# Volume of a Region defined by Inequalities [duplicate]

I am trying to compute the volume of a 3-dimensional region defined by 5 inequalities. Mathematica takes too long to compute it. In fact, I haven't ever seen it complete the execution of the below code:

theta = Pi/5;
dh = 0.5;
din = 1.016;
dout = 1.27;
zz = z * Cos[theta] - x*Sin[theta];
xx = z * Sin[theta] + x*Cos[theta];
yy = y;
region1 =
z^2 >= (dh/2 + 0.0254)^2 - (dout/2)^2*
x^2/(x^2 + y^2);
region2 = (din/2)^2 <
x^2 + y^2 < (dout/2)^2;
region3 = -1 <= z <= 1;
region4 = zz^2 <= (dh/2)^2 - (dout/2)^2*
xx^2/(xx^2 + yy^2);
region5 = (din/2)^2 < xx^2 + yy^2 < (dout/2)^2;
region = ImplicitRegion[{region1, region2, region3, region4, region5}, {x, y, z}];
Volume[region]


Any reason why it takes so long (or does not complete)? Any alternative method to compute the volume (that does not involve discretization)?

EDIT: Calculating the volume of even two of the regions similarly takes too long:

theta = Pi/5;
dh = 0.5;
din = 1.016;
dout = 1.27;
zz = z * Cos[theta] - x*Sin[theta];
xx = z * Sin[theta] + x*Cos[theta];
yy = y;
regionA = zz^2 <= (dh/2)^2 - (dout/2)^2 * xx^2/(xx^2 + yy^2);
regionB = (din/2)^2 < xx^2 + yy^2 < (dout/2)^2;
region = ImplicitRegion[{regionA, regionB}, {x, y, z}];
Volume[region]

• I do not recommend the use of either D[] (the differentiation operator) nor subscripts in your code; consider eliminating them. – J. M. will be back soon Jul 13 '16 at 16:20
• @J.M. Edited. Thanks – Shivanand Jul 13 '16 at 16:24
• @Feyre But Mathematica must at least show that the volume is zero. I want to compute this for different values of theta, dh, din and dout and not all of them necessarily give zero volume – Shivanand Jul 13 '16 at 16:28
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• For some reason, it won't discretize the region if region1 is included, but: r2 = DiscretizeRegion[region, {{-1, 1}, {-1, 1}, {-1, 1}}] Volume[r2] 0.033056 – Feyre Jul 13 '16 at 16:42

You should discretize your region, taking care to avoid singularities, here I leave out region 1:

region = ImplicitRegion[{region2, region3, region4, region5}, {x, y,
z}];
r2 = DiscretizeRegion[region, {{-1, 1}, {-1, 1}, {-1, 1}}] Now you can calculate the volume:

AbsoluteTiming[Volume[r2]]


{0.000382, 0.0330559}

• I need to add region1 too. Do you know why it cannot be discretized the way other regions can be? – Shivanand Jul 13 '16 at 19:10
• @Shivanand All I know is that when region1 and 4 come together, i get an error which is unknown to google. TriangulateMesh::tmfail: TriangulateMesh failed to triangulate the mesh. >> – Feyre Jul 13 '16 at 20:39
• I know for sure that the intersection is very small. Each region is very important for the final result – Shivanand Jul 13 '16 at 21:06