# Homework Help: Problem Deriving Volume of a Cylindrical shell

1. Apr 16, 2010

### My4rk89

I have seen the derivation for the formula:

h2(pi)rdr used in math textbooks. However, earlier today I had a physics problem where I needed to use the volume of a cylindrical shell of inner radius r and outer radius dr+r and length h. without remembering the formula I tried to derive it starting by determining the area of the base minus the area of the gap:
A=(pi)r^2-(pi)(r+dr)^2

and then multiplying by h

I could not get this in the form h2(pi)rdr starting from where I did.

I understand the derivation given here:
http://www.stewartcalculus.com/data...texts/upfiles/3c3-Volums-CylinShells_Stu .pdf

But I'm just frustrated that I can't understand why the method I tried doesn't reduce to the correct formula. Its simply the area of the base*height!

Why can't that formula be arrived at by my method???

2. Apr 16, 2010

### Staff: Mentor

Assuming that dr is positive, what you'll get from your formula is a negative number.

So instead, you should have A = pi(r + dr)^2 - pi*r^2 = pi*( (r + dr)^2 - r^2).

Expand the (r + dr)^2 part and combine like terms. What do you get?

3. Apr 16, 2010

### My4rk89

When I expand I get r^2 +2rdr + dr^2

so the r^2 drops out. but what happens to the the dr^2 ?????

4. Apr 17, 2010

### Staff: Mentor

The idea here is that if dr is small in comparison to r, the (dr)^2 will be very much smaller and can be discarded.

For example, if r = 1 and dr = .01, then (r + dr)^2 = (1.01)^2 = 1.0201.

If I expand (r + dr)^2, I get r^2 + 2r*dr + (dr)^2 = 1 + 2(.01) + (.0001). If I omit the (dr)^2 term, I get 1.02, which is pretty close to 1.0201.

The smaller dr is in comparison to r, the closer (r + dr)^2 is to r^2 + 2r*dr.

5. Apr 17, 2010

### My4rk89

Alright that's what I thought I just couldn't find any verification!

I never really encountered a differential element squared before.

Thanks so much!