This problem was posted at Kitamuki Kannon temple of Ueda City in Nagano Prefecture in 1828. Here is its entry at the [url=https://sangaku-archive.org/works/1828_Nagano_Kitamuki_965]Sangaku Archive[/url].[br][br]A [size=100]right circular cylinder with radius [i]r[/i] height [i]h[/i] is intersected by three planes. The first plane, P[sub]1[/sub], is parallel to the cylinder’s axis and intersects it in a rectangle with vertices B, C, D, and E. Let [math]\large d=|PQ|[/math], the distance from P[sub]1[/sub] to the point Q shown. Plane P[sub]2[/sub] passes through B and D, and P[sub]3[/sub] passes through C and E; both of these planes are perpendicular to P[sub]1[/sub]. Planes P[sub]2[/sub] and P[sub]3[/sub] intersect the cylinder in two curves, which bisect each other on the cylinder’s surface. Let R be bowtie-shaped region on the cylinder’s surface bounded by these two curves and the vertical edges of the rectangle, BE and CD. What is the area, [i]A[/i], of this region in terms of [i]r[/i], [i]h[/i], and [i]d[/i]?[/size]

Warning: this problem requires calculus!

The solution below has been adapted from the one in Fukagawa and Rothman’s [i]Sacred Mathematics[/i] (Princeton, 2008), which outlines an 1844 solution from [i]Sanpō Kyūseki Tsu-ko[/i]; while the author is given as Uchida Kyūmei, scholars credit the content of this treatise to his teacher, Hasegawa Hiromu (1810–1887). The approach should look familiar to anyone comfortable with integral calculus: we will imagine slicing the region vertically into thin strips, then add the areas of those strips together. If we suppose the (slightly curved) widths of these strips tend toward zero, then the sum of the (now infinitely-many) strip areas will equal the total area of the region. Slicing-and-summing techniques like this appeared in the early 18th century in Japan—applied, for example, in Takebe’s studies of circular arc length (an early objective of the nascent discipline of [i]enri[/i], or “circle principle”), and in deriving the formula for spherical volume—and gradually evolved into what we now call [i]definite integration[/i]. While the Chinese discovered a version of Cavalieri’s principle for slicing solids centuries earlier, they didn’t use definite integrals until Europeans introduced them in the mid-19th century.

Our first goal is to find an expression for the area of a typical strip in terms of a variable along some axis. The [i]x[/i]-axis shown above is a convenient one. Each strip is approximately rectangular; denote the height and width by [math]\large z(x)[/math] and [math]\large \Delta s[/math], respectively. Viewing from the side, in the direction perpendicular to plane P[sub]1[/sub], we can see that, by similar triangles,[br][center][math]\large[br]\frac{z(x)}{h}=\frac{x}{k},[br][/math][/center][br]where [i]k[/i] is half the width of BC.

Looking down from above, we make two observations. First, by the Pythagorean theorem, [br][center][math]\large k=\sqrt{r^2-(r-d)^2},[/math][/center][br]which we will use for a substitution in the final step. Second, notice that the orange and purple triangles below are similar. This fact, in conjunction with the Pythagorean theorem, gives us the approximate (curved) width of each strip in terms of [i]x[/i]:[br][center][math]\large \Delta s \approx \ell = \frac{r}{\sqrt{r^2-x^2}}\, \Delta x,[/math][/center][br]where [math]\large \Delta s \rightarrow \ell[/math] as [math]\large \Delta s \rightarrow 0[/math].

The area of each thin strip is then[br][center][math]\large[br]\begin{align}[br]\Delta A &\approx z(x) \cdot \Delta s \\[br]&\approx z(x) \cdot \frac{r}{\sqrt{r^2-x^2}} \, \Delta x.[br]\end{align}[br][/math][br][/center][br][br]In the limit, as [math]\large \Delta s \rightarrow 0,[/math][br][center][math]\large [br]dA = z(x) \cdot \frac{r}{\sqrt{r^2-x^2}}\, dx.[br][/math][/center]

The area of R is thus given by the integral[br][center][math]\large[br]\begin{align}[br]A&=\phantom{2} \int_R dA \\[br]&=2\int_0^k z(x) \cdot \frac{r}{\sqrt{r^2-x^2}}\, dx \\[br]&=2\int_0^k \frac{hr}{k} \cdot \frac{x}{\sqrt{r^2-x^2}}\, dx.[br]\end{align}[br][/math][/center]

Here is where a modern mathematician and a [i]wasanka[/i] diverge. Today we would evaluate this integral using the fundamental theorem of calculus, as below:[br][center][math]\large[br]\begin{align}[br]A &= \left. -\frac{2hr}{k} \sqrt{r^2-x^2} \, \right|_0^k \\[br]&= -\frac{2hr}{k} \left(\sqrt{r^2-k^2} -r \right) \\[br]&= \frac{2hrd}{\sqrt{r^2-(r-d)^2}} \\[br]&= \frac{2hr}{\sqrt{2r/d - 1}}.[br]\end{align}[br][/math][/center][br]Notice that the area of R is greatest when [math]\large d=r[/math], in which case [math]\large A=2hr[/math]; surprisingly, this is [math]\large 1/\pi[/math] times the surface area of the cylinder.

