Chapter 9, Problem 1P

The profile of a gear tooth shown in Fig. P9.1 (a) is approximated by the quadratic equation $y\left(x\right)=-\frac{4k}{l}x\left(x-l\right)$.

(a) Estimate the area A using six rectangles of equal width $\left(\Delta x=l/6\right)$, as shown in Fig. 9.1(b).

(b) Calculate the exact area by evaluating the definite integral, $A={\displaystyle \underset{0}{\overset{l}{\int }}y\left(x\right)dx}$.

To determine

(a)

The estimated area of given function $y\left(x\right)$ by rectangle method.

Answer to Problem 1P

The estimated area of given function by rectangle method is $0.643k{l}^2$.

Explanation of Solution

Given:

The expression of profile of gear tooth is given as:

  $y\left(x\right)=-\frac{4k}{l}x\left(x-l\right)$ ..... (1)

Number of rectangles given are 6.

Width of the rectangle is given as:

  $\Delta x=\frac{l}{6}$

Concept used:

The expression of area is given as:

  $A={\displaystyle \sum _{i=1}^{n}y\left(x_i\right)}\Delta x$ ..... (2)

Here, $\Delta x$ is the width of rectangle, $n$ are the number of rectangles and $y\left(x_i\right)$ is the value of given function at $x_i$.

Calculation:

For $x_1=\frac{l}{6}$ ,

Substitute $\frac{l}{6}$ for $x$ in equation (1).

  $\begin{array}{c}y\left(\frac{l}{6}\right)=-\frac{4k}{l}\left(\frac{l}{6}\right)\left(\frac{l}{6}-l\right)\\ =-\frac{4k}{l}\left(\frac{l}{6}\right)\left(\frac{-5l}{6}\right)\\ =0.55kl\end{array}$

Therefore,

  $y\left(x_1\right)=0.55kl$

For $x_2=\frac{l}{3}$ ,

Substitute $\frac{l}{3}$ for $x$ in equation (1).

  $\begin{array}{c}y\left(\frac{l}{3}\right)=-\frac{4k}{l}\left(\frac{l}{3}\right)\left(\frac{l}{3}-l\right)\\ =-\frac{4k}{l}\left(\frac{l}{3}\right)\left(\frac{-2l}{3}\right)\\ =0.88kl\end{array}$

Therefore,

  $y\left(x_2\right)=0.88kl$

For $x_3=\frac{l}{2}$ ,

Substitute $\frac{l}{2}$ for $x$ in equation (1).

  $\begin{array}{c}y\left(\frac{l}{2}\right)=-\frac{4k}{l}\left(\frac{l}{2}\right)\left(\frac{l}{2}-l\right)\\ =-\frac{4k}{l}\left(\frac{l}{2}\right)\left(\frac{-l}{2}\right)\\ =kl\end{array}$

Therefore,

  $y\left(x_3\right)=kl$

For $x_4=\frac{2l}{3}$ ,

Substitute $\frac{2l}{3}$ for $x$ in equation (1).

  $\begin{array}{c}y\left(\frac{2l}{3}\right)=-\frac{4k}{l}\left(\frac{2l}{3}\right)\left(\frac{2l}{3}-l\right)\\ =-\frac{4k}{l}\left(\frac{2l}{3}\right)\left(\frac{-l}{3}\right)\\ =0.88kl\end{array}$

Therefore,

  $y\left(x_4\right)=0.88kl$

For $x_5=\frac{5l}{6}$ ,

Substitute $\frac{5l}{6}$ for $x$ in equation (1).

  $\begin{array}{c}y\left(\frac{5l}{6}\right)=-\frac{4k}{l}\left(\frac{5l}{6}\right)\left(\frac{5l}{6}-l\right)\\ =-\frac{4k}{l}\left(\frac{5l}{6}\right)\left(\frac{-l}{6}\right)\\ =0.55kl\end{array}$

Therefore,

  $y\left(x_5\right)=0.55kl$

  $x_6=l$

Substitute $l$ for $x$ in equation (1).

  $\begin{array}{c}y\left(l\right)=-\frac{4k}{l}\left(l\right)\left(l-l\right)\\ =-\frac{4k}{l}\left(l\right)\left(0\right)\\ =0\end{array}$

Therefore,

  $y\left(x_6\right)=0$

Substitute $6$ for $n$ and $\frac{l}{6}$ for $\Delta x$ in equation (2).

  $A={\displaystyle \sum _{i=1}^{6}y\left(x_i\right)}\frac{l}{6}$ ..... (3)

Equation (3) can be written as:

  $A=\frac{l}{6}\left(\begin{array}{l}y\left(x_1\right)+y\left(x_2\right)+y\left(x_3\right)\\ +y\left(x_4\right)+y\left(x_5\right)+y\left(x_6\right)\end{array}\right)$ ..... (4)

Substitute $0.55kl$ for $y\left(x_1\right)$, $0.88kl$ for $y\left(x_2\right)$, $kl$ for $y\left(x_3\right)$, $0.88kl$ for $y\left(x_4\right)$, $0.55kl$ for $y\left(x_5\right)$, $0$ for in equation (4).

  $\begin{array}{c}A=\frac{l}{6}\left(\begin{array}{l}0.55kl+0.88kl+kl+0.88kl\\ +0.55kl+0\end{array}\right)\\ =\frac{l}{6}\left(3.86kl\right)\\ =0.643k{l}^2\end{array}$

Conclusion:

Thus, the estimated area of the given function by rectangle method is $0.643k{l}^2$.

To determine

(b)

The exact area of given function $y\left(x\right)$ by integration method.

Answer to Problem 1P

The exact area of given function by integration method is $0.67k{l}^2$.

Explanation of Solution

Concept used:

The expression of area is given as:

  $A={\displaystyle {\int }_a^{b}y\left(x\right)dx}$ ..... (5)

Here, $y\left(x\right)$ is the given function.

  $a$, $b$ are the lower and upper limits.

Calculation:

Substitute $0$ for $a$, $l$ for $b$ and $-\frac{4k}{l}x\left(x-l\right)$ for $y\left(x\right)$ in equation (5).

  $\begin{array}{c}A={\displaystyle {\int }_0^l\left(-\frac{4k}{l}x\left(x-l\right)\right)dx}\\ ={\displaystyle {\int }_0^l\left(-\frac{4k}{l}{x}^2+4kx\right)dx}\\ ={\left(-\frac{4k}{l}\left(\frac{x^3}{3}\right)+4k\left(\frac{x^2}{2}\right)\right)}_0^l\\ ={\left(-\frac{4k{x}^3}{3l}+2k{x}^2\right)}_0^l\end{array}$

Further simplify for $A$.

  $\begin{array}{c}A=\left(-\frac{4k{\left(l\right)}^3}{3l}+2k{\left(l\right)}^2\right)-\left(-\frac{4k{\left(0\right)}^3}{3l}-2k{\left(0\right)}^2\right)\\ =\left(-\frac{4k{l}^2}{3}+2k{l}^2\right)-\left(0\right)\\ =\left(\frac{-4k{l}^2+6k{l}^2}{3}\right)\\ =0.67k{l}^2\end{array}$

Conclusion:

Thus, the exact area of given function by integration method is $0.67k{l}^2$.