Trirectangular Tetrahedron

Author

Eric W. Weisstein
February 13, 2012

This notebook downloaded from http://mathworld.wolfram.com/notebooks/SolidGeometry/TrirectangularTetrahedron.nb.

For more information, see Eric's MathWorld entry http://mathworld.wolfram.com/TrirectangularTetrahedron.html.

©2011 Wolfram Research, Inc. except for portions noted otherwise

Figure

pts={{0,0,0},{a,0,0},{0,b,0},{0,0,c}};

sub=Subsets[pts,{3}]

OrientedQ[#,{.1,.1,.1}]&/@(Polygon/@sub/.{a->1,b->2,c->3})

Graphics3D[(poly=Polygon/@{
{{0,0,0},{0,b,0},{a,0,0}},
{{0,0,0},{a,0,0},{0,0,c}},
{{0,0,0},{0,0,c},{0,b,0}},
{{a,0,0},{0,b,0},{0,0,c}}
})/.{a->1,b->2,c->3}]

Properties

FullSimplify[And@@PolyhedronInequalities[poly,{x,y,z}],a>0&&b>0&&c>0]

ineq[x_,y_,z_]:=z≥0&&y≥0&&x≥0&&b c x+a c y+a b z≤a b c

Block[{a=1,b=2,c=3},RegionPlot3D[z≥0&&y≥0&&x≥0&&b c x+a c y+a b z≤a b c,{x,0,a},{y,0,b},{z,0,c},BoxRatios->Automatic]]

Centroid

Integrate[{x,y,z}Boole[ineq[x, y, z]], {x, -∞, ∞}, {y, -∞, ∞}, {z, -∞, ∞}, Assumptions -> a > 0 && b > 0 && c > 0]

(1/((a b c)/6)){(1/24) (a^2) b c,(1/24) a (b^2) c,(1/24) a b (c^2)}

Volume

Integrate[Boole[ineq[x, y, z]], {x, -∞, ∞}, {y, -∞, ∞}, {z, -∞, ∞}, Assumptions -> a > 0 && b > 0 && c > 0]

InertaiTensor

Integrate[InertiaTensor[{x,y,z}]Boole[ineq[x, y, z]], {x, -∞, ∞}, {y, -∞, ∞}, {z, -∞, ∞}, Assumptions -> a > 0 && b > 0 && c > 0]

(1/((a b c)/6)){{(1/60) a b c ((b^2)+(c^2)),-(1/120) (a^2) (b^2) c,-(1/120) (a^2) b (c^2)},{-(1/120) (a^2) (b^2) c,(1/60) a b c ((a^2)+(c^2)),-(1/120) a (b^2) (c^2)},{-(1/120) (a^2) b (c^2),-(1/120) a (b^2) (c^2),(1/60) a b ((a^2)+(b^2)) c}}//Simplify

GenertalizedDiameter

FullSimplify[Max[#.#&/@(Subtract@@@Subsets[pts,{2}])],a>0&&b>0&&c>0]