(1 point) Find the volume of the solid that lies within the sphere $x^{2}+y^{2}+z^{2}=25$, above the $x y$ plane, and outside the cone $z=2 \sqrt{x^{2}+y^{2}}$.

Step 1 ：Convert the given conditions into spherical coordinates. The sphere $x^{2}+y^{2}+z^{2}=25$ becomes $\rho^{2}=25$, or $\rho=5$. The cone $z=2 \sqrt{x^{2}+y^{2}}$ becomes $\rho \cos \phi = 2 \rho \sin \phi$, or $\tan \phi = 1/2$.

Step 2 ：Since we are above the $xy$ plane, $\phi$ ranges from $\arctan(1/2)$ to $\pi/2$. $\theta$ ranges from $0$ to $2\pi$.

Step 3 ：The volume element in spherical coordinates is $\rho^{2} \sin \phi \, d\rho \, d\theta \, d\phi$.

Step 4 ：Integrate this volume element over the region defined by the given conditions. The limits of integration for $\rho$ are $0$ to $5$, for $\theta$ are $0$ to $2\pi$, and for $\phi$ are $\arctan(1/2)$ to $\pi/2$.

Step 5 ：The volume of the solid that lies within the sphere $x^{2}+y^{2}+z^{2}=25$, above the $x y$ plane, and outside the cone $z=2 \sqrt{x^{2}+y^{2}}$ is $\boxed{74.535599249993\pi}$.