Calculating area of an annulus

[ShizukaSm]

My physics book when providing proof of the electric potential of a disk tells me that the area of a ring with radius R' and width dR' is 2 \pi R'dR'

The problem is, I have no idea how he arrives at this conclusion. Here is my attempt:
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'
Which is, of course, different from what my book says:
A = \pi dR'^2 + 2 \pi R' dR' \neq 2 \pi R'dR'

 

[dipole]

Notice that dR'^2 is the square of an infinitesimal quantity, which we take to be zero. So your derivation works out.

 

[Office_Shredder]

When doing infinitesimal proofs like this any squared terms are negligible - for example your dR'^2 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.

 

[LCKurtz]

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.

 

[ShizukaSm]

Ohh, I see ! Thanks to you three, that perfectly cleared my doubt.

 

[HallsofIvy]

A disk of radius R has area \pi R^2. A disk of radius R+ dR has area \pi(R+ dR)^2= \pi R^2+ 2\pi RdR+ \pi(dR)^2. The area of the annulus between them is 2\pi RdR+\pi (dR)^2. If dR is much smaller than R, that is approximatelyu 2\pi RdR.