009B Sample Final 2, Problem 4

A city bordered on one side by a lake can be approximated by a semicircle of radius 7 miles, whose city center is on the shoreline. As we move away from the center along a radius the population density of the city can be approximated by:

$\rho (x)=25000e^{-0.15x}$

people per square mile. What is the population of the city?

ExpandFoundations:
Many word problems can be confusing, and this is a good example.
We know that we are going to integrate over a half-disk of radius 7, but how do we construct the integral?
One key could be the expression of our density,
$\rho (x)=25,000e^{-0.15x}$
where  $x$  is the distance from the center.
Any slice along a radius gives us a cross section.
If we were revolving ALL the way around the center, this would be typical solid of revolution,
and we could find the volume of revolving the center by the usual shell formula
$V\ =\ \int _{x_{1}}^{x_{2}}2\pi R\cdot h\,dx.$
What changes, since we are only doing half of a disk?
Also, this particular problem will require integration by parts:
$\int u\,dv=uv-\int v\,du.$

Solution:

ExpandStep 1:
We can treat this as a solid of revolution, and use the shell method.
We are working on a half disk of radius 7, so we can integrate a cross-section where  $x$  goes from 0 to 7
and the height at each  $x$  is our density function,  $\rho (x).$
Normally  $2\pi R$  represents once around a circle of radius  $R,$
but in this case we only go half way around.
Therefore, we adjust our usual shell method formula to find the population as
${\begin{array}{rcl}\displaystyle {P}&=&\displaystyle {\int _{x_{1}}^{x_{2}}\pi R\cdot h\,dx}\\&&\\&=&\displaystyle {\int _{0}^{7}\pi x\cdot \rho (x)\,dx.}\end{array}}$

ExpandStep 2:
Let's plug in the actual formula for density and solve. We have
${\begin{array}{rcl}\displaystyle {P}&=&\displaystyle {\int _{0}^{7}\pi x\cdot 25,000e^{-0.15x}\,dx}\\&&\\&=&\displaystyle {25,000\pi \int _{0}^{7}xe^{-0.15x}\,dx.}\end{array}}$
To solve this, we need to use integration by parts.
Let  $u=x$  and  $dv=e^{-0.15x}dx.$
Then,  $du=dx$  and  $v=-{\displaystyle {\frac {e^{-0.15x}}{0.15}}.}$
Thus,         ${\begin{array}{rcl}\displaystyle {P}&=&\displaystyle {25,000\pi \int _{0}^{7}xe^{-0.15x}\,dx}\\&&\\&=&\displaystyle {25,000\pi \left[\left.-{\frac {xe^{-0.15x}}{0.15}}\right|_{0}^{7}+\int _{0}^{7}{\frac {e^{-0.15x}}{0.15}}\,dx\right]}\\&&\\&=&\displaystyle {25,000\pi \left[-{\frac {xe^{-0.15x}}{0.15}}-{\frac {e^{-0.15x}}{(0.15)^{2}}}\right]_{0}^{7}}\\&&\\&=&\displaystyle {25,000\pi \left[\left(-{\frac {7e^{-1.05}}{0.15}}-{\frac {e^{-1.05}}{(0.15)^{2}}}\right)+{\frac {1}{(0.15)^{2}}}\right]}\\&&\\&\approx &986,556.\end{array}}$
Note that in a calculator-prohibited test, no one would expect the actual numerical answer.
However, you would likely need the line above it to receive full credit.

ExpandFinal Answer:
${\displaystyle 25,000\pi \left[\left(-{\frac {7e^{-1.05}}{0.15}}-{\frac {e^{-1.05}}{(0.15)^{2}}}\right)+{\frac {1}{(0.15)^{2}}}\right]\ \approx \ 986,556}$

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