Differential Area of a Circular Hoop

Hello,

OK. So suppose you have a circular hoop of differential thickness dx that has a radius r. I have been told in multiple courses that the differential area of such a hoop is dA=2πr*dx. It is hard for me to see why this is.

Can somebody please explain how this derivation comes about? A proof or explanation would be appreciated.

Re: Differential Area of a Circular Hoop

Quote:

Originally Posted by

**thekonstant** Hello,

OK. So suppose you have a circular hoop of differential thickness dx that has a radius r. I have been told in multiple courses that the differential area of such a hoop is dA=2πr*dx. It is hard for me to see why this is.

Can somebody please explain how this derivation comes about? A proof or explanation would be appreciated.

There are two ways to look at this. If you were to "cut" the hoop and straighten it out, it would be a rectangle of width dx and length equal to the **circumference** of the circle, $\displaystyle 2\pi r$. The area is, then, $\displaystyle 2\pi r dx$.

Or: think of the hoop as the area **between** two circles, one of radius r and the other of radius r+ dx. The larger circle has area $\displaystyle \pi (r+ dx)^2= \pi(r^2+ 2rdx+ dx^2)$ and the smaller area $\displaystyle \pi r^2$. The difference is $\displaystyle \pi (2rdx+ dx^2)$ but since dx is a "differential", we ignore its square leaving just $\displaystyle 2\pi r dx$.

Re: Differential Area of a Circular Hoop

Fantastic Response. I tended to think of the hoop the second way you mentioned, but couldn't quite get past the squared differential.

Thanks for clearing things up.