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I have the question: 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 $$H(r)=0.119{e^{-2.1r}}$$

Write a definite integral (with independent variable r) giving the total volume of soot deposited within 5 kilometers of the incinerator each month.

I wrote that the integral was $$\int_0^5 \ {2{pi}r}{0.119{e^{-2.1r}}}\,dr.$$ and after evaluating, I said the volume of the soot was 0.0005404606479 km^3. I don't think this answer is correct but I can't seem to see where I'm going wrong. Is the integral set up correctly?

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What is the role of the $dr$ in the expression for $H(r)$? –  User58220 Jan 29 '13 at 18:50
    
Sorry that was a typo. –  Gabrielle Jan 29 '13 at 19:02
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2 Answers

up vote 1 down vote accepted

In the integrand, you seem to be multiplying $r$ and $dr$ in kilometers, and then multiplying by $H(r)$, the depth in millimeters. The limits of the integration, and the coefficient in the exponential are also in kilometers. Change the depth to kilometers by dividing $H(r)$ by $10^6$, and the result will be in $km^3$.

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So technically I could just move the decimal place over in my answer 6 times and it would be correct ? –  Gabrielle Jan 29 '13 at 19:30
    
Six places to the left, for $km^3$ –  User58220 Jan 29 '13 at 20:08
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What you have done is correct. This means 1/2 m^3 of soot each month.

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Doesn't one convert $km^3$ to $m^3$ by multiplying by $10^9$? –  User58220 Jan 29 '13 at 19:11
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