# Calculate the population within a 6 mile radius of the city center, if the radial population...

## Question:

Calculate the population within a 6 mile radius of the city center, if the radial population density is {eq}\rho (r) = 6 (5 + r^2)^{\dfrac 1 3} {/eq} (in thousands per square mile). (Use decimal notation. Give your answer to two decimal places.)

## Integration:

A mathematical quantity which used, to sum up, the all small or infinitesimal data or function to provide absolute value or function to analyse the entire system is known as integration. It is part of the calculus.

## Answer and Explanation:

Given Data:

• The radial population density is: {eq}\rho \left( r \right) = 6{\left( {5 + {r^2}} \right)^{\dfrac{1}{3}}}\;{\rm{thousands}}\;{\rm{per}}\;{\rm{square}}\;{\rm{mile}} {/eq}
• The radius of city center is: {eq}r = 6\;{\rm{mile}} {/eq}

The radial population density is function of radius of city center.

Lets take small area of region of radius {eq}dr {/eq}

The expression for area of small region is

{eq}dA = 2\pi rdr {/eq}

The expression for population within small region of radius is

{eq}dP = dA\rho \left( r \right) {/eq}

Substitute and integrate the above expression with radius of the city center is

{eq}\begin{align*} \int {dP} &= \int_0^r {2\pi rdr\left( {6{{\left( {5 + {r^2}} \right)}^{\dfrac{1}{3}}}} \right)} \\ P &= 12\pi \int_0^r r \left( {6{{\left( {5 + {r^2}} \right)}^{\dfrac{1}{3}}}} \right)dr \cdots\cdots\rm{(I)} \end{align*} {/eq}

Let {eq}5 + {r^2} = v \cdots\cdots\rm{(II)} {/eq}

Substitute and solve the above expression with r = 0

{eq}\begin{align*} 5 + {0^2} &= v\\ v &= 5 \end{align*} {/eq}

Substitute and solve the expression (II) with {eq}r = 6 {/eq}

{eq}\begin{align*} 5 + {\left( 6 \right)^2} &= v\\ v &= 41 \end{align*} {/eq}

The domain is

{eq}5 \le v \le 41 {/eq}

Differentiate the expression (II)

{eq}\begin{align*} \dfrac{{d\left( {5 + {r^2}} \right)}}{{dr}} &= \dfrac{{dv}}{{dr}}\\ 2rdr &= dv\\ dr &= \dfrac{{dv}}{{2r}} \end{align*} {/eq}

Substitute and solve the expression (I)

{eq}\begin{align*} P &= 12\pi \int_5^{41} {r\left( {\dfrac{{dv}}{{2r}}} \right)\left( {{{\left( v \right)}^{\dfrac{1}{3}}}} \right)} \\ &= 6\pi \int_5^{41} {{{\left( v \right)}^{\dfrac{1}{3}}}} dv\\ &= 6\pi \left[ {\dfrac{3}{4}\left( {{v^{\dfrac{4}{3}}}} \right)} \right]_5^{41}\\ &= \dfrac{{18\pi }}{4}\left[ {{{\left( {41} \right)}^{\dfrac{4}{3}}} - {{\left( 5 \right)}^{\dfrac{4}{3}}}} \right]\\ &= 1877.792\;{\rm{thousands}} \end{align*} {/eq}

Thus the population within the radius of city center is {eq}1877.792\;{\rm{thousands}} {/eq}