Volume of a spherical box without integral. A spherical box is the solid defined by $\rho_0\le \rho \le \rho_1,\ \theta_0\le \theta \le \theta_1,\ \phi_0\le \phi\le \phi_1$ in spherical coordinates. It looks like 

Picture credit: KhanAcademy. 
Is there an elementary way to compute the volume of this spherical box? When I used integrals, I have found that the volume is 
$$\frac13 (\rho_1^3-\rho_0^3)(\theta_1-\theta_0)(\cos(\phi_1)-\cos(\phi_0)).$$
 A: Assuming $\phi$ to be the polar angle, a constant value of it corresponds
to a circular cone with apex at the origin, which intercepts a spherical cap of surface area
$$
A\left( {r,\phi } \right) = 2\pi r^{\,2} \left( {1 - \cos \phi } \right)
$$
which gives a ratio wrt the total surface of the sphere of
$$
\omega \left( \phi  \right) = {{A\left( {r,\phi } \right)} \over {4\pi r^{\,2} }} = {1 \over 2}\left( {1 - \cos \phi } \right)
$$
and the ratio comprised between $\phi_0$ and $\phi_1$ will be
$$
\omega \left( {\phi _{\,0} ,\phi _{\,1} } \right) =  - {1 \over 2}\left( {\cos \phi _{\,1}  - \cos \phi _{\,0} } \right)
$$
The dihedral angle $\theta_1 - \theta_0$ will then clearly intercept the fraction
$$
\omega \left( {\phi _{\,0} ,\phi _{\,1} ,\theta _{\,0} ,\theta _{\,1} } \right) = {{\left( {\theta _{\,1}  - \theta _{\,0} } \right)} \over {2\pi }}\omega \left( {\phi _{\,0} ,\phi _{\,1} } \right)
 =  - {{\left( {\theta _{\,1}  - \theta _{\,0} } \right)} \over {4\pi }}\left( {\cos \phi _{\,1}  - \cos \phi _{\,0} } \right)
$$
which in turn corresponds to the fraction of the volume intercepted between the two spheres
$$
\eqalign{
  & V = \omega \left( {\phi _{\,0} ,\phi _{\,1} ,\theta _{\,0} ,\theta _{\,1} } \right){{4\pi } \over 3}\left( {R^{\,3}  - r^{\,3} } \right) =   \cr 
  &  =  - \left( {R^{\,3}  - r^{\,3} } \right){1 \over 3}\left( {\theta _{\,1}  - \theta _{\,0} } \right)\left( {\cos \phi _{\,1}  - \cos \phi _{\,0} } \right) \cr} 
$$
