Use a triple integral to find the volume of the solid bounded below by the cone $z=\sqrt{x^{2}+y^{2}}$ and bounded above by the sphere $x^{2}+y^{2}+z^{2}=392$.

Step 1 ：Convert the given equations into spherical coordinates. The cone $z=\sqrt{x^{2}+y^{2}}$ corresponds to $\phi = \frac{\pi}{4}$ in spherical coordinates, and the sphere $x^{2}+y^{2}+z^{2}=392$ corresponds to $\rho = 14$.

Step 2 ：Set up the triple integral to find the volume of the solid. We need to integrate over $\rho$ from 0 to 14, over $\phi$ from 0 to $\frac{\pi}{4}$, and over $\theta$ from 0 to $2\pi$. The volume element in spherical coordinates is $dV = \rho^{2}\sin\phi d\rho d\phi d\theta$.

Step 3 ：Perform the integration. The integrand is $\rho^{2}\sin(\phi)$.

Step 4 ：The result of the integration is a numerical value that represents the volume of the solid. This is the final answer.

Step 5 ：The volume of the solid is \(\boxed{\frac{2\pi}{3}(2744 - 1372\sqrt{2})}\).