# Homework Help: Volume Integral homework

1. Oct 17, 2011

### 1MileCrash

1. The problem statement, all variables and given/known data

The soot produced by a garbage incinerator spreads out in a circular pattern. The depth H(r) in millimeters, of the soot deposited each month at distance r kilometers from the incinerator is given by H(r) = 0.115e^(-2r)

Write the definite integral for the amount of soot deposited per month in a 5 kilometer radius.

2. Relevant equations

3. The attempt at a solution

Looks easy, but something is confusing me.

I integrate from 0 to 5 kilometers and multiply H(r) by 2pir, but since these functions give out different units, one in millimeters and one in kilometers, how can I just straight multiply? I divided H(r) by 10e^6 to put the millimeters into kilometers, but the book doesn't do that. The book just multiplies the two functions and integrates from 0 to 5.

Why is no unit conversion nessesary?

2. Oct 17, 2011

### QuarkCharmer

You have to convert the units. Your book might be wrong. Are you sure it didn't multiply/divide a unit conversion factor in there somewhere you didn't notice?

3. Oct 17, 2011

### 1MileCrash

I am sure. If I ran their integral as is, would I get a correct volume in cubic meters, or just a nonsense result?

4. Oct 17, 2011

### dacruick

noo you don't have to convert the units. H(r) is in millimetres and r is in kilometres. The equation already takes into account that r is in km, and spits out an answer in mm.

5. Oct 17, 2011

### 1MileCrash

What? But I have to multiply that by the area, 2pir, which is surely in kilometers..

6. Oct 17, 2011

### dacruick

Oh I'm sorry I think I've misunderstood what's going on. Let me know if this is right: H(r)dr will give you the depth of the soot at each r value along a radius. you want to get the volume of this soot so you multiply the H(r)dr by 2pir, but you're pretty sure that there has to be a unit conversion. It sounds to me like you're right.

EDIT: oooo they give you an answer in cubic metres? Sounds to me like they did a little unit conversion under your nose. If they did km * mm, they might not do a unit conversion because km/1000 = metres and mm*1000 = metres. So they might have just skipped it and settled on metres.

7. Oct 17, 2011

### 1MileCrash

No, they gave me nothing. Im asking if evaluating their integral that has no conversion will give a correct volume in cubic meters, since one unit is 1000 times smaller, and one is 1000 times bigger.

8. Oct 17, 2011

### dacruick

ohh thats a question? Well this is confusing.

9. Oct 17, 2011

### dacruick

I feel like you need two integrals for this problem. I think that you need one to integrate across the radius to find depth. And the second integral to find the volume of that slice rotated 360 degrees.

10. Oct 17, 2011

No..