# Certain volume formula?

get the mathematical expression for the volume formula. known coordinate system x, y, z receive .stl model  rb1 = 28.5;
r00 = 32;
PhiEnd = 0.318;
PsiEnd = 0.6144;
(***********************************)
x = rb1*Sin[Psi] - (rb1*Psi - r00*(1 - Cos[Phi]))*Cos[Psi];
y = rb1*Cos[Psi] + (rb1*Psi - r00*(1 - Cos[Phi]))*Sin[Psi];
z = r00*Sin[Phi];
(***********************************)
func1 = {x, y, z};
func = {{x, y, z}, {Phi, 0, PhiEnd}, {Psi, 0,
PsiEnd}};
(*a=ParametricPlot3D[{x, y,z}, {Phi, 0, PhiEnd}, {Psi, \
0,PsiEnd},MaxRecursion\[Rule]4,PlotStyle\[Rule]{Orange,Specularity[\
White,10]},Axes\[Rule]None,Mesh\[Rule]None]*)
a = ParametricPlot3D @@ func ;(*Parametric function definition*)
(***********************************)
Off[NIntegrate::slwcon]; (*Disable warning when \
integration*)
(***********************************)
ar = Area @@
func ;(*Calculation of surface area embedded numerical method*)
SetDirectory[
NotebookDirectory[]]; (*Set the directory for exporting the current c\
location of this file*)
exp = Export["a.dxf",
a];
(***********************************)

(*Cross[D[#1,Phi],D[#1,Psi]]&@@func *) (*Calculation of the area through \
intgeral formula*)

(***************************************************)
(*Steps*)
d1 = D[func1, Phi];(*Derivative Phi*)
d2 = D[func1, Psi];(*Derivative Psi*)
cr = Cross[d1, d2];(*Derivative intersection vector d1 и d2*)
(*Formula
Cross[{a,b,c},{x,y,z}]
{-c y+b z,c x-a z,-b x+a y}
*)
(* 1024=32*32
912=28,5*32 *)
nr = Norm[cr]; (*Vector length cr*)
nr] ;(*Derivation of the sub-expression in the traditional form*)
int1 = Hold[
Integrate[\[Sqrt](Abs[-1024 cos(Psi) cos^2(Phi) -
912. Psi cos(Psi) cos(Phi) + 1024 cos(Psi) cos(Phi) +
0.]^2 + Abs[
912. Psi sin(Phi) cos^2(Psi) +
1024 cos(Phi) sin(Phi) cos^2(Psi) -
1024 sin(Phi) cos^2(Psi) + 912. Psi sin(Phi) sin^2(Psi) +
1024 cos(Phi) sin(Phi) sin^2(Psi) -
1024 sin(Phi) sin^2(Psi) + 0.]^2 + Abs[
1024 sin(Psi) cos^2(Phi) + 912. Psi sin(Psi) cos(Phi) -
1024 sin(Psi) cos(Phi) + 0.]^2), {Phi, 0, PhiEnd}, {Psi, 0,
PsiEnd}]];
(*The withdrawal of integral in traditional form*)
int2 = Integrate[nr, {Phi, 0, PhiEnd}, {Psi, 0, PsiEnd}];
(***************************************************************)
(*One expression*)
abs = Evaluate[
Norm[Cross[D[#1, Phi], D[#1, Psi]] & @@
func]];(*Integral area formula*)
int = Integrate[abs, {Phi, 0, PhiEnd}, {Psi, 0,
PsiEnd}]; (*Area calculation using the integral formula*)
(*Timing[Integrate[Evaluate@Norm@Cross[D[#1,Phi],D[#1,Psi]],##2]&@@\func]*)
conv = int/ ar ;


The surface area was calculated..how to calculate surface volume ...? by coordinate equations, the surface was obtained by integration, and a mathematical expression was obtained

Please help me to get the expressions of the volume of the figure, similar to the calculation of the surface area, according to the specified conditions

• Ah, this looks better now. – Henrik Schumacher Apr 11 '19 at 15:18
• These many sin(Phi), cos^2(Psi)m etc. might cause you some trouble as this is not correct Mathematica syntax. Rember: Built-in symbols always start with a capital letter and function arguments ought to be embraced by brackets ([ ]), not by parentheses (( )). – Henrik Schumacher Apr 11 '19 at 15:20
• @Alexsey S Your question is not clear how to calculate surface volume?. The surface has no volume. Apparently there is a question about the volume of the 3D region bounded by the surface. But from your data it is not clear how the 3D region is built. Try to use R=DiscretizeRegion[ ParametricRegion[{x, y, z} /. rb1 -> r, {{Phi, 0, PhiEnd}, {Psi, 0, PsiEnd}, {r, 28.5, 33.5}}]] and RegionMeasure[R]`. (Out[]=115.637) – Alex Trounev Apr 12 '19 at 12:20
• Cross-posted community.wolfram.com/groups/-/m/t/1658725 – Michael E2 Apr 13 '19 at 3:33