Find the volume between $z=\sqrt{x^{2}+y^{2}}$ and $x^2+y^2+z^2=2$ in spherical cordinates I am asking to find the volume of the volume trap above the cone $z=\sqrt{x^{2}+y^{2}}$ and below the sphere   $x^2+y^2+z^2=2$
When I checked the solution I noticed that it was writen as $$V=\int_{0}^{2 \pi} \int_{0}^{\frac{\pi}{4}} \int_{0}^{\sqrt{2}} r^{2} \sin \theta \,d r \,d \theta \,d \varphi$$ and my question is why the boundries of $\theta$ is between $0$ to $\frac{\pi}{4}$ and not $0$ to $\pi$.
why  $0$ to $\pi$ is wrong? I just can't imagine the scenerio in my head

Intersection of $x^2+y^2+z^2=2$ and $z=\sqrt{x^{2}+y^{2}}$ on $OXY$ plane gives $x^2+y^2=1$ so on Cartesian coordinates we have $$\int\limits_{-1}^{1}\int\limits_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}}\int\limits_{\sqrt{x^{2}+y^{2}}}^{\sqrt{2-x^2-y^2}}dxdydz = 4\int\limits_{0}^{1}\int\limits_{0}^{\sqrt{1-x^2}}\int\limits_{\sqrt{x^{2}+y^{2}}}^{\sqrt{2-x^2-y^2}}dxdydz$$
Now let's take spherical coordinates:
\begin{array}{}
x = r \sin \phi \cos \theta;  \\ 
y = r \sin \phi \sin \theta; \\
z = r \cos \phi
\end{array}
From $z$ coordinate bounds
$$r \sin \phi \leqslant r \cos \phi \leqslant \sqrt{2-r^2 \sin^2 \phi}$$
Left inequality gives $\sin \phi \leqslant \cos \phi$, from which we can obtain $\phi \leqslant \frac{\pi}{4}$. Right inequality gives $r \leqslant \sqrt{2}$.
So for integral in spherical coordinates we have
$$4\int\limits_{0}^{\frac{\pi}{2}}\int\limits_{0}^{\frac{\pi}{4}}\int\limits_{0}^{\sqrt{2}} r^{2} \sin \theta d r  d \phi d \theta$$
