Triple Integral boundaries and solution

Question

So I'm working on triple integrals, specifically on those where I have a region bounded by various surfaces. (In other words, the ranges for each integrals are not given, but must be calculated.)

I know how to do the integral once I have the boundaries of each integral. I also know how to use regionplot and implicitregion to graph my surface. Mathematica then shows a nice picture of the bounded region.

What I don't know how to do is tell Mathematica to immediately calculate the integral. I've found a couple of examples online but in each case they seemed to have already calculated the boundaries and then just entered them into the Integrate function.

I need to figure out the interval. How do I figure out the boundaries or just immediately tell Mathematica to calculate the integral (with the boundaries being part of the effort)?

Thanks.

Here's an example of the code I found in another document. It all makes sense and is replicable until you get to the Integrate function. It has boundaries. Are these done by hand? Or is there a way to have Mathematica create them?

Here is the code that doesn't seem to work. I did follow Bob's method below, and I recreated his example, but the following doesn't work. Mathematica says "Volume = undefined" and the integral is zero.

Clear["Global`*"]

reg = ImplicitRegion[    y==4-z&&y==-x^2&&y==0&&z==0, {x, y, z}];

Region[reg, Axes -> True, Boxed -> True]

volume[reg]

Integrate[1, {x, y, z} ∈ reg]

Here's an example I found online. But again the limits are entered in here.

Answer 1

Clear["Global`*"]

reg = ImplicitRegion[
   z <= 4 - 4 (x^2 + y^2) && (z >= (x^2 + y^2)^2 - 1), {x, y, z}];

Region[reg, Axes -> True, Boxed -> True]

The volume can be calculated by any of the following methods:

Volume[reg]

(* (8 π)/3 *)

RegionMeasure[reg]

(* (8 π)/3 *)

Integrate[1, {x, y, z} ∈ reg]

(* (8 π)/3 *)

EDIT:

Alternatively,

Integrate[
  Boole[z <= 4 - 4 (x^2 + y^2) && (z >= (x^2 + y^2)^2 - 1)],
 {x, -Infinity, Infinity}, {y, -Infinity, Infinity},
 {z, -Infinity, Infinity}]

(* (8 π)/3 *)

Answer 2

New Edition

One way is calculate all the possible measure of regions.

BTW, the questioner's equations can't enclose any finite regions.Here we change the sign of the second equation.

equations = {y == 4 - z, y == x^2, y == 0, z == 0};
expressions = (First[#] - Last[#]) & /@ equations;
inequalities = 
  And @@@ Tuples@
    Outer[Construct[#2, #1] &, 
     expressions, {GreaterEqual[#, 0] &, LessEqual[#, 0] &}];
regions = ImplicitRegion[#, {x, y, z}] & /@ inequalities;
PositionIndex[Volume /@ regions]

regions[[9]] // Volume
(* 256/15 *)

regions[[9]] // 
 RegionPlot3D[#, PlotPoints -> 100, MaxRecursion -> 2] &

Original

equations = {y == 4 - z, y == x^2, y == 0, z == 0};
expressions = (First[#] - Last[#]) & /@ equations;
inequalities = 
  And @@@ Tuples[
    Thread[{GreaterEqual[#, 0] & /@ expressions, 
      LessEqual[#, 0] & /@ expressions}]];
regions = ImplicitRegion[#, {x, y, z}] & /@ inequalities;
PositionIndex[RegionMeasure /@ regions]
inequalities[[9]]
inequalities[[11]]

<|∞ -> {1, 2, 3, 5, 6, 7, 10, 12, 13, 14, 15, 16}, 0 -> {4, 8}, 256/15 -> {9}, 4 -> {11}|>

-4 + y + z <= 0 && -x^2 + y >= 0 && y >= 0 && z >= 0

-4 + y + z <= 0 && -x^2 + y >= 0 && y <= 0 && z >= 0