I did not see a duplicate here so here is my explanation: I am OK with derivation of arc length and even worked out a simple example using a line!

Graph of y =mx + b and plug into equation from the bounds of 0 to 10 and made my slope = 1 . Works great using arc length formula , the length is 10* square root of 2 ! BUT

An integral solves the quadrature for that particular curve so using the integral as an area according to the first theorem of fundamental calculus or the second theorem for that matter don't I end up with an area? But the length of that line certainly isn't an area under a curve, and in this case the curve just happens to be a straight line. I guess I need a little help unifying the two concepts before moving on.


I think the answer you are looking for is determined by the dimensions of what you are looking at. Let's imagine that $x$ and $y$ are measured in meters. Then, from $y=mx+b$ we expect that $m$ is dimensionless and $b$ is is meters. Now let's look at those integrals for arc length, $s$ and area, $A$.

$$ s=\int\sqrt{1+y'^2}~dx\\ A=\int y~dx $$

$y'=m$ is dimensionless and $dx$ is meters, hence $s$ has the dimensions of meters. Similarly, $y$ has the dimensions of meters, so $A$ must be in meters squared.

  • $\begingroup$ humm..I see your point. , are you saying that the arc is not solving the quadrature in the sense that it's just a straight line and the length is a constant ? but can I reconcile this idea with triple integrals where in a sense you are solving the quadrature with a unit less dimension of 1 so you end up with solving 3 separate integrals and each iteration adds one dimension, which is different than the double integral case that does not use a dimensionless quadrature since it can calculate volume also unless you make the function above the region equal to 1. A 2D example of your 1D idea ? $\endgroup$ – Sedumjoy Oct 4 '17 at 15:53
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    $\begingroup$ You will always find that my dimensional arguments hold. For example, take the area as $A=\iint dx~dy$, or the volume, $V=\iiint dx~dy~dz$, or a centroid of an area example, $R=\frac{1}{A}\iint x~dx~dy$. It will always be so, or something is wrong. $\endgroup$ – Cye Waldman Oct 4 '17 at 16:00
  • $\begingroup$ I think I can unravel it from here. if you can see it I should be able to see it also. thank you $\endgroup$ – Sedumjoy Oct 4 '17 at 17:33
  • $\begingroup$ got it now!!!!!!!!!! a read again did it $\endgroup$ – Sedumjoy Oct 4 '17 at 19:00

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