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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 a distance $r$ kilometers from the incinerator is given by $H (r) = 0.116 e^{-1.5 r}$.

I'm supposed to get the integral? I found that integral from $0$ to $5$ is $2pi*r*0.116*e^{(-1.5*r)/1000000}$ km$^3$ …Unit and I have to include the units along with my answer which keep giving me wrong.

I have tried Km$^3$, m$^3$, mm$^3$ and it's all wrong

Can anyone help with it?

Vladimir Vargas
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Ahmed Al.Mutaiwea
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    First, what question were you asked? The volume? The average depth? Second, this is not a shell method problem, it is integration in polar coordinates. – André Nicolas Nov 22 '14 at 02:06
  • Write a definite integral (with independent variable r) giving the total volume of soot deposited within 5 kilometers of the incinerator each month. volume. That's the question – Ahmed Al.Mutaiwea Nov 22 '14 at 05:16

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Better keep $r$ in kilometers as in the problem statement--and therefore you must not put a conversion factor into the exponent of $e.$ The exponential should be $e^{-1.5r}$ with nothing else in the exponent.

To make units of height and width consistent so that you can measure the volume in $\mbox{km}^3,$ you could observe that $H(r) = 0.116 e^{-1.5 r}\, \mbox{mm} = (0.116 \times 10^{-6}) e^{-1.5 r}\, \mbox{km}.$

If you would rather the answer be in $\mbox{mm}^3,$ you can make a substitution, $r =10^6 u,$ so that $r\, \mbox{km} = u\, \mbox{mm}.$ Just make sure to follow all the usual rules of substituting a variable in a definite integral.

David K
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