Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
What is the role of the $dr$ in the expression for $H(r)$? – DJohnM Jan 29 '13 at 18:50
Sorry that was a typo. – Gabrielle Jan 29 '13 at 19:02
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 $km^3$.

share|cite|improve this answer
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 $km^3$ – DJohnM Jan 29 '13 at 20:08

What you have done is correct. This means 1/2 m^3 of soot each month.

share|cite|improve this answer
Doesn't one convert $km^3$ to $m^3$ by multiplying by $10^9$? – DJohnM Jan 29 '13 at 19:11

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.