0
$\begingroup$

For a function y=f(x) ,the area under the curve is given by $\int y(x) dx$.

This must mean that the infinitesimal area is $da = y(x)dx$ where y(x) is a finite value, i.e., not an infinitesimal.

However, in solving physics problems involving integration, I usually think "Oh, it's an area integral so there should be two infinitesimal quantities multiplied..." and then I go on to choose the right coordinate system and find the area:

For example:

$da= dr rd\theta$ or $da=dx dy$ and so on, where there are two infinitesimals clearly.

Is there a gap in my learning about infinitesimal areas? How is it that the infinitesimal area is first degree infinitesimal in one and a second degree infinitesimal in the other? (Hoping the word "degree" is used properly)

$\endgroup$
3
  • $\begingroup$ Perhaps @NiharKarve but I would like a physics person to answer, without too much rigour or abstractness $\endgroup$
    – Sidarth
    Jun 16 at 6:13
  • $\begingroup$ Integral in $dy$ is trivial: it is just $\int_0^{f(x)} dy = f(x)$ $\endgroup$
    – nwolijin
    Jun 16 at 8:00
  • $\begingroup$ Regarding the word "degree", it's better to use "order" here. $\endgroup$
    – sleepy
    Jun 16 at 9:03
0
$\begingroup$

Imagine you want to calculate the area of a rectangle on the $x-y$ plane. In the spirit of the infinitesimal analysis, what you can do is either think of this rectangle as consisting of very small "dots" and then sum the areas of each dot, or think of it as consisting of very narrow "stripes" and sum the areas of each stripe. In the first case, the area of each dot is $dxdy$ and to get the entire area, you have to calculate a two-dimensional integral $\int dxdy$ (remember, an integral is just a sum). In the second case, the area of each stripe is $y(x)dx$ (or vice versa, $x(y)dy$), and you calculate a one-dimensional integral $\int y(x)dx$. Of course the result will be the same.

This might be mathematically not 100% accurate, but should give a physical feeling about it. So the order of "infinitesimality" you're talking about just shows how many integrations you have to perform. Of course, the same holds in the 3d space, so a volume of a small "cube" will be $dxdydz$, a narrow "rod" $z(x,y)dxdy$, and so on. Differentials $dq$ are of the same physical dimensions as non-infinitesimal values, so the product will be the same physical quantity.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.