Spherical bread slices area With integration you can prove that if a sphere is cut into $n$ paralel slices of equal width, then those slices have the same external area.
It is often presented as "a spherical loaf of bread is cut $n-1$ times with equidistant paralel cuts, thus leaving $n$ slices of equal width. Those slices have the same amount of crust".
Is is possible to get to the same conclusion through a classical geometrical aproach?
 A: as a surface of revolution a sphere may be viewed as the disjoint union of infinitesimal frustra created as slices between planes perpendicular to a line passing through its centre. if we make this line our x-axis, and use a circle of radius $r$ centered at the origin, then the radius of the frustrum at a particular value of $x$ is just $y$ whereas the length of the infinitesimal line element is $ds$. thus between suitably chosen values $x=a$ and $x=b$ we have:
$$
A = \int_{x=a}^{x=b} dA = \int_{x=a}^{x=b} 2\pi yds \tag{1}
$$
however 
$$
ds = \sqrt{dx^2+dy^2} = \sqrt{1 + \left(\frac{dy}{dx}\right)^2}dx = \frac{rdx}{y}\tag{2}
$$
(because $\frac{dy}{dx}=-\frac{x}{y}$ and $x^2+y^2=r^2$). now substituting (2) into (1) we obtain the all-important cancellation of $y$, leaving
$$
A =   2\pi r \int_{x=a}^{x=b} dx = 2\pi r (b-a)
$$
although expressed here in the more recent language of calculus and coordinate geometry,  the all-important cancellation is a mere matter of classical geometry once the principle of infinitesimals is applied. certainly this principle - e.g. as Archimedes' exhaustions - was known to the later Greek geometers.
