TORRICELLI'S SOLUTION TO THE AREA OF CYCLOID
When a circle rolls along line M (without slipping) a fixed red point on it traces a cycloid. The area under one arch equals three areas of the rolling circle.
Galileo was the first to try to find its area. He cut a cycloid from a metal sheet and weighed it. His conclusion was: It's almost three rolling circles, but not exactly. He was wrong. His pupil Torricelli solved the problem as follows.
At each moment draw the vertical diameter of the circle MN, and drop on it a perpendicular from that tracing red point. Let A be it's foot.
Because of "back and forth" symmetry of the whole motion, the locus of points A cosists two centrally symmetric curves. In fact, is a sine curve, which is obvious if we project the rotating radius on the vertical line and note that its angle of turn is linear with the distance rolled along the line M.
Because of that symmetry, the areas under and above that curve are halves that of the rectangle, which in turn equals 4 areas of the rolling circle.
Now by Cavalieri's Principle, which was known to Archimedes in a more general form (who credits Democriites for its many uses), Torricelli finds that the area between the sine curve and the cycloid is exactly the area of the rolling circle.
Thus add one to two and get three areas of the rolling circle under Cycloid.
PS. See also my solution at: www.its.caltech.edu/~mamikon/calculus.html
WREN'S SOLUTION TO THE LENGTH OF CYCLOID
The English architect Wren used the following property of the cycloid.
The involute of a cycloid is a cycloid of the same size. This means, if you wrap a red string over the blue cycloid and start unwrapping it the end of the string will trace the green cycloid which is exactly the same size but shifted as shown above.
Now take two circles one exactly above the other. Both roll independently along "own" horizontal lines M and N, and trace the cycloids. Then obviously the string AB which is tangent to blue cycloid will end up on the green cycloid and it will be bisected by line M. So the segment AO is half the length of AB. The reason is the (central) symmetry of the two circles with respect to their touching point.
But AB itself was a srting wrapped on the blue arc AM. Therefore we conclude the arclength AM is twice the length of the tangent chord AO of the rolling circle.
When the point A is at the cusp C, the tangent chord is CD, which is twice the diameter of the rolling circle. Thus (this was the result formulated by Wren):
Total arclength of a cycloid equals four times the diameter of the rolling circle.
The Involute of a Cycloid is a Cycloid
Wren's solution for cycloidal length fully relies on the fact that the involute of cycloid is a cycloid of the same size, as in picture, which is known from calculus, or differential geometry, prior to arc lengths, as a more basic (?) property. It allous us to avoid integration. But to call it elementary is not correct. I have known this before, and have created for my own satisfaction the following verification, based on the definition of the involute as locus of centers of curvature (which in turn can be understood intuitively on "elementary" level (acceptance of the limiting process). Here is my interpretation, which could have been known to Wren, but I don't have any reference.
Start rolling the upper circle along M. The bottom pint O is obviously the center of instantaneous rotation. This means that the point B, which traces the cycloid, moves at that moment in the direction perpendicular to the line BO. Draw the second circle exactly under neat of the upper circle, with point O being its uppermost point. The line BO, if continued through the second circle, will intersect it at some point A.
Because of the central symmetry of the picture with respect to point O, the line connecting A with the bottom point of the first circle will be perpendicular to OA, thus being tangent to the second cycloid, traced by A. If the second circle rolls on the line N exactly under the first circle (independently) with the same angular speed, but with a phase shift by 180 degrees of the tracing point A relative to the point B tracing the first cycloid.
This whole picture suggests that although the point O is the center of instantaneous rotation of the first circle, i.e. for the tracing point B, it is NOT the center of curvature of the cycloid traced by B. The latter one is defined as the intersection of two normals to the cycloid at two closed points. And it seems "obvious' that the point A where these two normals intersect (almost being parallel) is the center of curvature of the upper cycloid. This is true and intuitively convincing. .
Similar argument works for epi-and hypo-cycloids as well as for trochoids.