How to prove $\int_0^{2r} 2h \sqrt{r^2 -(x-r)^2}dx = \pi r^2 h$ without geometry?

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.

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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! – André Nicolas May 19 '11 at 4:57

2 Answers

Put $(x-r) = r \sin\theta$ then you have $x = r + r \sin\theta$. Which says $\mathrm{dx}=r\cos\theta \rm{d}\theta$. When $x= 0$ we have $-r = r\sin\theta$ which says that $\theta = -\frac{\pi}{2}$. And when $x =2r$ we have $r=r\sin\theta$ which says that $\theta = \frac{\pi}{2}$. So your integral is now,

\begin{align*} \int\limits_{0}^{2r}\sqrt{r^{2}-(x-r)^{2}} \ dx &=\int\limits_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \sqrt{r^{2}-r^{2}\sin^{2}\theta} \cdot r\cos\theta \ \text{d}\theta \\ &= r^{2} \int\limits_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \cos^{2}\theta \ \text{d}\theta \end{align*}

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I believe the difficulty comes down to finding $$\int\sqrt{a^2-x^2}\,dx$$ The standard technique is to substitute $x=a\sin\theta$, $dx=a\cos\theta\,d\theta$. Can you take it from there?

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But it is more like $\int \sqrt{a^2 +bx - x^2} dx$. – tutorscomputer May 19 '11 at 4:28

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@tutorscomputer, can you complete the square in $a^2+bx-x^2$? or, in the original, first substitute $u=x-r$, $du=dx$? – Gerry Myerson May 19 '11 at 4:32
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