How Does Simpson's Rule Estimate the Volume of an Artificial Lake?

[Pete1]

An artificial lake is made up of 5m width and 100m length in dimension. The depth of the lake varies every 20m length as recorded in the following table. Use Simpson's rule approximation to estimate the volume of the water in the lake.

Distance (m) 0 20 40 60 80 100 120
Depth (m) 2.0 1.6 1.8 2.1 2.1 1.9 2.2

 

[HOI]

Okay, what difficulty are you having? The problem says "use Simpson's rule". Do you know what "Simpson's rule" is? Simpson's rule is a numerical method for approximating an integral:
$$\int_a^b f(x)dx= \frac{b- a}{3n}\left(y_0+ 4y_1+ 2y_2+ \cdot\cdot\cdot+ 2y_{n-2}+ 4y_{n-1}+ y_n\right)$$
The interval from a to b is divided into n sub-intervals, each of length (b- a)/n. Each $$y_i$$ is the value of the function at $$x_i$$, the endpoint of the ith sub-interval.
Note that coefficients "4" and "2" alternate except for the first and last terms. And that n must be even (an integer number of "4" and "2" terms plus the first and last) so that the number of points must be odd (we start counting with "0"). In this problem we are given 7 points so y values will be labeled $$y_0$$, $$y_1$$, $$y_2$$, $$y_3$$, $$y_4$$, $$y_5$$, and $$y_6$$.

Or is the difficulty setting up the integral? You are told that the lake has width 5 m and length 100 m. You are given the depth at 20 m intervals of the length. Nothing is said about the depth along the width so the only thing you can do is assume it is constant along the width. You are told, for example, that the depth at 40 m is 1.8 meters. The width is 5 m and the distance to the next given depth, at 60 m, is 20m. That is a 1.8 by 5 by 20 rectangular solid so has volume (1.8)(5)(20)= 180. That is your "$$y_2$$". Calculate $$y_0$$ as (2.0)(5)(20), $$y_1$$ as (1.6)(5)(20), $$y_2$$ as (1.8)(5)(20), $$y_3$$as (2.1)(5)(20), $$y_4$$ as (2.1)(5)(20), $$y_5$$ as (1.9)(5)(20), and $$y_6$$ as (2.2)(5)(20).