int_{0}^{\frac{\pi}{2}}{ \int_{0}^{\sin{s} + \frac{1}{\sqrt{2}}} r^3 \cos{s} \mathrm dr\mathrm ds \mathrm dt}} = \frac{1}{2}\left(\frac{29}{20} + \sqrt{2}\right)\pi \approx 4.49$$
Thanks.