This post is related to the question. There were proposed two ways of finding the volume of the cutted part of a sphere in the form of a wedge. Here the procedure is presented that shows the rest of the sphere. Parameters procedure: **R** - radius of the sphere, **H1** - the distance the first cutting plane to the plane xOy, **H2** - the distance the second cutting plane to the plane zOy. Necessary conditions: **R>0**, **H1>=0**, **H2>=0**,** H1^2+H2^2<R^2 **. For clarity, different surfaces are painted in different colors.

**restart;**

**Pic := proc (R::positive, H1::nonnegative, H2::nonnegative)**

**local A, B, C, E, F;**

**if R^2 <= H1^2+H2^2 then error "Should be H1^(2)+H2^(2)<R^(2)" end if;**

**A := plot3d([R*sin(theta)*cos(phi), R*sin(theta)*sin(phi), R*cos(theta)], phi = arctan(sqrt(-H1^2-H2^2+R^2), H2) .. 2*Pi-arctan(sqrt(-H1^2-H2^2+R^2), H2), theta = 0 .. Pi, color = green);**

**B := plot3d([R*sin(theta)*cos(phi), R*sin(theta)*sin(phi), R*cos(theta)], phi = -arctan(sqrt(-H1^2-H2^2+R^2), H2) .. arctan(sqrt(-H1^2-H2^2+R^2), H2), theta = 0 .. arccos(sqrt(R^2-H2^2-H2^2*tan(phi)^2)/R), color = green);**

**C := plot3d([R*sin(theta)*cos(phi), R*sin(theta)*sin(phi), R*cos(theta)], phi = -arctan(sqrt(-H1^2-H2^2+R^2), H2) .. arctan(sqrt(-H1^2-H2^2+R^2), H2), theta = arccos(H1/R) .. Pi, color = green);**

**E := plot3d([r*cos(phi), r*sin(phi), H1], phi = -arccos(H2/sqrt(R^2-H1^2)) .. arccos(H2/sqrt(R^2-H1^2)), r = H2/cos(phi) .. sqrt(R^2-H1^2), color = blue);**

**F := plot3d([H2, r*cos(phi), r*sin(phi)], phi = arccos(sqrt(-H1^2-H2^2+R^2)/sqrt(R^2-H2^2)) .. Pi-arccos(sqrt(-H1^2-H2^2+R^2)/sqrt(R^2-H2^2)), r = H1/sin(phi) .. sqrt(R^2-H2^2), color = gold);**

**plots[display](A, B, C, E, F, axes = none, view = [-1.5 .. 1.5, -1.5 .. 1.5, -1.5 .. 1.5], scaling = constrained, lightmodel = light4, orientation = [60, 80]);**

**end proc:**

Example of use:

**Pic(1, 0.5, 0.3);**

** **