Volume between a sphere and 4 planes Given the attached figure, we are interested in the volume between the sphere with radius R centered at $O$ given by the sphere equation $x^2 +y^2+z^2=R^2$ and the four planes as described: 
Plane AOC with equation $a_{1}x+b_{1}y+c_{1}z = D_{1}$
Plane COB with equation $a_{2}x+b_{2}y+c_{2}z = D_{2}$
Plane AOB with equation $a_{3}x+b_{3}y+c_{3}z = D_{3}$
Plane ABC with equation $a_{4}x+b_{4}y+c_{4}z = D_{4}$
The resulting volume is the volume between the plane ABC and the spherical triangle. How can be extracted?
Volume subtended by 3 points of sphere and the corresponding plane
 A: Hint:  
Once you have the tetrahedron $O,A,B,C$ then you can find the solid angle
 it subtends in $O$ by applying the formula given in the link.   
From there you have the volume of the "spherical pyramid", and the rest follows easily.
Since you want an alternative approach as a check, the points $A,B,C$ lie on great circles, as they also lie on planes through the origin.
 They are therefore a Spherical Triangle, 
of which you know all the parameters once you have the vectors from the origin: vertices, sides, angles, etc.
 So you can use the formulas of spherical trigonometry to check.
Finally, if you want to go through integration, then use Cartesian coordinates.
Take , e.g., $OA$ as the $z$ axis. Take the longitudinal plane bisecting $OAB, OAC$ as the $xz$ plane, and get the relevant symmetric projections$B',C'$ of $B,C$ on the $xy$ plane.
Project the great circle on $OBC$ on the $xy$ plane: you get an ellipse with semi-axes $R \cos \theta, R$.
Integrate over $(x,y)$ laying inside the ellipse, and within the lines $OB', \,OC'$, and with $z$ ranging from the plane $OBC$ to the sphere.
