How can I rewrite my code to make it simpler?

Question

Two months ago,I written a function IRobot to caculate the $I_{xx},I_{yy},I_{zz}$ and $I_{xy},I_{xz},I_{yz}$,shown as below:

  IRobot[IStyle_String, ρ_, VaribleRange_List] := Module[
   {IRobotResult},
    IRobotResult =
    Which[
        IStyle == "xx", Integrate[ρ (y^2 + z^2),
          {z, VaribleRange[[1, 1]], VaribleRange[[1, 2]]}, 
          {y, VaribleRange[[2, 1]], VaribleRange[[2, 2]]},
          {x, VaribleRange[[3, 1]], VaribleRange[[3, 2]]}],
        IStyle == "yy", Integrate[ρ (x^2 + z^2),
          {z, VaribleRange[[1, 1]], VaribleRange[[1, 2]]}, 
          {y, VaribleRange[[2, 1]], VaribleRange[[2, 2]]}, 
          {x, VaribleRange[[3, 1]], VaribleRange[[3, 2]]}],
        IStyle == "zz", Integrate[ρ (x^2 + y^2),
          {z, VaribleRange[[1, 1]], VaribleRange[[1, 2]]},
          {y, VaribleRange[[2, 1]], VaribleRange[[2, 2]]},
          {x, VaribleRange[[3, 1]], VaribleRange[[3, 2]]}],
        IStyle == "xy", Integrate[ρ x y,
           {z, VaribleRange[[1, 1]], VaribleRange[[1, 2]]},
           {y, VaribleRange[[2, 1]], VaribleRange[[2, 2]]}, 
           {x, VaribleRange[[3, 1]], VaribleRange[[3, 2]]}],
        IStyle == "xz", Integrate[ρ x z,
           {z, VaribleRange[[1, 1]], VaribleRange[[1, 2]]}, 
           {y, VaribleRange[[2, 1]], VaribleRange[[2, 2]]},
           {x, VaribleRange[[3, 1]], VaribleRange[[3, 2]]}],
       IStyle == "yz", Integrate[ρ y z,
           {z, VaribleRange[[1, 1]], VaribleRange[[1, 2]]}, 
           {y,VaribleRange[[2, 1]], VaribleRange[[2, 2]]},
           {x,VaribleRange[[3, 1]], VaribleRange[[3, 2]]}]
    ]
  ]

And I can use it correctly.

 IRobot["xx", ρ, {{-h1, 0}, {-r1, r1}, {-L1 - Sqrt[r1^2 - y^2], Sqrt[r1^2 - y^2]}}]
 (*===>*)
 1/12 h1 r1 (4 h1^2 (2 L1 + \[Pi] r1) + r1^2 (8 L1 + 3 \[Pi] r1)) ρ

However,I would like to simplify my code by Similarity,my trail as below:

 Flatten /@Thread@List[{z, y, x},
   Apply[Part[VaribleRange, ##] &, Table[{i, j}, {i, 1, 3}, {j, 1, 2}], {2}]]
(*==>*)
 {{z, VaribleRange[[1, 1]], VaribleRange[[1, 2]]},
  {y, VaribleRange[[2, 1]], VaribleRange[[2, 2]]},
  {x, VaribleRange[[3, 1]], VaribleRange[[3, 2]]}}

Trial 1

IRobot[IStyle_String, ρ_, VaribleRange_List] := Module[
   {IRobotResult,intlist},
   intlist=
    Flatten /@Thread@List[{z, y, x},
   Apply[Part[VaribleRange, ##] &, Table[{i, j}, {i, 1, 3}, {j, 1, 2}], {2}]];
    IRobotResult =
    Which[
        IStyle == "xx", Integrate[ρ (y^2 + z^2),intlist],
        IStyle == "yy", Integrate[ρ (x^2 + z^2),intlist],
        IStyle == "zz", Integrate[ρ (x^2 + y^2),intlist],
        IStyle == "xy", Integrate[ρ x y,intlist],
        IStyle == "xz", Integrate[ρ x z,intlist],
        IStyle == "yz", Integrate[ρ y z,intlist]
    ]
  ]

Unfortunately,it failed.

Trail2

 IRobot[IStyle_String, ρ_, VaribleRange_List] := Module[
   {IRobotResult,intlist},
   intlist[z_,y_,x_,range_]:=
    Flatten /@Thread@List[{z, y, x},
   Apply[Part[range, ##] &, Table[{i, j}, {i, 1, 3}, {j, 1, 2}], {2}]];
    IRobotResult =
    Which[
        IStyle == "xx", Integrate[ρ (y^2 + z^2),intlist[z,y,x,VaribleRange]],
        IStyle == "yy", Integrate[ρ (x^2 + z^2),intlist[z,y,x,VaribleRange]],
        IStyle == "zz", Integrate[ρ (x^2 + y^2),intlist[z,y,x,VaribleRange]],
        IStyle == "xy", Integrate[ρ x y,intlist[z,y,x,VaribleRange]],
        IStyle == "xz", Integrate[ρ x z,intlist[z,y,x,VaribleRange]],
        IStyle == "yz", Integrate[ρ y z,intlist[z,y,x,VaribleRange]]
    ]
  ]

Also it ends in failure!

