# How can I tell the difference if my integral is measuring arc length or area?

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.
• 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. – Cye Waldman Oct 4 '17 at 16:00