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Sorry for off the beaten path question. Struggling to know how integration can distinguish between a length and an area so I will frame the question in a peculiar manner.

I have a line $y = x.$ and $x = 5$. Is the arc length $5?$ From $0$ to $5?$ Is the area $25?$ Let's use integration. How can you tell the difference between arc length and area? If a straight line is not an arc length the question still holds: how do you keep the dimensions synchronized so one is a square giving area and the other a simple length ?

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    $\begingroup$ Integration always gives you the "area under the curve" between the two points of evaluation. With the formula for arc length, the formula has been written in such a way that the "area under the curve" (that is, the curve that you're integrating) is equal to the arc length --- you don't integrate the arc curve itself, but instead you integrate some other curve that tells you how long the arc in question is. $\endgroup$
    – Bilbottom
    May 7, 2018 at 16:48
  • $\begingroup$ The arc length of what? The line $y=x$? That’s not equal to $5$, as you can see by applying the Pythagorean theorem. $\endgroup$
    – amd
    May 7, 2018 at 16:54
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    $\begingroup$ @BillWallis That ought to be an answer instead of a comment. $\endgroup$
    – amd
    May 7, 2018 at 16:55
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    $\begingroup$ @amd Thanks for the suggestion, I have added it as an answer. $\endgroup$
    – Bilbottom
    May 7, 2018 at 16:58
  • $\begingroup$ You say $x=5,$ but your further comments make me suspect you meant $0\le x\le 5. \qquad$ $\endgroup$ May 7, 2018 at 23:58

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The easiest way to distinguish is to look at the integrals once you've assembled them, and do dimensional analysis. Here's the rule for dimensions using the two main calculus operations of differentiation and integration. A word about notation: $[x]$ means "the units of $x$". \begin{align*} \left[\frac{dy}{dx}\right]&=\frac{[y]}{[x]} \\ \left[\int y\,dx\right]&=[y]\,[x]. \end{align*} This ought to be taught in every single calculus course, but alas, it seems to be vastly neglected.

So let's take two examples: an arc length integral, and an area integral - the area integral first because it's more straight-forward.

Example 1: We want the area under the curve $y=x^2$ from $x=0$ to $x=2$. Let's say we're measuring both $x$ and $y$ in meters. No problem: $$A=\int_0^2y\,dx=\int_0^2x^2\,dx=\left(\frac{x^3}{3}\right)\bigg|_0^2=\frac83.$$ What are the units? Using the rule from above, $[A]=[y]\,[x]=\text{m}\cdot\text{m}=\text{m}^2$, so it's an area. Don't be confused because $y=x^2$. The units of $y$ are still just meters.

Example 2: We want the arc length of $y=x^2$ from $x=0$ to $x=2$ (hopefully this should look familiar). The formula is $$L=\int_0^2\sqrt{1+\left(\frac{dy}{dx}\right)^{\!\!2}}\,dx.$$ Let's check out the units according to our rules above: \begin{align*}[L]&=\left[\int\sqrt{1+\left(\frac{dy}{dx}\right)^{\!\!2}}\,dx\right]\\ &=\left[\sqrt{1+\left(\frac{dy}{dx}\right)^{\!\!2}}\,\right][x]\\ &=\sqrt{\frac{[y]^2}{[x]^2}}\,[x]\\ &=\frac{[y]}{[x]}\,[x]\\ &=[y]. \end{align*} Here, you'll notice we assumed that under the square root symbol, the units of the $1$ match $[y]^2/[x]^2$, or else we couldn't have done the addition. You can only add things together that have identical units. In any case, we wound up with units of length: $[y]=\text{m}$, so this result is a length, not an area.

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    $\begingroup$ ok ...I thinkI got it.......it works out.....the trick was the dimensions ...the arc length drops the dimensions on x. In my example dy = 0 and you are still left with a single dimension of length. $\endgroup$
    – Sedumjoy
    May 7, 2018 at 22:07
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    $\begingroup$ I applaud your attention to units, but I would suggest mentioning that if one is taking $y$ and $x$ to each have units of meters, it's technically correct to write $y = x^2/1\ \mathrm{m}$ (rather than $y = x^2$) so that the units work out. $\endgroup$
    – David Z
    May 7, 2018 at 22:16
  • $\begingroup$ @DavidZ: I'm not sure I agree. What if it was $y=x^3?$ Or far worse: $y=\sin(x)$. Now you have an infinite series expansion in powers of $x$. We're simply measuring $y$ in one direction, with certain units, and there happens to be a functional relationship between $x$ and $y$. In my opinion, the units are best left off. $\endgroup$ May 8, 2018 at 0:40
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    $\begingroup$ @AdrianKeister You would use $y=x^3/ 1\ \mathrm m^2$ and $y=1\ \mathrm m\ \times \sin(x/1\ \mathrm m)$. David Z is right here: if you really want to be consistent with the units, you cannot just write $y=x^2$ (well, you could if you were to declare that all factors of $1\ \mathrm m$ are implicit, but that would only obscure the issue...) $\endgroup$ May 8, 2018 at 0:59
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Integration is the art of summing lots of small things. In the case of "area under the graph" you're summing up the areas of many small regions. In the case of arc-length, you're summing up the lengths of many short segments.

