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If you take the circumference of the circle and stretch it into a rope and then multiply it by the height(radius) why don't you get the area of a circle?

I ask this question when using the shell method revolved around the $x$ axis on $x^2$ from $[0,2] $ instead of the disk method.

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    $\begingroup$ Think of it as the average circumference from radius 0 to radius r $\endgroup$ – Don Thousand Oct 14 '18 at 4:56
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Because if you join the ends of a small piece of your rope to the centre, you get a triangle, not a rectangle. A triangle has half the area of a rectangle with the same base and height.

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    $\begingroup$ Thank you, I realize I was looking at it completely wrong. I can't bend a ruler into a circle without parts of it squeezing into itself. $\endgroup$ – user603992 Oct 14 '18 at 5:08
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You are stretching both the circumference and the center to make a rectangle of height$R$

By turning the circumference of the circle into a rope and keeping the center fixed you are going a triangle not a rectangle so the area is $$ (1/2)(2\pi R)(R) = \pi R^2 $$

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Here I'd not argue with lots of formulas.
Rather I'd provide the picture proof from Wikipedia:
cf. https://en.wikipedia.org/wiki/Area_of_a_circle#/media/File:CircleArea.svg

--- rk

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In your logic, you are saying that unraveling a circle gives you many small rectangles. How does that make any sense when you try to roll it back?

Instead, it comes out as many triangles of height $r$, and sum of all the base lengths would be $2\pi r$(the circumference). Hence their total area is $0.5\times r\times2\pi r=\pi r^2$

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