Volume of a sphere "corner" I would like to find the formula of the volume of the "corner" of a sphere of radius R, more specifically the volume delimited in a sphere by the intersection of two perpendicular planes, one parallel to the (Oxz) plane at a distance from it, and the other parallel to the (Oxy) plane also at a distance D from it.
This might not be extremely clear without a drawing, so I will explain where this problem originates from:
Consider a cube with edge $2D$, and a sphere of radius $R$ with the center of the sphere at the center of the cube. The sphere grows (e.g. with time) and we would like to determine the volume of the sphere lying inside the cube.
At first, the sphere is totally inside the cube, so the answer is easy and that volume is
$$V = \frac{4}{3} \pi \times R^3$$
This formula holds up to the point where the sphere is inscribed in the cube, ie $R = D$.
Then, the sphere keeps growing. We thus need to remove from the formula above the volume of each spherical cap above each face to obtain the formula of the sphere volume inside the cube:
$$V = \frac{4}{3}\pi \times R^3 - 6 \times V_\text{cap} $$ with
$$ V_\text{cap} = \frac{\pi h^2}{3} \times (3R - h) $$
where $ h = R-D $ is the height of the spherical cap. This formula holds until the spherical caps start overflowing over the edges and intersect with the spherical cap from the adjacent face (at $ R = D \sqrt 2$ ). for $ R > D \sqrt 2$, the formula above actually removes twice the volume of the intersection of each pair of spherical caps (one pair per edge), while we only need to remove it once. Hence the actual formula is:
$$ V = V = \frac{4}{3} \pi \times R^3 - 6 \times V_\text{cap} + 12 \times V_\text{cap-intersection} $$
Now the tricky part (where I need help) is to actually find the formula for the volume of the intersections of each pair of spherical caps. If you draw it, it actually looks like the corner of a sphere (the space stuck between the two perpendicular planes intersecting and the edge of the sphere). I was not able to compute correctly that volume using integral calculus (probably due to some mistakes in the long and tedious calculus), hence my question. 
Any help or lead to solve this problem would be greatly appreciated.

We want to calculate the volume of the set
$$ C = \{ (x,y,z)\in\mathbb{R}^3: x^2+y^2+z^2\leq R^2, x\geq s, y\geq t\},$$
that is the Ball of radius $R$ (not sphere!) intersected with the halfspace in direction $(1,0,0)$ at distance $s>0$ and the halfspace in direction $(0,1,0)$ at distance $t>0$.
By Fubini we may write
$$ \mathrm{vol}(C) = \int_{s}^{\sqrt{R^2-t^2}} \int_{t}^{\sqrt{R^2-\alpha^2}} \int_{-\sqrt{R^2-\alpha^2-\beta^2}}^{\sqrt{R^2-\alpha^2-\beta^2}} \, d\gamma\, d\beta\, d\alpha.$$
or
$$\mathrm{vol}(C) = \int_{s}^{\sqrt{R^2-t^2}} \int_{t}^{\sqrt{R^2-\alpha^2}} 2\sqrt{R^2-\alpha^2-\beta^2}\, d\beta\,d\alpha.$$
