Chapter 15.3, Problem 29E

To determine

Find the volume of rock moved to make a tunnel $100feetlong$ .

Answer to Problem 29E

  $8000cubicfeet$

Explanation of Solution

Given information:

The vertical cross sections of a tunnel can be modeled by the parabola $y=6-0.06x^2$ . $xandy$ are measured in feet.

  

Calculation:

Here, we will calculate the area of parabolic tunnel:

  $\begin{array}{l}{\displaystyle \underset{-10}{\overset{10}{\int }}ydx}={\displaystyle \underset{-10}{\overset{10}{\int }}6-0.06x^2}\\ a=-10,b=10\\ \Delta x=\frac{10+10}{n}=\frac{20}{n}and{x}_i=-10+i\frac{20}{n}\end{array}$

  $\begin{array}{l}Threfore,\\ {\displaystyle \underset{-10}{\overset{10}{\int }}6-0.06x^2}=\underset{n\to \infty }{{\displaystyle \lim }}{\displaystyle \sum _{i=1}^n\left(6-0.06x_i{}^2\right)}\frac{20}{n}\end{array}$

  ${\displaystyle \underset{-10}{\overset{10}{\int }}6-0.06x^2}=\underset{n\to \infty }{{\displaystyle \lim }}{\displaystyle \sum _{i=1}^n\left(6-0.06{\left(-10+i\frac{20}{n}\right)}^2\right)}\frac{20}{n}$

  ${\displaystyle \underset{-10}{\overset{10}{\int }}6-0.06x^2}=\underset{n\to \infty }{{\displaystyle \lim }}{\displaystyle \sum _{i=1}^n\left(6-0.06\left(100-i\frac{400}{n}+i^2\frac{400}{n^2}\right)\right)}\frac{20}{n}$

  $\begin{array}{l}{\displaystyle \underset{-10}{\overset{10}{\int }}6-0.06x^2}=\underset{n\to \infty }{{\displaystyle \lim }}{\displaystyle \sum _{i=1}^n\left(6-6+24\frac{i}{n}-24\frac{i^2}{n^2}\right)}\frac{20}{n}\\ {\displaystyle \underset{-10}{\overset{10}{\int }}6-0.06x^2}=\underset{n\to \infty }{{\displaystyle \lim }}\frac{20}{n}{\displaystyle \sum _{i=1}^n\left(\frac{24i}{n}-\frac{24i^2}{n^2}\right)}\end{array}$

  $\begin{array}{l}{\displaystyle \underset{-10}{\overset{10}{\int }}6-0.06x^2}=\underset{n\to \infty }{{\displaystyle \lim }}\frac{20}{n^3}{\displaystyle \sum _{i=1}^n\left(24i{n}^2-24i^2\right)}\\ {\displaystyle \underset{-10}{\overset{10}{\int }}6-0.06x^2}=\underset{n\to \infty }{{\displaystyle \lim }}\frac{20}{n^3}24\left(n^2{\displaystyle \sum _i^{n}i}-{\displaystyle \sum _i^n{i}^2}\right)\end{array}$

  ${\displaystyle \underset{-10}{\overset{10}{\int }}6-0.06x^2}=\underset{n\to \infty }{{\displaystyle \lim }}\frac{20}{n^3}24\left(n^2\left(\frac{n\left(n+1\right)}{2}\right)-\left(\frac{n\left(n+1\right)\left(2n+1\right)}{6}\right)\right)$

  ${\displaystyle \underset{-10}{\overset{10}{\int }}6-0.06x^2}=\underset{n\to \infty }{{\displaystyle \lim }}\frac{20}{n^3}24\left(n^3\left(\frac{1+1/n}{2}\right)-n^3\left(\frac{\left(1+1/n\right)\left(2+1/n\right)}{6}\right)\right)$

  $\begin{array}{l}{\displaystyle \underset{-10}{\overset{10}{\int }}6-0.06x^2}=480\left(\frac{1}{2}-\frac{1}{3}\right)\\ {\displaystyle \underset{-10}{\overset{10}{\int }}6-0.06x^2}=80\end{array}$

  $\begin{array}{l}\therefore Volume=area\times length\\ Volume=80\times 100\\ Volume=8000\end{array}$

Hence, the volume is $8000cubicfeet$ .