Why is the area of infinite small annulus 2πr dr?

[bolzano95]

Homework Statement

When I have a disk with radius r then naturally the area is πr^2. Then I want to do this by calculus and my first step is simply taking πrdr. But the correct way is to take 2πrdr. To me this is really confusing, because I would never take 2πr dr (circumference x width).

Homework Equations

The Attempt at a Solution

Well, wiki says when you integrate you get this you get the same solution as without using calculus. So the way to remember this is kind of reversed way- you know what is the solution, but you still have to put something into the integral to calculate?

 

[phyzguy]

It's not the area of a disk that is 2π r dr, it is the area of an annulus with a very small width. The area of a disk is π r^2. The area of an annulus is π ro^2 - π ri^2, where ro is the outer radius and ri is the inner radius. If the width of the annulus is very thin, then you can take ro = r, and ri = r - dr.
Then A = π (r^2 - (r-dr)^2) = π (r^2 - (r^2 - 2r dr + dr^2)) = 2π r dr - π dr^2. Neglecting the term in dr^2 gives 2π r dr.

 

[drvrm]

Then I want to do this by calculus and my first step is simply taking πrdr.

so, if you want to do using calculus what do you get ?

why not take rate of change of area with respect to change in r .

then write change in area...

 

[bolzano95]

Thank you, phyzguy, this really helped me! :)