You know which one you're doing when you're setting up the integral, and when you interpret the result, but the actual calculation between those two steps doesn't really care one way or the other.

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  • $\begingroup$ Integration is also commonly used to sum up many small vectors, for a variety of applications. $\endgroup$
    – Arthur
    May 7, 2018 at 16:55
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This is a supplementary answer to Bill Wallis' comment to really flesh out the idea. Informally, we can reformulate the task of finding arc length into some infinitesimally small lengths. If we zoom in on a small section of a function's graph, we can see that the section of the graph somewhat resembles a triangle:

enter image description here

Source.

You can see that with some small step in $x$, $\mathrm dx$ a corresponding step is $y$, $\mathrm dy$ is created. By using the Pythagorean theorem, we can find the hypotenuse, a rough approximation of the curve itself. And with integration, we want to sum an infinite number of these minute hypotenuses to create an exact measurement of arc length.

As it turns out, the area under the graph of $\sqrt{\left(\mathrm dx\right)^2 + \left(\mathrm dy\right)^2}$ is the arc length of the original curve as Bill Wallis describes. We have created a function whose area under its curve gives us the arc length of the original curve. Thus another intuition for the use of the integral in the task of arc length finding.

If you go further, you can observe that if we are finding the arc length with respect to $x$, we find $\mathrm dx$ and $\mathrm dy$ with respect to $x$. This leads us to the following formulas usually presented to students:

$$\text{Arc Length w.r.t } x = \int_a^b \sqrt{\left({\mathrm dx\over \mathrm dx}\right)^2 + \left({\mathrm dy\over \mathrm dx}\right)^2} \,\mathrm dx = \int_a^b \sqrt{1 + \left({\mathrm dy\over \mathrm dx}\right)^2} \,\mathrm dx$$ $$\text{Arc Length w.r.t } y = \int_a^b \sqrt{\left({\mathrm dx\over \mathrm dy}\right)^2 + \left({\mathrm dy\over \mathrm dy}\right)^2} \,\mathrm dy = \int_a^b \sqrt{\left({\mathrm dx\over \mathrm dy}\right)^2 + 1} \,\mathrm dy$$

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As per my comment, integration always gives you the "area under the curve" between the two points of evaluation. With the formula for arc length, the formula has been written in such a way that the "area under the curve" (that is, the curve that you're integrating) is equal to the arc length --- you don't integrate the arc curve itself, but instead you integrate some other curve that tells you how long the arc in question is.


Consider the 'curve' given by $y = x$. Suppose the 'arc length' is wanted from $x = 0$ to $x = a$ where $a$ is some positive real number. By drawing this out, you can easily see that the length of the 'curve' between $x = 0$ and $x = a$ is $a\sqrt{2}$ by using Pythagoras' Theorem. Another way to have found this value could be with the following integral: $$ \int_{0}^{a} \sqrt{2} \,\,\mathrm{d}x. $$ This integral gives the area under the curve $y = \sqrt{2}$. This is clearly a different curve to above. However, the integral gives the value $a\sqrt{2}$, which is the area under the curve $y = \sqrt{2}$, but it is also the length of the arc $y = x$ between $x = 0$ and $x = a$.

We integrate something that gives us an area, but that area corresponds to the length that we want.


If you've done an statistics, this is similar to a Cumulative Density Function (CDF) that gives you an area when you plug in one value, with this CDF corresponding to some Probability Density Function (PDF).

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  • $\begingroup$ that's all good and fine but that does not answer the question. "Integration" doesn't know what you are summing. The fact is an arc length has ONE dimension. and an area has TWO dimensions ...when you integrate you should always use the same technique so how is it that in one case you end up with one dimension and the other case you end up with two dimensions. One of the answers above which I accepted cancels one of the dimensions out..although a little skeptical I believe he is correct. $\endgroup$
    – Sedumjoy
    May 8, 2018 at 15:36
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    $\begingroup$ @Sedumjoy You're getting confused by the dimensions and putting too much emphasis on them. Integration (in one variable) always gives you the area, which is two dimensional. The difference is what that area means with respect to the problem at hand. In the case of the arc length, the function that you integrate gives you an area that is equal to the arc length. $\endgroup$
    – Bilbottom
    May 8, 2018 at 17:03
  • $\begingroup$ @Sedumjoy See my update. $\endgroup$
    – Bilbottom
    May 9, 2018 at 15:23
  • $\begingroup$ Bill Wallis : That would solve the dilemma for me Bill...you mind showing that with a simple example? $\endgroup$
    – Sedumjoy
    May 9, 2018 at 19:33
  • $\begingroup$ @Sedumjoy Is the example in the updated answer not sufficient? $\endgroup$
    – Bilbottom
    May 9, 2018 at 19:35

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