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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

$2\pi(R-r)$

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

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    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.2012-01-02

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@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!

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

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    @Korgan: Rotate that picture up so that you’re looking squarely at the trapezoid and don’t see any of the elliptical surface of the cut. The base of the trapezoid will then appear to lie on the $x$-axis, and $h$ will be as much *less* than the height in the middle as the righthand edge is *more* than the height in the middle. Thus, you don’t actually need to calculate $h$. You can, though: $h=\pi(r-\sqrt{r^2-y^2})$, $\pi$ being the slope of the oblique plane.2012-01-03