Chapter 6.5, Problem 29E

To determine

To calculate: The cross section area using Simpson’s Rule.

Answer to Problem 29E

The cross section area is $466.67{\text{inch}}^2$ .

Explanation of Solution

Given information:

The function:

  

Calculation:

Simpson's rule is used to find the value of integral.

To approximate ${{\displaystyle \int }}_a^{b}f(x)dx,$ , use

  $S=\frac{h}{3}\left(y_0+4y_1+2y_2+4y_3+\dots .+2y_{n-2}+4y_{n-1}+y_n\right)$ .

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

Now, apply the Simpson's rule with

  $\begin{array}{l}n=6\hfill \\ \left[a,b\right]=\left[0,24\right]\hfill \end{array}$

So,

  $\begin{array}{cc}h& =\frac{b-a}{n}\\ & =\frac{24-0}{6}\\ & =4\end{array}$

$\begin{array}{cc}h& =\frac{b-a}{n}\\ & =\frac{24-0}{6}\\ & =4\end{array}$ . x		$y=f\left(x\right)$
$x_0$	0	$y_0$	0
$x_1$	4	$y_1$	18.75
$x_2$	8	$y_2$	24
$x_3$	12	$y_3$	26
$x_4$	16	$y_4$	24
$x_5$	20	$y_5$	18.75
$x_6$	24	$y_6$	0

Now, substitute the values from given table in this equation-

  ${{\displaystyle \int }}_0^{24}f(x)dx=$ cross-section area of body

  $\begin{array}{l}=\frac{h}{3}\left(y_0+4y_1+2y_2+4y_3+2y_4+4y_5+y_6\right)\hfill \\ =\frac{4}{3}(0+4(18.75)+2(24)+4(26)+2(24)+4(18.75)+0)\hfill \\ \approx 466.67{\text{inch}}^2\hfill \end{array}$