[center][icon]/images/ggb/toolbar/mode_parallelplane.png[/icon][/center]

The fundamental theorem of calculus, developed in 17th-century Europe, provided a unified theory linking differentiation and integration, which, on a practical level, allowed the symbolic computation of integrals using antiderivatives. Edo period mathematicians didn't systemically study differentiation, and thus never developed a general theory of antiderivatives that could be broadly applied to integrals using something like the fundamental theorem.[br][br]Nevertheless, [i]wasanka[/i] knew the values of some simple definite integrals. For example, they were familiar with the general fact that[br][center][math]\large[br]\int_0^k x^n \, dx = \frac{k^{n+1}}{n+1},[br][/math][/center][br]known in the West as Cavalieri's quadrature formula. First-year calculus students today learn that, for small [i]n[/i], these integrals do not require antiderivatives. Rather, using some well-known results about sums of integers, they can be computed directly from the expression that sums the areas of thin (almost-)rectangles under the curve [math]\large x^n.[/math]

In order to evaluate more complicated integrals, Japanese mathematicians relied on power series. [i]Wasanka[/i] found power series expansions for many functions, including[br][center][math]\large[br]\sqrt{r^2-x^2} = r - \frac{x^2}{2r} - \frac{x^4}{8r^3} - \frac{x^6}{16r^5} - \frac{5x^8}{128r^7} - \frac{7x^{10}}{256r^9} - \cdots[br][/math][/center][br]and[br][center][math]\large[br]\frac{1}{\sqrt{r^2-x^2}} = \frac{1}{r} + \frac{x^2}{2r^3} + \frac{3x^4}{8r^5} + \frac{5x^6}{16r^7} + \frac{35x^8}{128r^9} + \frac{63x^{10}}{256r^{11}} + \cdots.[br][/math][/center][br]The expressions above are identical to the Taylor series for these functions, and calculus students today would compute these terms using higher-order derivatives. However, differentiation is not necessary. If one simply [i]assumes[/i] that a given expression has a series expansion, as in[br][center][math]\large[br]\sqrt{r^2-x^2} = a_0+a_1x+a_2x^2+a_3x^3+a_4x^4+ \cdots,[br][/math][/center][br]one can apply some elementary algebraic and combinatorial techniques to determine what the coefficients [math]a_k[/math] must be. (Try squaring both sides of the equation above and setting the corresponding coefficients of [math]x^k[/math] equal to each other.) These are examples of what we now call the generalized binomial theorem; [i]wasanka[/i] figured out the power series expansions of other functions, too, including some trigonometric functions and their inverses.[br][br]Why would Edo period mathematicians have thought to assume that these series expansions existed in the first place? The Chinese learned about infinite series from the Jesuits in the mid-18th century, but Seki studied them earlier in Japan; he utilized the geometric series[br][center][math]\large[br]\frac{a}{1-r}=a+ar+ar^2+ar^3+ar^4+\cdots[br][/math][/center][br]with [math]\large r=1/4[/math] in a groundbreaking computation of pi that was correct to eleven decimal places. His disciple Takebe would take studies of series much further.

Armed with the formulas above, we can compute the definite integral for [i]A[/i] without directly knowing the antiderivative of the integrand. To do this, replace part of the integrand by its power series expansion, and then integrate term by term. (Integrating a power series term by term is not always kosher; the series must converge in a particular way to avoid problems. Outside of some simple cases---geometric series, for instance---there is little indication that Japanese mathematicians were concerned with questions of convergence and divergence.)[br][center][math] \large[br]\begin{align}[br]A &= 2\int_0^k \frac{hr}{k} \cdot \frac{x}{\sqrt{r^2-x^2}}\, dx \\ \\[br]&= \frac{2hr}{k} \int_0^k x \cdot \left[ \frac{1}{r} + \frac{x^2}{2r^3} + \frac{3x^4}{8r^5} + \frac{5x^6}{16r^7} + \frac{35x^8}{128r^9} + \cdots \right] dx \\ \\[br]&= \frac{2hr}{k} \int_0^k \left[ \frac{1}{r} x + \frac{1}{2r^3}x^3 + \frac{3}{8r^5}x^5 + \frac{5}{16r^7}x^7 + \frac{35}{128r^9}x^9 + \cdots \right] dx \\ \\[br]&= \frac{2hr}{k} \left[ \frac{1}{r} \cdot\frac{k^2}{2} + \frac{1}{2r^3}\cdot\frac{k^4}{4} + \frac{3}{8r^5}\cdot\frac{k^6}{6} + \frac{5}{16r^7}\cdot\frac{k^8}{8} + \frac{35}{128r^9}\cdot\frac{k^{10}}{10} + \cdots \right] \\ \\[br]&= \frac{2hr}{k} \left[ \frac{k^2}{2r} + \frac{k^4}{8r^3} + \frac{k^6}{16r^5} + \frac{5k^8}{128r^7} + \frac{7k^{10}}{256r^9} + \cdots \right][br]\end{align}[br][/math][/center][br]At this point, if we were given values for [i]h[/i], [i]r[/i], and [i]d[/i], a numerical answer would be possible; some [i]soroban[/i] calculations could provide the area to arbitrary precision. On the other hand, observe that the series in brackets looks very similar to the one for [math]\large \sqrt{r^2-x^2}[/math]. In fact, it is equal to[br][center][math]\large[br]\left. -\sqrt{r^2-x^2} \right|_{x=0}^{x=k}.[br][/math][/center][br]We thus arrive at the original conclusion, that the area of R is[br][center][math]\large[br]A = \frac{2hr}{\sqrt{2r/d - 1}}.[br][/math][/center]