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"Let $\mathcal S$={$\mathbf x \in \mathbb{R}^3 : ||\mathbf x||=1$}

Prove that the area of the part of $\mathcal S$ that lies between the two parallel planes given, say, by $x_3=a$ and $x_3=b$, is the same as the area of the part of the circumscribing cylinder (given by $x_1^2+x_2^2=1$) that lies between these two planes."

I would like a solution, if possible, that does not use calculus. I can do it by calculus myself, but this is taken from a chapter from a book that makes no reference to calculus, so I'm assuming it wants a non-calculus answer.

Obviously it's the area of the part of the sphere between these planes that I'm struggling to find, I have tried splitting it up into lunes and triangles, as we use these throughout the exercise up to this point, but I can't find the angles and it just gets generally messy.

Thank you

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This is known as Archimedes' Hat-Box Theorem. Archimedes developed the methods for solving such problems many centuries before the invention of calculus, so I suppose they would suffice for a "non-calculus" answer.

There is a discussion of this theorem (along with some nice three-dimensional diagrams) on Zachary Abel's Math Blog, under the heading "Spherical Surfaces and Hat Boxes."

The method described on that page uses the fact that if we have some region $A$ on the surface of a sphere, the volume of the interior part of the sphere between the sphere's center and $A$ is $\frac13$ the radius of the sphere multiplied by the area of $A$. This can be shown through the method of exhaustion by triangulating $A$ with finer and finer meshes and considering the volumes of triangular pyramids (determined by the meshes) whose bases are inside or outside the sphere.

I think it is a bit simpler to prove the theorem if we assume initially that one of the planes is tangent to the sphere; that is, the part of $\mathcal S$ that we measure is a spherical cap. (To get the more general result, if $-1 < a < b < 1$ subtract the spherical cap above the plane $x_3 = b$ from the spherical cap above $x_3 = a$.)

For a plane $x_3=a$ cutting a unit sphere $\mathcal S$, we construct two three-dimensional regions. The first region, $U$, is bounded by the spherical cap cut off by the plane and the cone whose vertex is the center of the sphere and whose base is the circular boundary of the spherical cap. The second region, $V$, is the region bounded by the cylinder $x_1^2+x_2^2=1$, by the cone $x_1^2+x_2^2=x_3^2$, and by the cone $a^2 x_1^2 + a^2 x_2^2 = x_3^2$. That is, $V$ consists of all points between the center of the sphere and the part of the cylinder $x_1^2+x_2^2=1$ between $x_3 = a$ and $x_3 = 1$.

Now consider a horizontal plane $x_3 = z$, where $0 < z < 1$. The intersection of this plane with $U$ is a circle, while the intersection of the same plane with $V$ is an annulus (a disk with a circular hole). Moreover, a little geometry shows that for any given $z$, the area of the circle is always equal to the area of the annulus. Since this applies to every simultaneous cross-section of $U$ and $V$ by any plane perpendicular to the $x_3$ axis, by Cavalieri's principle, the volumes of $U$ and $V$ are equal.

We can show that the volume of $V$ is $\frac13$ of the area of the part of the cylinder $x_1^2+x_2^2=1$ between $x_3 = a$ and $x_3 = 1$ (one way is to add the volume inside the cylinder between the planes $x_3 = a$ and $x_3 = 1$ to the volume of the cone with vertex at $0$ and base $x_1^2+x_2^2\leq 1, x_3 = a$ and subtract the volume of the cone with vertex at $0$ and base $x_1^2+x_2^2\leq 1, x_3 = 1$).

But the volume of $U$ is $\frac13$ the area of the spherical cap (since the radius of the sphere is $1$). Since this is equal to the volume of $V$, the area of the spherical cap must equal the area of the cylinder $x_1^2+x_2^2=1$ between $x_3 = a$ and $x_3 = 1$.


An alternative method is to take the zone of the sphere between the two planes $x_3=z$ and $x_3=z+\delta$, approximate the area of this zone by the area of a frustum (truncated cone) between the circles $x_1^2+x_2^2=x_3^2, x_3z$ and $x_1^2+x_2^2=x_3^2, x_3=z+\delta$, and show that this area is the area of the cylinder $x_1^2+x_2^2=1$ between the planes $x_3=z$ and $x_3=z+\delta$. I think a little trigonometry is generally used to approximate the area of the frustum; the approximation over a finite number of frustums converges to the exact area of the spherical cap as we reduce the spacing between the parallel planes.


A possible third method would be to inscribe a spherical polygon inside a spherical cap and circumscribe another spherical polygon around the spherical cap. The areas of the polygons are the sums of areas of spherical triangles. The area of the cap would be the area that the circumscribed and inscribed polygons converge to, by principle of exhaustion. If you can find the area of intersection of a spherical cap and a lune with vertex at the center of the spherical cap, you can get the area of the cap directly, but while the intersection of the cap and lune is "triangular" in some sense, I would not consider it a true spherical triangle, since one of the "edges" does not lie along a great circle.

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