Or

I want to by

 Which@@(
 Flatten@{IStyle == #1, Integrate[ρ #2, intlist[z,y,x,VaribleRange]]} & @@@
 {{"xx", (y^2 + z^2)}, {"yy", (x^2 + z^2)}, {"zz", (x^2 + y^2)},
  {"xy", x y}, {"xz", x z}, {"yz", y z}})
  (*==>*)

to achieve

  Which[
        IStyle == "xx", Integrate[ρ (y^2 + z^2),intlist[z,y,x,VaribleRange]],
        IStyle == "yy", Integrate[ρ (x^2 + z^2),intlist[z,y,x,VaribleRange]],
        IStyle == "zz", Integrate[ρ (x^2 + y^2),intlist[z,y,x,VaribleRange]],
        IStyle == "xy", Integrate[ρ x y,intlist[z,y,x,VaribleRange]],
        IStyle == "xz", Integrate[ρ x z,intlist[z,y,x,VaribleRange]],
        IStyle == "yz", Integrate[ρ y z,intlist[z,y,x,VaribleRange]]
    ]

So my question is why and how to revise it.

Answer 1

Proposal

I recommend something like this:

IRobot2[IStyle_String, ρ_, VaribleRange_List] :=
  Block[{x, y, z},
    With[{expr = 
     Switch[IStyle,
       "xx", (y^2 + z^2),
       "yy", (x^2 + z^2), 
       "zz", (x^2 + y^2),
       "xy", x y,
       "xz", x z,
       "yz", y z
     ]},
     Integrate[ρ expr, ##] & @@ Join[{{z}, {y}, {x}}, VaribleRange, 2]
    ]
  ]

The Block is there to localize x, y, z to prevent these Symbols from incorrectly evaluating while they are manipulated, before passing them to Integrate using SlotSequence and Apply.

---

Code review

I first posted my recommendation above because it is faster for me to do so than to analyze and debug your attempts that did not work but as requested I will do the latter now.

The Trial 1 code throws this error when used:

Integrate::ilim: Invalid integration variable or limit(s) in {{z,-h1,0},{y,-r1,r1},{x,-L1-Sqrt[r1^2-Power[<<2>>]],Sqrt[r1^2-y^2]}}. >>

That should draw your attention to that part of the code, meaning intlist and how it is used in Integrate. The problem arises because intlist is (indeed) a list, whereas the parameters of Integrate need to be a sequence, i.e. 1, 2, 3 rather than {1, 2, 3}. In my own code above I used SlotSequence for this because it is more universal and I could not remember off-hand the way in which Integrate evaluated its arguments. However your Trial 1 code can be corrected by wrapping the existing right-hand-side definition of intlist in Sequence @@ ( . . . ), or a bit more concisely:

intlist = Sequence @@ (
  Flatten /@ Thread@{ {z, y, x}, Array[VaribleRange[[##]] &, {3, 2}] }
 )

I stated that using SlotSequence is more robust. Here is an example where it is needed:

iterSeq = Sequence[{x, 5}, {i, x}];
Table[x + i^2, iterSeq]              (*failure*)

iterList = {{x, 5}, {i, x}} ;
Table[x + i^2, ##] & @@ iterList     (*success*)

(Attributes[Table] includes HoldAll and iterSeq is not evaluated.)

Trial 2 can be corrected in the same manner.

The second attempt at Trial 2 was clever but you didn't quite get it right. Which needs input in the form of:

Which[
  test1, value1,
  test2, value2,
  . . .
]

While your code results in:

Which[
  {test1, value1},
  {test2, value2},
  . . .
]

Also you are not doing anything to Hold the "value" expressions, therefore Integrate will evaluate prematurely.

To correct these problems we need to Flatten the entire expression, not the sub-lists, and we need to add expression holding:

Which @@ Flatten[
 {IStyle == #1, Unevaluated @ Integrate[ρ #2, intlist[z, y, x, VaribleRange]]} & @@@
   {{"xx", (y^2 + z^2)},
    {"yy", (x^2 + z^2)},
    {"zz", (x^2 + y^2)},
    {"xy", x y},
    {"xz", x z},
    {"yz", y z}}
]

Unevaluated should suffice here (I did not actually test this section of code) but if not you could use Hold in its place, then add a ReleaseHold outside of Which.