Why does Mathematica give an incorrect answer to this multiple integral?

Question

This is not a new problem but I would like to understand why Mathematica gives the result that it does. (Volume of a hypersphere)

In[4]:= Integrate[
  Boole[x1^2 + x2^2 + x3^2 + x4^2 + x5^2 + x6^2 < r^2], {x1, -Infinity,
  Infinity}, {x2, -Infinity, Infinity}, {x3, -Infinity, 
  Infinity}, {x4, -Infinity, Infinity}, {x5, -Infinity, 
  Infinity}, {x6, -Infinity, Infinity}, Assumptions -> {r > 0}]

Out[4]= -(1/6) \[Pi]^3 r^6

A negative answer is obviously incorrect. Replace r^2 by 4

In[3]:= Integrate[
  Boole[x1^2 + x2^2 + x3^2 + x4^2 + x5^2 + x6^2 < 4], {x1, -Infinity, 
  Infinity}, {x2, -Infinity, Infinity}, {x3, -Infinity, 
  Infinity}, {x4, -Infinity, Infinity}, {x5, -Infinity, 
  Infinity}, {x6, -Infinity, Infinity}, Assumptions -> {r > 0}]

Out[3]= (32 \[Pi]^3)/3

This is correct.

If we reduce the number of variables, then we obtain a correct answer:

In[5]:= Integrate[
  Boole[(x1^2 + x2^2 + x3^2 + x4^2 ) < r^2], {x1, -Infinity, 
  Infinity}, {x2, -Infinity, Infinity}, {x3, -Infinity, 
  Infinity}, {x4, -Infinity, Infinity}, Assumptions -> {r > 0}]

Out[5]= (\[Pi]^2 r^4)/2

Answer 1

As an aside, the multiple integrals can be generalized as

vol[n_Integer?Positive] :=
 Module[{x, var},
  var = Array[x, n];
  Assuming[
   {Element[var, Reals], r > 0},
   Integrate[
     Boole[var.var < r^2],
     Sequence @@
      Evaluate[
       {#, -Infinity, Infinity} & /@ var]] //
    Simplify]]

Timing[vol /@ Range[5]]

(* {96.016268, {2*r, Pi*r^2, (4*Pi*r^3)/3, (Pi^2*r^4)/2, (8*Pi^2*r^5)/15}} *)

The timing is somewhat improved by changing the limits
of integration from $\{-\infty ,\infty \}$ to $\{-r, r\}$

vol[n_Integer?Positive] :=
 Module[{x, var},
  var = Array[x, n];
  Assuming[
   {Element[var, Reals], r > 0},
   Integrate[
     Boole[var.var < r^2],
     Sequence @@
      Evaluate[
       {#, -r, r} & /@ var]] //
    Simplify]]

Timing[vol /@ Range[5]]

(* {64.282676, {2*r, Pi*r^2, (4*Pi*r^3)/3, (Pi^2*r^4)/2, (8*Pi^2*r^5)/15}} *)

Bob Hanlon