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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.

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  • $\begingroup$ Yes, this works, at least in the constant-thickness case. See math.stackexchange.com/questions/11735/… $\endgroup$ Commented Jan 20, 2013 at 0:01
  • $\begingroup$ Okay, what if the thickness is not constant ? $\endgroup$ Commented Jan 20, 2013 at 0:07
  • $\begingroup$ If the thickness is not constant, you can try to estimate the average thickness and use that. It will be close. $\endgroup$ Commented Jan 20, 2013 at 0:16
  • $\begingroup$ You mean take measurements at various heights and taking the average to be used as the overall height ?? How accurate will the answer be ? $\endgroup$ Commented Jan 20, 2013 at 0:19
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    $\begingroup$ 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 $\pm 10\%$. $\endgroup$ Commented Jan 20, 2013 at 3:16

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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.

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  • $\begingroup$ What if the line has varied thickness across it ?? What will happen then ? $\endgroup$ Commented Jan 20, 2013 at 0:08
  • $\begingroup$ @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 $\Delta A = h(\ell) \Delta\ell$. $\endgroup$ Commented Jan 20, 2013 at 0:36
  • $\begingroup$ @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 ? $\endgroup$ Commented Jan 20, 2013 at 0:39
  • $\begingroup$ @Nled What's your gut feeling? $\endgroup$ Commented Jan 20, 2013 at 0:44
  • $\begingroup$ @FlybyNight Not sure what you're trying to get it at. $\endgroup$ Commented Jan 20, 2013 at 0:44

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