Calculating volume [closed]

Question

I'm new in mathematica and I'm stucked on how I can get the volume of a solid created by the inequation:

(radFld[Grp, "Bx", {x, y, z}] - Btotal)/Btotal <= Value

where radFld is a function from the plug-in Radia and it does calculs given an x and z to find a magnetic field of the configuration of magnetics given by Grp. In other words: radFld[Grp, "Bx", {x, y, z}] - Btotal)/Btotal = f(x,y,z)

Also, "vmax" is a constant.

To get the volume, I tried this:

Integrate[Boole[(radFld[Grp, "Bx", {x, y, z}] - Btotal)/Btotal <= vmax], {x, -3, 3}, {y, -3, 3}, {z, -10, 10}]

but my results seems not to be correct, as it goes from 0 to 540 and does not assume any other value between this range, whatever is "vmax".

Also, I tried this to make sure the region has been modified with "vmax":

RegionPlot3D[(radFld[Grp, "Bx", {x, y, z}] - Btotal)/Btotal] <= vmax , {x, -3, 3}, {y, -3, 3}, {z, -10, 10}]

and I got different regions for different "vmax".

Is there any other suggestion to get the volume or of what I am doing wrong?

Thanks in advance

Answer 1

This is too long for a comment, but I can remove this answer later. I don't really understand your sentence

but my results seems not to be correct, as it goes from 0 to 540 and does not assume any other value between this range, whatever is "vmax".

Maybe you can try to rephrase this, so that it will become crystal clear, what you mean ;-)

Additionally, the approach you already mentioned should work. Let's assume a simple example: a sphere. We can define a sphere in the same way you defined your function

f[v_] := Sqrt[v.v]
RegionPlot3D[f[{x, y, z}] < 1, {x, -2, 2}, {y, -2, 2}, {z, -2, 2}]

Now you can use NIntegrate in the same you already showed with Integrate

NIntegrate[Boole[f[{x, y, z}] < 1], {x, -2, 2}, {y, -2, 2}, {z, -2, 2}]

and you get 4.18879 which happens to be $$\frac{4}{3}\pi$$ which is of course the correct solution. You can try other radii and you'll see that the result is correct.