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If I want an area differential of a circle, it goes: $$dA=d(\pi \ r^2)=2\pi r dr$$ This is very useful. But What about a square?

$$dA=d(l^2)=2l\cdot\ dl$$

Is this valid? Well, if I simply integrate it, of course it gives me the square's area value, but when you look my first example, it makes sense that on a circle the product between dr and the respective perimeter comes up to the area. I want to know if the square's area differential form makes sense as well, because it is not so intuitive like the circle's example.

Vitor Aguiar
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  • $2\pi r dr$, is what I think you mean. – Alex Jones Jun 21 '17 at 23:27
  • @AlexanderJ93 yes, thanks – Vitor Aguiar Jun 21 '17 at 23:28
  • I think this intuition works... imagine splitting the square into two right triangles. The areas of both of the triangles is just $l$ integrated. – u8y7541 Jun 21 '17 at 23:39
  • Think of the square as $dA = l\cdot dl + l\cdot dl$. Geometrically, when $l$ increases by $dl$, you add a rectangle of dimensions $l \times dl$ on the right, one of dimensions $dl \times l$ on top, and a tiny square of $dl\times dl$ in the upper right. And you ignore the tiny square because the ratio of its area to $dl$ goes to zero as $dl$ goes to zero. – Mark Fischler Jun 21 '17 at 23:39
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    In the case of a circle, if you expand uniformly by a distance of $dx$, then that corresponds to increasing the radius by $dx$. But for the case of a square, if you expand it uniformly by a distance of $dx$, that increases the side length by $2, dx$, so $dA = 2l , dl = 2l (2 , dx) = 4l \cdot dx$, which is indeed the perimeter times the expansion distance. – Daniel Schepler Jun 21 '17 at 23:43
  • @MarkFischler

    what does $ l \cdot\ dl $ mean? Is it a dl on the bottom versus the whole lenght on the vertical? or vice-versa.

    By the way, In rectangles, if I say its area is $A=l \cdot\ w $, then $ dA=l \cdot\ dw + w \cdot\ dl $, then if you integrate it comes to $2 \cdot\ l \cdot\ w = 2 \cdot\ A$ , which is impossible.

     – Vitor Aguiar Jun 22 '17 at 01:51

1 Answers1

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Think about a square from $(0, 0)$ to $(\ell, \ell)$. Now increase $\ell$ a little bit. The square's area increases by a thin strip on the right (length $\ell$) and a thin strip on top (also length $\ell$), and each strip has width "$d\ell$". You've double-counted a small square of size $d\ell^2$, but we can ignore that.

John Hughes
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  • so it is incorrect? I remember I did this thing on one exercise of Faraday's law, and it gave me wrong because I couldn't say $dA=2l dl$ – Vitor Aguiar Jun 21 '17 at 23:54
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    I think that unless you're willing to make a formal definition for what differentials 'mean', it doesn't really matter whether this is right or not -- roughly speaking, it's neither true nor false until you nail down the definitions, and then state and prove the theorems relating these definitions to integrals (or other places where you want to apply them). But for intuition's sake, it's a reasonable thing to think about. – John Hughes Jun 22 '17 at 00:27
  • I still have 2 questions that bother me. First, what does $ l \cdot\ dl $ mean? Is it a dl on the bottom versus the whole lenght on the vertical? or vice-versa.

    In rectangles, if I say its area is A=lw, then $ dA=ldw + wdl $, then if you integrate it comes to 2lw = 2A , which is impossible.

     – Vitor Aguiar Jun 22 '17 at 01:44
  • The dot means "multiply" -- I put it in there to keep letters separated. If your new square goes from $(0,0)$ to $(\ell+a, \ell+a)$, you can divide it into four pieces: the original, with area $\ell^2$; the right strip, with area $a\ell$, the top strip, with area $a\ell$, and the top right corner, with area $a^2$. Now if you call the extra length "$d\ell$ instead of $a$, you get the formula I wrote (except for the $a^2$, which folks treat as "negligible"). As for whether this "means" anything...that's where writing out the definitions and theorems comes in. So right now, the answer is "no". – John Hughes Jun 22 '17 at 09:35