Chapter 6.1, Problem 28E

A cross-section of an airplane wing is shown. Measurements of the thickness of the wing, in centimeters, at 20-centimeter intervals are 5.8, 20.3, 26.7, 29.0, 27.6, 27.3, 23.8, 20.5, 15.1, 8.7, and 2.8. Use the Midpoint Rule to estimate the area of the wing’s cross-section.

To determine

To estimate: The area of cross section of airplane wing

Answer to Problem 28E

The area of cross section of airplane wing is $\underset{\_}{4,232{\text{cm}}^{\text{2}}}$.

Explanation of Solution

Given information:

The thickness is measured each 20 cm interval.

The general expression for Midpoint Rule is shown below:

${\displaystyle \underset{a}{\overset{b}{\int }}f\left(x\right)dx}\approx {\displaystyle \sum _{i=1}^{n}f\left({\overline{x}}_i\right)\Delta x}=\Delta x\left(f\left({\overline{x}}_1\right)+\mathrm{...}+f\left({\overline{x}}_n\right)\right)$

The interval width is $\Delta x=\frac{b-a}{n}$ (1)

The midpoint is  ${\overline{x}}_i=\frac{1}{2}\left(x_{i-1}+x_i\right)$ (2)

Here, width of the subinterval is $\Delta x$, the upper limit is b, the lower limit is a, the number of subinterval is n, and the midpoint of $\left(x_{i-1},x_i\right)$ is ${\overline{x}}_i$.

Calculation:

Find the width of the subinterval $\left(\Delta x\right)$ using Equation (1) as shown below.

Consider the number of subintervals as 5.

Substitute 5 for n, 200 cm for b, and 0 for a in Equation (1).

$\begin{array}{c}\Delta x=\frac{200-0}{5}\\ =40\text{cm}\end{array}$

Let the distance from the left end of the wing be x.

The height of the wing at x is $w$.

The expression to find the area of the pool for the width $x=0$ to $x=200\text{cm}$ with $n=5$ using midpoint rule is shown below.

$\begin{array}{c}\text{Area}={\displaystyle \underset{0}{\overset{200}{\int }}wdx}\\ =\Delta x\left[w\left({\overline{x}}_1\right)+w\left({\overline{x}}_2\right)+w\left({\overline{x}}_3\right)+w\left({\overline{x}}_4\right)+w\left({\overline{x}}_5\right)\right]\end{array}$ (3)

Here, the height of the wing at distance ${\overline{x}}_1$ is $w\left({\overline{x}}_1\right)$, the height of the wing at distance ${\overline{x}}_2$ is $w\left({\overline{x}}_2\right)$, the height of the wing at distance ${\overline{x}}_3$ is $w\left({\overline{x}}_3\right)$, the height of the wing at distance ${\overline{x}}_4$ is $w\left({\overline{x}}_4\right)$, and the height of the wing at distance ${\overline{x}}_5$ is $w\left({\overline{x}}_5\right)$.

The end points of the distance intervals are 0, 40 cm, 80 cm, 120 cm, 160 cm, and 200 cm.

The midpoints of the intervals are 20 cm, 60 cm, 100 cm, 140 cm, and 180 cm.

Substitute 40 cm for $\Delta x$, 20 cm for ${\overline{x}}_1$, 60 cm for ${\overline{x}}_2$, 100 cm for ${\overline{x}}_3$, 140 cm for ${\overline{x}}_4$, and 180 cm for ${\overline{x}}_5$ in Equation (3) as shown below.

$\text{Area}=40\left(w\left(20\right)+w\left(60\right)+w\left(100\right)+w\left(140\right)+w\left(180\right)\right)$ (4)

Substitute 20.3 cm for $w\left(20\right)$, 29.0 cm for $w\left(60\right)$, 27.3 cm for $w\left(100\right)$, 20.5 cm for $w\left(140\right)$, and 8.7 meter for $w\left(200\right)$ in Equation (4) as shown below.

$\begin{array}{c}\text{Area}=40\left(20.3+29.0+27.3+20.5+8.7\right)\\ =40\left(105.8\right)\\ =4,232{\text{cm}}^{\text{2}}\end{array}$

Therefore, the area of cross section of airplane wing is $\underset{\_}{4,232{\text{cm}}^{\text{2}}}$.