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How can I make Mathematica calculate a volume given a set of inequalities over three dimensional space?

I know how to plot the 3D region of the volume that I need to calculate, like this:

RegionPlot3D[
  x^2 + y^2 + z^2 <= 1 && y >= Abs[x]/Sqrt[3] && y <= 1/2, 
  {x, -1, 1}, {y, -1, 1}, {z, -1, 1}]

It yields as output the figure:

Output 1

But how can I calculate its volume? Do I have to set up an integral? Is that the only way?

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This can be done in one line of that Mathematica code

 Volume[ImplicitRegion[x^2 + y^2 + z^2 <= 1&&y>= RealAbs[x]/Sqrt[3] && y<= 1/2, {x, y, z}]]

$\frac{17 \pi }{72}$

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  • $\begingroup$ Love it. There is no reference of it in the documentation of Volume[] though. Thanks a lot. $\endgroup$
    – zurg
    Feb 22 at 17:14
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    $\begingroup$ @zurg: See that reference for Volume in 3D. $\endgroup$
    – user64494
    Feb 22 at 17:22
  • 1
    $\begingroup$ BTW, In arbitrary dimension, you can use RegionMeasure to achieve the same end result $\endgroup$ Feb 22 at 19:31

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