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Homework Help: Integrating the area of a disc

  1. Mar 14, 2008 #1
    1. The problem statement, all variables and given/known data

    Finding the area of a disc by integration of rings.

    2. Relevant equations

    A ring of radius r and thickness dr has an area of [tex]2 \pi rdr[/tex].

    3. The attempt at a solution

    Why isn't it [tex] \pi (2rdr + (dr)^2)[/tex]?
  2. jcsd
  3. Mar 14, 2008 #2


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    I don't understand - why do you think it might be? :confused:

    What would [tex] \pi (dr)^2[/tex] represent?
  4. Mar 14, 2008 #3


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    The exact area of the ring is pi*((r+dr)^2-r^2)=pi(2*r*dr+dr^2), indeed. But in the integration we are taking the limit of a sum of these where dr->0. The limit of the terms involving dr^2 will go to zero. You can ignore corrections involving higher powers of 'infinitesimals' like dr.
  5. Mar 14, 2008 #4
    Oh dear... i posted too quickly.

    I just realised why it isn't.

    My old reasoning was that the ring's area = the area enclosed by the outside circumference - the area enclosed by the inside circumference.

    [tex] A = \pi (r+dr)^2 - \pi r^2
    = \pi (r^2 + 2rdr + (dr)^2 - r^2)
    = \pi (2rdr + (dr)^2) [/tex]

    I now realise that the ring's radius is measured from the centre of the width of the ring.

    [tex] A = \pi (r+ \frac{1}{2} dr)^2 - \pi (r- \frac{1}{2} dr)^2
    =\pi (r^2 +rdr + \frac{1}{4} (dr)^2 - r^2 + rdr - \frac{1}{4} (dr)^2)
    = 2 \pi rdr [/tex]

    EDIT: Oh you posted before i saw... We both have different reasons. Who is right?
  6. Mar 14, 2008 #5


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    If r and dr are the same, then those are the areas of two different rings. If dr is a finite number then you'd better pay attention to which is which. But,either calculation works for integration once you ignore higher powers of dr.
  7. Mar 15, 2008 #6
    This reasoning is correct, in my opinion.
    When I was in school, the explanation I got from my teacher was that while doing integration we add infinitesimally small quantities. The quantity dr is very small and so [tex]dr^2[/tex] will be even smaller......like [tex]0.000001^2=0.000000000001[/tex] , so we neglect the [tex]dr^2[/tex] thing from that expression.
  8. Mar 15, 2008 #7


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    The simplest way to do the original problem is to say that the area of a small ring, of inner radius r and thickness dr is approximately [itex]2\pi r[/itex], the length of the ring, times dr, the thickness. That would be exactly correct if it were a rectangle of length [itex]2\pi r[/itex] and width dr. The point is that in the limit, as we change from Riemann sum to integral, that "approximation" becomes exact (the "[itex]dr^2[/itex] in your and Dick's reasoning) goes to 0 : the area is given by [itex]\int \pi r dr[/itex].

    Can someone explain to my why this is a physics problem and not a mathematics problem?
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