I want to find the volume of a torus

torus =
 RevolutionPlot3D[{2 + Cos[t], Sin[t]}, {t, 0, 2 Pi}]

enter image description here

dtorus =
 DiscretizeGraphics[Cases[Normal @ torus, _GraphicsGroup, -1][[1]]]

enter image description here

The Documentation for RegionMeasure states:

"RegionMeasure is also known as count (0D), length (1D), area (2D), volume (3D)..."

{Area @ dtorus, RegionMeasure @ dtorus, Volume @ dtorus}

{78.6557, 78.6557, 0}

Next, I want to find the volume of the torus' bounding cuboid

bounds = RegionBounds[dtorus]

{{-3., 3.}, {-3., 3.}, {-1., 1.}}

cuboid =
  Graphics3D[{Green, Opacity @ 0.2, Cuboid @@ Transpose[bounds]}];

Show[torus, cuboid, Boxed -> False, Axes -> False]

enter image description here

Now find the volume of the cuboid

dcuboid =
 DiscretizeGraphics @ cuboid

enter image description here

{Area @ dcuboid, RegionMeasure @ dcuboid, Volume @ dcuboid}

{Infinity, 72., 72.}


How can it be that the Volume of the bounding cuboid, 72, is lower than the "Volume" of the torus, 78 .6557?

What do I overlook here?

What other options do I have to find the volume of my torus?

  • $\begingroup$ But isn't the volume of your torus 0? {Area @ dtorus, RegionMeasure @ dtorus, Volume @ dtorus}->{78.6557, 78.6557, 0}. As I understand, RevolutionPlot3D plots a surface. A surface itself has 0 volume, for instance, a sphere. $\endgroup$ Sep 17, 2014 at 12:39
  • $\begingroup$ @GregoryRut A sphere has volume 4/3 Pi * radius^3, hasn't it? $\endgroup$
    – eldo
    Sep 17, 2014 at 12:48
  • $\begingroup$ No, it's somehing different. A ball has a volume. You could define an interior volume of a sphere. $\endgroup$ Sep 17, 2014 at 12:49
  • 5
    $\begingroup$ If you do RegionDimension[dtorus], you'll see why. In this case RegionMeasure is giving you the surface area of the torus since you have a 2D region. $\endgroup$
    – RunnyKine
    Sep 17, 2014 at 14:43
  • 6
    $\begingroup$ @RahulNarain, the truth is, DiscretizeGraphics will almost always give a surface discretization regardless of how the graphics was generated, which will result in a 2D region in general. $\endgroup$
    – RunnyKine
    Sep 17, 2014 at 20:31

2 Answers 2


As I have mentioned in my comment, in order to define a torus with a volume, you need an inequality. Just like you would define the interior of a sphere: $x^2+y^2+z^2<R^2$. The interior of the torus is defined by an inequality $\left(R - \sqrt{x^2 + y^2}\right)^2 + z^2 < r^2 $, where R and r are the major and minor radii, respectively.

tor[R_, r_, x_, y_, z_] := (R - Sqrt[x^2 + y^2])^2 + z^2 < r^2

   tor[2, 1, x, y, z], {x, y, z}], {{-3, 3}, {-3, 3}, {-3, 3}}]

The formula for interior volume of a torus reads $ V = 2 \pi^2 R r^2.$ In our case the volume would be roughly $V\approx $39.5.

It seems that Volume could be applied for 'derived regions' (like tor, for instance) yet it seems that it works only for selected types of areas.

  • $\begingroup$ Why do you use 6 instead of 2 Pi ? 2 Piwould give 305.657 $\endgroup$
    – eldo
    Sep 17, 2014 at 12:59
  • $\begingroup$ There was an error with def. of tor function (R^2 instead of R). It's fixed now. The range of variables has been modified as well. $\endgroup$ Sep 17, 2014 at 13:18

I'm not 100% sure on this, but I think you can construct a torus as a Cartesian product of a disk and a circle. Unfortunately you can't visualise it because the embedding dimension is 4, but the volume seems to come out correct:

Volume @ RegionProduct[Disk[{0, 0}, 1], Circle[{0, 0}, 2]]

(*  4 π^2  *)

or with symbolic radii:

Volume @ RegionProduct[Disk[{0, 0}, r], Circle[{0, 0}, R]]

(* 2 π^2 r^2 R *)

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.