4
$\begingroup$

I wanted to find the volume of a cylinder, radius r, height h by slicing it in to rectangles:

I placed the cylinder on the x-axis, one corner of the base diameter at (0,0) the opposite at (2r, 0). I have found that an area of a cross-section perpendicular to the x-axis is $A(x) = 2h \sqrt{r^2 -(x-r)^2}$ so:

$V=\int_0^{2r} 2h \sqrt{r^2 -(x-r)^2}dx$

I have tested this and it gives the volume correctly for various r and h. But how to show:

$\int_0^{2r} 2h \sqrt{r^2 -(x-r)^2}dx = \pi r^2 h$ without just saying "we know $V= \pi r^2 h$"?

This problem was my own device, perhaps it is not possible to do this.

  • 0
    As a general heuristic, it would have been nicer to put one "corner" of the base diameter at $(-r,0)$ and the opposite one at $(r,0)$. Symmetry is almost always helpful!2011-05-19

2 Answers 2