-2
$\begingroup$

I've seen many derivations in which Integration is used. But I don't understand the fact that why after going to a distance like $y$ or $x$, we take an element $dy$ or $dx$? Instead can't we take any other element of a different width say $d\theta$ after going to $y$?

What does this mean?

Improve this question
$\endgroup$
1
  • 1
    $\begingroup$ If you are taking an element $dx$ or $dy$ , it means you are working in cartesian coordinates. If you are taking an element $dr$ or $d\theta$ ,it means you are working in polar coordinate. An infinitesimal element depend on which coordiante you are working . You should choose the coordinate which is more convenient. $\endgroup$
    – Himanshu
    Commented Sep 17, 2020 at 14:02

1 Answer 1

Reset to default
3
$\begingroup$

If you want to integrate some attribute $f$ with respect to $\theta$ - and if you think it is physically meaningful to do that - then you an certainly do so. But you will need to find an expression for $f$ as a function of $\theta$. In other words $\int f(\theta) \space d \theta$ makes sense, but $\int f(x) \space d \theta$ does not unless you know how $x$ varies with $\theta$.

Improve this answer
$\endgroup$
3
  • $\begingroup$ Okay. But why we need to go to y to take an element dy can't we take that element at the origin and name it something like dA or dx or dy or anything? Why is is that we take ab element dy after going to a distance y? $\endgroup$
    – Vibhu Mishra
    Commented Sep 17, 2020 at 14:12
  • $\begingroup$ @VibhuMishra To find the area of a circle you chop it into slices and add up the areas of each slice. You don’t have to “go to” $y$ to find the area of the slice at $y$ - you could imagine moving all the slices to the origin instead - but the area of each slice will still depend on the value that $y$ had where that slice came from. $\endgroup$
    – gandalf61
    Commented Sep 17, 2020 at 14:20
  • $\begingroup$ Thanks! I got it now, I was so confused in the derivations $\endgroup$
    – Vibhu Mishra
    Commented Sep 17, 2020 at 14:27

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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