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I will describe in words a proof which might as well be illustrated without words (I would provide it in such visual form instead had I only the skills to make diagrams any nicer than MS Paint scrawls):

Imagine a sphere as the Earth, oriented in the usual way with the North pole on top. [This makes no difference except to my linguistic convenience, of course]

Consider an infinitesimal "square" patch of the Earth's surface, whose sides are oriented along lattitudinal and longitudinal lines, and the distortion this square undergoes when projected horizontally outward to the cylinder circumscribing the Earth (with the polar axis as its axis).

The ratio of the square's horizontal distance from the polar axis to the radius of the Earth is equal to the ratio of the square's vertical span to the length of its longitudinally oriented sides. (These are both the cosine of the angle between the square's position and the equator (equivalently, the angle between the square's orientation and vertical)).

Accordingly, the factor by which the square's lattitudinally oriented sides are stretched in our cylindrical projection is equal to the factor by which its longitudinally oriented sides are squashed.

Thus, our cylindrical projection is area-preserving, from which we have that the area of the entire sphere is the same as the side area of its circumscribing cylinder.

This, it seems to me, is a perfectly "directly geometric" account of the desired fact.

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I will describe in words a proof which might as well be illustrated without words:

Imagine a sphere as the Earth, oriented in the usual way with the North pole on top. [This makes no difference except to my linguistic convenience, of course]

Consider an infinitesimal "square" patch of the Earth's surface, whose sides are oriented along lattitudinal and longitudinal lines, and the distortion this square undergoes when projected horizontally outward to the cylinder circumscribing the Earth (with the polar axis as its axis).

The ratio of the square's horizontal distance from the polar axis to the radius of the Earth is equal to the ratio of the square's vertical span to the length of its longitudinally oriented sides. (These are both the cosine of the angle between the square's position and the equator (equivalently, the angle between the square's orientation and vertical)).

Accordingly, the factor by which the square's longitudinally lattitudinally oriented sides are squashed stretched in our cylindrical projection is equal to the factor by which its lattitudinally longitudinally oriented sides are stretchedsquashed.

Thus, our cylindrical projection is area-preserving, from which we have that the area of the entire sphere is the same as the side area of its circumscribing cylinder.

This, seems to me, a perfectly "directly geometric" account of the desired fact.

2 added 43 characters in body

I will describe in words a proof which might as well be illustrated without words:

Imagine a sphere as the Earth, oriented in the usual way with the North pole on top. [This makes no difference except to my linguistic convenience, of course]

Consider an infinitesimal "square" patch of the Earth's surface, whose sides are oriented along lattitudinal and longitudinal lines, and the distortion this square undergoes when projected horizontally outward to the cylinder circumscribing the Earth (with the polar axis as its axis).

The ratio of the square's horizontal distance from the polar axis to the radius of the Earth is equal to the ratio of the square's vertical span to the length of its longitudinally oriented sides. (These are both the sine cosine of the angle between the square's position and the equator (equivalently, the angle between the square's orientation and vertical)).

Accordingly, the factor by which the square's longitudinally oriented sides are squashed in our cylindrical projection is equal to the factor by which its lattitudinally oriented sides are stretched.

Thus, our cylindrical projection is area-preserving, from which we have that the area of the entire sphere is the same as the side area of its circumscribing cylinder.

This, seems to me, a perfectly "directly geometric" account of this the desired fact.

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