Chapter 6.5, Problem 30E

To determine

the length of the tank.

Answer to Problem 30E

The length of the tank is 10.63ft

Explanation of Solution

Given information:

The tank must now 5000 lb of gasoline, which has a density of 42 lb/fl³. Estimate the length of the tank.

  

Formula used:

  $S=\frac{h}{3}\left(y_0+4y+2y_2+4y_3+\mathrm{....}+2y_{n-2}+4y_{n-1}+y_n\right)$

Calculation:

To approximate ${\displaystyle \underset{a}{\overset{b}{\int }}f\left(x\right)}dx,$ use

  $S=\frac{h}{3}\left(y_0+4y+2y_2+4y_3+\mathrm{....}+2y_{n-2}+4y_{n-1}+y_n\right)$

Where [a, b] is partitioned into an even number n of subintervals of equal length $h=\frac{b-a}{n}$

Now,

Now, apply the Simpson’s rule with

  $\begin{array}{l}n=6\\ \left[a,b\right]=\left[0,6\right]\end{array}$

So,

  $\begin{array}{l}h=\frac{b-a}{n}\\ =\frac{6-0}{6}\\ h=1\end{array}$

  

Now, substitute the values from given table in this equation

  $\begin{array}{l}{\displaystyle \underset{0}{\overset{6}{\int }}f\left(x\right)}dx=\text{cross-section area of tank}\\ \text{=}\frac{h}{3}\left(y_0+4y_1+2y_2+4y_3+2y_4+4y_s+y_6\right)\\ =\frac{1}{3}\left(1.5+4\left(1.6\right)+2\left(1.8\right)+4\left(1.9\right)+2\left(2.0\right)+4\left(2.1\right)+2.1\right)\\ {\displaystyle \underset{0}{\overset{6}{\int }}f\left(x\right)}dx=11.2f{t}^2\end{array}$

Now, calculate the length of tank

  $\begin{array}{l}\text{length of tank}=\text{volume}\times \text{cross-section area}\\ \text{=}\frac{\text{mass}}{\text{density}}\times 11.2f{t}^2\\ =\frac{5000\text{lb}}{42{\text{lb/ft}}^3}\times 11.2{\text{ft}}^2\\ \text{length of tank}=10.63\text{ft}\end{array}$

Conclusion:

The length of the tank is 10.63ft