Calculating Soot Deposition in a Circular Pattern

[1MileCrash]

Homework Statement

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.

Homework Equations

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?

 

[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?

 

[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?

 

[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.

 

[1MileCrash]

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

 

[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.

 

[1MileCrash]

No, they gave me nothing. I am 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.

 

[dacruick]

ohh that's a question? Well this is confusing.

 

[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.

 

[1MileCrash]

No..