# Unable to compute the measure of a region in four dimensions

I need to be able to calculate an integral with this implicit region:

ImplicitRegion[
16 π^2 -
Sqrt[32 π^2 + Abs[Cos[θ1] r1]^2 +
Abs[r1 Sin[θ1]]^2] Sqrt[
Abs[Cos[θ1] r1]^2 +
Abs[Cos[φ1] r1 Sin[θ1]]^2 +
Abs[r1 Sin[φ1] Sin[θ1]]^2] + Cos[θ1]^2 r1^2 +
Cos[φ1] r1^2 Sin[θ1]^2 <= 0.5 &&
16 π^2 -
Sqrt[32 π^2 + Abs[Cos[θ1] r1]^2 +
Abs[r1 Sin[θ1]]^2] Sqrt[
Abs[Cos[θ1] r1]^2 +
Abs[Cos[φ1] r1 Sin[θ1]]^2 +
Abs[r1 Sin[φ1] Sin[θ1]]^2] + Cos[θ1]^2 r1^2 +
Cos[φ1] r1^2 Sin[θ1]^2 >= -0.5 &&
16 π^2 -
Abs[r2] Sqrt[32 π^2 + Abs[r2]^2 +
Abs[r1 Sin[φ1] Sin[θ1]]^2 +
Abs[r1 Sin[θ1] - Cos[φ1] r1 Sin[θ1]]^2] +
r2^2 <= 0.5 &&
16 π^2 -
Abs[r2] Sqrt[32 π^2 + Abs[r2]^2 +
Abs[r1 Sin[φ1] Sin[θ1]]^2 +
Abs[r1 Sin[θ1] - Cos[φ1] r1 Sin[θ1]]^2] +
r2^2 >= -0.5 && 160 <= r1 <= 300 && 160 <= r2 <= 300 &&
0 <= φ1 <= 2 π && 0 <= θ1 <= 2 π,
{r1, r2, φ1, θ1}]


I'm sorry if it's difficult to visualize, but I wouldn't know how to do it properly in a reasonable time here. Anyway, it's the same expression four times basically ( $$f(\theta1,\phi1, r1,r2) \leq 0.5)$$ and Mathematica tells me:

Unable to compute the dimension of region

and I don't understand what's wrong, apart from the fact that it might have problems because it's in 4D. How could I formulate it differently?

### Edit

I managed to simplify it by making some assumptions:

ImplicitRegion[
-633.655 <=
r1 (r1 (4. + 1. θ1^4 - 2. θ1^2 ϕ1^2) -
Sqrt[(128 π^2 + r1^2 (4 + θ1^4)) (4 + θ1^4 + θ1^2 ϕ1^4)]) <=
-629.655 &&
314.827 + 2. r2^2 <=
r2 Sqrt[128 π^2 + 4 r2^2 + r1^2 θ1^2 ϕ1^2 (4 + ϕ1^2)] &&
r2 Sqrt[128 π^2 + 4 r2^2 + r1^2 θ1^2 ϕ1^2 (4 + ϕ1^2)] <=
316.827 + 2 r2^2 &&
r1 >= 0 && r2 >= 0 && 160 <= r1 <= 200 && 160 <= r2 <= 200 &&
-0.1 <= θ1 <= 0.1 && -0.1 <= ϕ1 <= 0.1,
{r1, r2, θ1, ϕ1}]


Now it doesn't give that error anymore, but instead it never finishes calculating.

### Edit

I need to integrate a function over that region: NIntegrate[fn[r1, r2, θ1, ϕ1], {r1, r2, ϕ1, θ1} ∈ IntegrationZone, Method -> "MonteCarlo"] but what my problem is, is that NIntegrate[], or more precisely, Volume[], cannot calculate the area of that function. If I change the method to e.g. GlobalAdaptive, I get:

iCopyExpr() called on symbol.

so that's why I'm trying MonteCarlo.

• It is necessary to fix Pi instead of \pi, Sqrt[] instead of Sqrt(), theta1 instead of theta_1,phi1 instead of phi_1 . If everything is fixed in the code, then an error occurs during numerical integration: DiscretizeRegion::cdim: The region given at position 1 in DiscretizeRegion[ImplicitRegion[...,{r1,r2,phi1,theta1}]] is in dimension 4. DiscretizeRegion only supports dimensions 1 through 3. May 31, 2019 at 18:38
• Please replace in your WL code the LaTeX notation/code with proper WL symbols. (As pointed out in the previous comment.) May 31, 2019 at 19:16
• @AntonAntonov done May 31, 2019 at 19:44
• @MichaelE2 if I paste the fragment above (the first one) into M (v10.0), and try to calculate the Volume, I get the error Jun 1, 2019 at 15:01

You can use Boole and integrate numerically:

NIntegrate[
Boole[-633.65530 <= r1 (r1 (4+ θ1^4-2 θ1^2 ϕ1^2) - Sqrt[(128 π^2+r1^2 (4+θ1^4)) (4+θ1^4+θ1^2 ϕ1^4)]) <= -629.65530 && 314.82730+2 r2^2 <= r2 Sqrt[128 π^2+4 r2^2+r1^2 θ1^2 ϕ1^2 (4+ϕ1^2)] && r2 Sqrt[128 π^2+4 r2^2+r1^2 θ1^2 ϕ1^2 (4+ϕ1^2)] <= 316.82730+2 r2^2]
, {r1,160,200}, {r2,160,200}, {θ1,-1/10,1/10}, {ϕ1,-1/10,1/10}
, Method->"MonteCarlo", WorkingPrecision->30]


I get the result $$61.4(5)$$. You can probably reduce WorkingPrecision to machine precision and get the same result; I leave this to you.

• how do I integrate a(n) (explicit) function using the above region? That was the question actually.. Jun 1, 2019 at 17:15
• @LowFieldTheory Just introduce the function inside the integral: NIntegrate[a[r1,r2,θ1,ϕ1] Boole[...] ...]. Jun 1, 2019 at 17:21
• so simple and yet so powerful, I suppose this was it, thx Jun 1, 2019 at 17:29