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Estimate the total mass of water (in kg) in a model hurricane using the following assumptions, being mindful of units.

(a) The mass density of water $(g/m^3)$ is modeled as $$ p(r,z) =\frac{50}{r}e^{\frac{-z}{4}}. $$ z and r are measured in km.

(b) The hurricane occupies a cylindrical annulus that is 9 km deep. The eye of the hurricane (where there is no water) has a radius of 50 km and the hurricane's outer diameter is 1000 km.

I set up the triple integral for this as

$$ \int_0^9\int_0^{2\pi}\int_{50}^{1000}50e^{\frac{-z}{4}} dr d\theta dz $$

I got an answer of 1067979 when evaluating this integral. According to the question, this should be in units of grams, which would make the final answer 1068 kg. This seems really low for the mass of a hurricane. Did I make a mistake that I'm overlooking?

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    $\begingroup$ Maybe the mistake is just in unit conversion? I understood the wording in part a) to mean that z and r are measured in km, and input in km into p(r,z), which yields the density in $g/m^3$. Should I have converted these to meters first? $\endgroup$ – dibdub Jul 7 '16 at 7:00
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Use metres as the unit of length, then as the scale hight of the hurricane is $4\mbox{ km}=4000\mbox{ m}$, the mass in grams is: $$ \begin{aligned} \int_{\theta=0}^{2\pi}\int_{r=50000}^{1000000}\int_{z=0}^{9000}50\exp\left({\frac{-z}{4000}}\right) \;dz\; dr\; d\theta&= 178920.155 \int_{\theta=0}^{2\pi}\int_{r=50000}^{1000000}\;dr \;d\theta\\ &=1.06798\,{10}^{12}\mbox{ grams } \end{aligned} $$

which when converted to $\mbox{ kg}$ gives $1.06798\,{10}^{9}\mbox{ kg }$

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  • $\begingroup$ Where are you getting 4000 from? 4 is a constant, it doesn't have units. $\endgroup$ – dibdub Jul 7 '16 at 8:34
  • $\begingroup$ The $4$ is the scale height of the hurricane, $z/4$ in the original expression has to be dimensionless, as the height in that expression is in km when we switch to metres we also have to switch the scale height from km to m. Another way of looking at it is if $z=1$ km then the $z/4=1/4$, but now if we switch to m we want the argument of the exponential to be $-(1000/1000)/4=-1000/4000=z/4000$ where $z$ is now in metres. $\endgroup$ – Conrad Turner Jul 7 '16 at 8:48
  • $\begingroup$ Well then if the equation was set up for kilometers, why doesn't it work if you use kilometer values? $\endgroup$ – dibdub Jul 7 '16 at 8:49
  • $\begingroup$ It does work in km once you convert the density units to mass per cubic km. The density is in units of grams per cubic metre, you need to convert this into grams per cubic km (or kg per cubic km), which requires a factor of $10^9$ (or $10^6$ for kg). You have to use a consistent set of units through out $\endgroup$ – Conrad Turner Jul 8 '16 at 3:22

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