This is hard to explain but I'll do my best. I hope I'm clear.

Imagine you have a donut. You want to find the volume of it and the method you want to use is to imagine slicing one side of that donut and opening it out into a cylinder. Only, it's not exactly a cylinder as it has two pointed ends: one side of the cylinder has the length of the inner circumference of the donut, the other side has the length of the outer circumference.

The 'middle part' is a simple cylinder. I want to find the volume of each end part.

There's an easy trick to it, but that's not the solution I'm looking for; the easy trick being putting the two end parts together to make a smaller cylinder.

But, how I want to do this is to find the volume of one of these pointy cylindrical endparts with an integral. However, I can't seem to hit the right answer.

Let's say my stretched out donut has a left side of length


and a right side of length

$2\pi(R+r)$ and a radius of $r$. Let's say I slice the top and bottom pointed end parts off. I now have the middle part, a cylinder with a radius of $r$ and a height of $2\pi(R-r)$ and two pointed endparts. Each endpart has a radius of $r$, and a height of $2\pi r$.

If I cut an endpart into triangle wedge-shaped cross sections, each wedge will have a length of $2\sqrt{r^2-y^2}$, a height of $2\pi\sqrt{r^2-y^2}$ and a depth of $dy$. So the volume of each wedge is $2\pi(r^2-y^2)dy$

If I integrate this with limits $-r$ and $r$ I get

$2\pi\int_{-r}^r (r^2-y^2)dy = 2\pi (4r^3/3)$

Assuming that's right, I can double that and add it to the volume of my cylinder to get the donut volume. So:

$4\pi (\frac{4r^3}{3}) + 2\pi^2r^2(R-r) = \frac{16\pi r^3}{3} + 2\pi^2 r^2(R-r)$

The volume of the donut is actually $2\pi^2 r^2R$ (using the solid of revolution approach) so looks like I picked up some extra dough somewhere... $16\pi \frac{r^3}{3} - 2\pi^2 r^3$ cubic units of extra dough to be precise.

Is my arithmetic wrong? Is my method wrong? Can you spot what I messed up?

Sketch: enter image description here

  • $\begingroup$ I don’t think that there’s any way to slice the endcaps into parallel triangular cross-sections. If you slice in the direction that I think you’re using, I believe that you get trapezoids except in the very centre, where you do get a triangle as a degenerate trapezoid. $\endgroup$ Jan 2, 2012 at 22:50
  • $\begingroup$ Um, could you perhaps add some drawings or something? It's completely unclear to me what you're doing; I don't even know what it is you consider the "left" and "right" "sides" of your torus; or what "slicing one side" means. How can an exact cylinder arise? Where do "pointed ends" come from? $\endgroup$ Jan 2, 2012 at 23:28
  • $\begingroup$ @Henning: Slice the torus with a plane normal to the circle running through its centre, which is of radius $R$. Open it up so that the central circle becomes a straight line of length $2\pi R$, oriented vertically. The radius of the cross-section is $r$, so the inner circumference of the torus is $2\pi(R-r)$; when you open it up, orient it so that the straightened-out inner circumference is to the left, the straightened-out outer circumference of $2\pi(R+r)$ is to the right, with the straightened-out central circle halfway between them. The outline is a symmetric trapezoid. $\endgroup$ Jan 2, 2012 at 23:39
  • $\begingroup$ Exactly. Now the next step would be to cut out the cylinder from this shape which has height $2\pi (R-r)$. You have the top and bottom end remaining. So the question is, how do you find the volume of each of these end parts? $\endgroup$ Jan 2, 2012 at 23:45
  • 1
    $\begingroup$ Korgan, you don’t get that triangular wedge: the lefthand end of it, where it comes to a point, shouldn’t be there. It gets cut off by a vertical line before that point, because the cylinder wraps around. $\endgroup$ Jan 2, 2012 at 23:50

2 Answers 2


@Brian M. Scott. Ok, your info gave me what I needed to understand. I didn't understand why you said I didn't need to calculate h, so I went ahead and did the calculation with h and, after too much bad arithmetic, found the answer.

So the cross-sections are trapezoids, not triangles. Thinking that they might have been triangles before now seems ridiculous.

So the lengths of the sides of the trapezoids are $\pi(r-x)$ and $\pi(r+x)$ and the bottom is $2x$. The area of the trapezoid is then $2\pi rx$. The volume of the trapezoidal wedge is $2\pi rx dy$. The sum of the wedges is the integral $2\pi r\int_{-r}^{r} x dy$, where x is $\sqrt{r^2-y^2}$. The answer to that is $\pi^2r^3$.

Two of those plus the volume of the cylinder is:

$2\pi^2r^3 + \pi r^2(2\pi(R-r)) = 2\pi^2r^2R$

Which is the volume of the donut. Thanks for your help everyone!


When you slice the endcaps in the direction that you’ve chosen, you get trapezoidal cross-sections, not triangular ones: the triangle in your illustration should be cut off vertically at the lefthand end, because the cylinder wraps around. The mean height of each of the trapezoidal cross-sections is the mean height of the endcap, which is $\pi r$, and the depth is $2\sqrt{r^2-y^2}$, so the volume is $$4\pi r\int_0^r\sqrt{r^2-y^2}dy\;.$$ There’s no need to do the trig substitution, since the integral clearly gives the area of a quarter-circle of radius $r$, or $\pi r^2/4$, and the volume of the endcap is therefore $\pi^2 r^3$. Two of them give you exactly what you need: $$2\pi^2 r^2(R-r)+2\pi^2 r^3=2\pi^2 r^2 R\;.$$

  • $\begingroup$ Your answer is clearly correct, but I'm having a hard time imagining these trapezoidal cross-sections. Can anyone sketch it? $\endgroup$ Jan 3, 2012 at 0:17
  • $\begingroup$ Ok I see it :) Now I'm trying to figure out the height of the left side of the trapezoidal cross-section. You've gone ahead and used the mean height, but I'm not sure where that comes from yet, so. $\endgroup$ Jan 3, 2012 at 0:31
  • $\begingroup$ @Korgan: Look at it from the side, so that you see the endcap as a triangle. The slice in the centre really does look like that triangle. As you move the slice towards you, the lefthand side of the trapezoid moves to the right, and the righthand side moves to the left at the same rate. They meet in the middle when the slice is just tangent to the cylinder on the side towards you. At each stage the righthand edge extends as far above the horizontal midline of the triangle as the lefthand edge fails to reach the midline from below. $\endgroup$ Jan 3, 2012 at 0:38
  • $\begingroup$ @Korgan: Some of the artwork here may be helpful. $\endgroup$ Jan 3, 2012 at 0:40
  • $\begingroup$ Thanks but I did a sketch of it already. i695.photobucket.com/albums/vv314/korganwhitney/whatH.png Now I'm trying to figure out h in the sketch. It's getting late and I'm running low on caffeine, so this isn't happening as quickly as I'd like. $\endgroup$ Jan 3, 2012 at 0:44

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