# Small question

1. May 18, 2004

### Silverious

If I intergrate 2(pi)r I get (pi)r^2

If I integrate that I get 1/3 (pi)r^3, which is close to the volume of a sphere. But where do I get a 4/3(pi)r^3 ?

2. May 18, 2004

### arildno

Differentiate the volume expression, and find a suitable interpretation of the result.

Last edited: May 18, 2004
3. May 18, 2004

### matt grime

beacause disks are not the infinitesimal shells that add up to a solid sphere

4. May 18, 2004

### Silverious

Would it work if I used half of the area forumula, and used the disk method to rotate it about an axis? Should that give me 4/3(pi)r^3?

I would find out for myself but I'm a little busy.

Thanks for the replies.

5. May 18, 2004

### matt grime

that's a different technique: add up the shells of surface area of a sphere

6. May 19, 2004

### Silverious

OKay I'm starting to understand.

So I use the surface area formula(for a sphere). Integrate from 0 to r? :uhh:

Edit: Oh my god. I feel so stupid...

You see, I didn't know the formula for the surface area of a sphere. ......

Oh well. Anyways, uum. So then, what is 1/3(pi)r^3? Without looking it up, making a wild guess...is it the volume of a cylinder?

Edit2: I really need to stop thinking.... I have no clue what I'm talking about. :yuck: So is 1/3(pi)r^3 just nonsense?

Edit3: Interesting that a cone's volume is 1/3(pi)r^2 h. Good I have much time to think about it.

Last edited: May 19, 2004
7. May 19, 2004

### matt grime

the surface area of a sphere is 4(pi)r^2