1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Calculating area of an annulus

  1. May 30, 2013 #1
    My physics book when providing proof of the electric potential of a disk tells me that the area of a ring with radius [itex]R'[/itex] and width [itex]dR'[/itex] is [itex]2 \pi R'dR'[/itex]
    The problem is, I have no idea how he arrives at this conclusion. Here is my attempt:
    [itex]A = \pi (R'+dR')^2 - PiR'^2\\ A = \pi(R'^2 + dR'^2 + 2R'dR') - \pi R'^2 \\ A = \pi dR'^2 + 2 \pi R' dR'[/itex]
    Which is, of course, different from what my book says:
    [itex]A = \pi dR'^2 + 2 \pi R' dR' \neq 2 \pi R'dR' [/itex]
  2. jcsd
  3. May 30, 2013 #2
    Notice that [itex]dR'^2[/itex] is the square of an infinitesimal quantity, which we take to be zero. So your derivation works out.
  4. May 30, 2013 #3


    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    When doing infinitesimal proofs like this any squared terms are negligible - for example your [itex] dR'^2[/itex] should be thrown out.

    An alternate way to see the result is if you cut the ring and lay it out flat it's a rectangle (almost) with side lengths easily calculated.
  5. May 30, 2013 #4


    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    I will drop the primes so as not to confuse with the derivative. What your text is using is the differential approximation to the annulus area. You have a circle of area ##A=\pi R^2##. The approximate change in area when ##R## is changed by an amount ##dR## is the differential ##A'dR = 2\pi R dR## which approximates the exact change, which you have calculated. For small ##dR## you can ignore the ##dR^2## term.

    [Edit] Wow! Two other answers just while I was typing. That's life in PF.
  6. May 30, 2013 #5
    Ohh, I see ! Thanks to you three, that perfectly cleared my doubt.
  7. May 31, 2013 #6


    User Avatar
    Science Advisor

    A disk of radius R has area [itex]\pi R^2[/itex]. A disk of radius R+ dR has area [itex]\pi(R+ dR)^2= \pi R^2+ 2\pi RdR+ \pi(dR)^2[/itex]. The area of the annulus between them is [itex]2\pi RdR+\pi (dR)^2[/itex]. If dR is much smaller than R, that is approximatelyu [itex]2\pi RdR[/itex].
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted