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Area of a ring element

  1. Jan 6, 2016 #1
    The circumference of the shaded ring is 2πρ however I am struggling to understand how the area, dA, of the ring is equal to (2πρ)dρ? I mean the circumference varies depending on the value of ρ so surely we can't multiply by dρ to yield the entire area of the shaded ring? If we decided to go by the method followed in the diagram above, then the area of the circle with radius ρ would be circumference*thickness = 2πρ(ρ) = 2πρ2 but this isn't correct as the area should be πρ2?

    I know this is supposed to be simple but I am having a brain freeze and it just isn't clicking atm.
  2. jcsd
  3. Jan 6, 2016 #2


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    The ring is assumed to be of infinitesimal thickness. Here, thickness dρ is the infinitesimal change is the radius. Hence, as dρ→0, i.e.as dρ tends to 0, the inner circumference of the ring→the outer circumference i.e. area of the ring=2πρ⋅dρ.
  4. Jan 6, 2016 #3


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    This method is useful only when the "change" is very small, to be precise, "infinitesimal"(→0). This is a very common technique in calculus.
  5. Jan 6, 2016 #4


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    [itex]\int 2\pi \rho d\rho = \pi \rho^2[/itex]
  6. Jan 8, 2016 #5
    yes, you have to notice in the diagram that the width ['thickness'] of the ring is illustrated as dp.
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