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Take a look at this curve.

Imagine that I have drawn this on a paper, and that I want to find the area of it. (The thickness of the line is assumed to be constant).

Assume that I use a string to trace the edge of the figure, then compare that to a ruler. Then I measure the height using the same method. I will get values for the width and height, so can I just say Area = Width $\times$ Height? I will be extending that curvy shape into a rectangle. So in this sense, could I not assume that the area is like that of a rectangle, namely Area = Width $\times$ Height?

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EDIT

If this is applicable, I'm really interested to know if this also works with a curvy shape that has non-constant thickness.

Fly by Night
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NLed
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  • Yes, this works, at least in the constant-thickness case. See http://math.stackexchange.com/questions/11735/how-to-prove-that-a-torus-has-the-same-volume-as-a-cylinder-with-the-height-equ/11738#11738 – user7530 Jan 20 '13 at 00:01
  • Okay, what if the thickness is not constant ? – NLed Jan 20 '13 at 00:07
  • If the thickness is not constant, you can try to estimate the average thickness and use that. It will be close. – Ross Millikan Jan 20 '13 at 00:16
  • You mean take measurements at various heights and taking the average to be used as the overall height ?? How accurate will the answer be ? – NLed Jan 20 '13 at 00:19
  • @RossMillikan sorry forgot to tag your name to get a notification – NLed Jan 20 '13 at 00:29
  • Are you looking for an approximate value? What is meant by "thickness", I mean in mathematical terms? – Maesumi Jan 20 '13 at 00:30
  • @Maesumi The height the of the curved line basically. The various thickness means that the height of the figure changes at various points. – NLed Jan 20 '13 at 00:31
  • It will be quite accurate if your average is. If you take lots of measurements, or can take ones that represent the range well, it will be very good. – Ross Millikan Jan 20 '13 at 01:14
  • @RossMillikan Would your method work with a figure that looks like this ? http://imageshack.us/a/img593/2648/imagermqy.jpg – NLed Jan 20 '13 at 01:24
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    I would say horizontally across the middle is a little wider than average. I would take 80% of that as the average width, take the length of the centerline, and multiply them. You can check by cutting it out of paper and weighing against a square, or by plotting it on graph paper and counting squares. I would bet on $\pm 20%$ and think it is probably within $\pm 10%$. – Ross Millikan Jan 20 '13 at 03:16

1 Answers1

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If you measure the length in the middle of the (thickened) line, then indeed the area is just this length times thickness - at least as long as the curve is smooth enough and does not bend back on itself.

Hagen von Eitzen
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  • What if the line has varied thickness across it ?? What will happen then ? – NLed Jan 20 '13 at 00:08
  • @NLed Then you would need to use integration. If $\ell$ is the length along the top and $h$ is the height then you'll need to calculate $\Delta A = h(\ell) \Delta\ell$. – Fly by Night Jan 20 '13 at 00:36
  • @FlybyNight is that Height x Length x Integration of Length ? Integrate from what value to what value though ? Also, can I take the average of different heights and use that as an approximation to solve in A=WxH ? – NLed Jan 20 '13 at 00:39
  • @Nled What's your gut feeling? – Fly by Night Jan 20 '13 at 00:44
  • @FlybyNight Not sure what you're trying to get it at. – NLed Jan 20 '13 at 00:44
  • @NLed I'm saying that I've given you a clue, and that for your own pleasure and satisfaction it's probably best that you spend a little while thinking about it yourself. If I just tell you the answer then you won't get any satisfaction and you'll probably forget within a few weeks. Have a think and let us know what you arrive at. – Fly by Night Jan 20 '13 at 00:47
  • Would've preferred a more sound explanation from you @FlybyNight rather than clues and puzzles ... – NLed Jan 20 '13 at 14:28