I would like to see how the projection of the flat surface of a truncated sphere (as shown in the image) change when rotating the sphere. I would also want to know the area of the projection. Could anyone tell me how to do this? Thanks! truncated sphere

  • 2
    $\begingroup$ Please post your code. $\endgroup$
    – cvgmt
    Commented Aug 30, 2022 at 2:53
  • 1
    $\begingroup$ Welcome to the Mathematica Stack Exchange. Please provide code for the presented image. There are two pieces of information missing in this post: (1) The equation for the disk in 3D space and (2) the equation for the plane where you want to project the image of this disk. $\endgroup$
    – Syed
    Commented Aug 30, 2022 at 4:12

1 Answer 1


The projection can be done by simply setting one of the coordinates to a fixed value.

The area of the projection is more difficult to calculate. The projection of the whole sphere is a circle. The circular cut projects to an ellipse. We first subtract the area of the circular segment of the projection that is replaced by half an ellipse and then add the area of the ellipse.

To make things simpler, we choose a sphere of radius: 1:

acs[h_, r_] = 
  r^2 ArcCos[1 - h/r] - (r - h) Sqrt[
     2 r h - h^2];(*area circular segment*)
ael[phi_, dphi_, r_] = 
  Pi  r  (Sin[phi + dphi] - Sin[phi]);(*area ellipse*)
 rm = RotationMatrix[phi, {0, 1, 0}];
 sp = ParametricPlot3D[
   rm . {Sin[p] Sin[t], Cos[p] Sin[t], Cos[t]}, {p, 0, 2 Pi}, {t, 
    dphi, Pi}, PlotPoints -> 20, 
   PlotLabel -> StringForm["Area= ``", Pi +
      Which[Pi/2 > phi + dphi || Pi/2 < phi - dphi , 0,
       phi <= Pi/2, 1/2 ael[phi, dphi, 1] - acs[1 - Sin[phi], 1],
       True, 1/2 ael[Pi - phi, dphi, 1] - acs[1 - Sin[Pi - phi], 1]
 proj = ParametricPlot3D[{#[[1]], #[[2]], -1} &@(rm . {Sin[p] Sin[t], 
       Cos[p] Sin[t], Cos[t]}), {p, 0, 2 Pi}, {t, dphi, Pi}, 
   PlotRange -> {{-1, 1}, {-1, 1}, {-1, 1}}, PlotPoints -> 20];
 Show[sp, proj, PlotRange -> {{-1, 1}, {-1, 1}, {-1, 1}}]
 , {phi, 0, Pi}, {{dphi, 0.8}, 0, Pi}]

enter image description here

  • $\begingroup$ Thanks a lot! That's very helpful. Actually, the projection I am interested in is the circular cut part (the flat disk). May I ask how to revise the code to show the area and projection of the disk (it is shown as a hole in the current code) as the sphere rotates(with the same manipulation parameter phi and dphi)? It seems to be easier than the current case, could you give me some hints? Thanks! $\endgroup$
    – Nick W
    Commented Aug 31, 2022 at 0:34
  • $\begingroup$ The half axes of the ellipse are: r ( Sin[ phi + dphi] - Sin[phi ]) and r Sqrt[1-Sin[phi]^2] and the center of the projection: {r Sin[phi],0,-1} $\endgroup$ Commented Aug 31, 2022 at 8:11

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