A circumparabolic region is found between a circle and a parabola, with the circle being chosen to include the vertex and x-intercepts of the parabola used, with the circle, to define the two circumparabolic regions for a given parabola-circle pair. There are four such regions shown above, rather than only two, because two parabolas are used above. The formulae for the parabolas, as well as the circle, are shown.

A puzzle which I will not be solving, I suspect, until I learn more integral calculus: what fraction of the circle’s area is shown in yellow?

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Nice information, I worked on it, and I made…

A double picture : https://www.facebook.com/gspteachdrawanim/photos/a.831603383632614.1073741843.700447180081569/927386047387680/?type=3

And a video :

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Impressive work!

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Thank you Robert…

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So the answer is approx 15% (I won’t spoil the exact one). Do you know of a proof that does not require integration?

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I do not . . . do you?

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No, not really

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Well, formally speaking, you can find the area between a parabola and the x-axis by exhaustion, so that’s 4/3ds times the area of the triangle bounded by the points (-1, 0), (0, 1), and (1, 0). And you know the area of the circle. But that’s just using integration by a softer name.

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The yellow area is pi minus eight thirds of square unit.

This approx. = 0.47492

Archimedes of Syracuse needed no calculus. RSVP

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Thank you . . . but I am still definitely keeping “learn more calculus” on my “to-do” list!

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I would strongly recommend it since taking an anti-derivative of a simple polynomial was

2nd semester stuff ( back in 1974 anyway!)

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