Estimating Area of Kidney-Shaped Pool Using Simpson's Rule

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Homework Help Overview

The problem involves estimating the area of a kidney-shaped swimming pool using Simpson's Rule, based on measured widths at 2-meter intervals. The participants are tasked with applying the rule to a set of given width measurements.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the application of Simpson's Rule, with some attempting to calculate the area using the provided widths. Questions arise regarding the correct grouping of intervals and the handling of the midpoint values.

Discussion Status

There are various attempts to apply Simpson's Rule, with some participants expressing confusion over the formula and the number of points provided. Suggestions for grouping the intervals differently are being explored, and there is acknowledgment of ambiguity in the problem setup.

Contextual Notes

Participants note the odd number of width measurements and question whether to consider the ends of the pool where the width is zero. There is also mention of the requirement to round the final area to the nearest square meter.

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Homework Statement



The widths (in meters) of a kidney-shaped swimming pool were measured at 2 meter intervals as indicated in the figure. Use Simpson's Rule to estimate the area of the pool. Please round the answer to the nearest square meter.

7-7-030alt.gif


a = 5.8
b = 6.6
c = 5.4
d = 5.8
e = 5.2
f = 4.8
g = 5.2


Homework Equations



I know that the general formula for Simpson's rule is [tex]\frac{b-a}{6}[/tex][f(a)+4f[tex]\frac{a+b}{2}[/tex]+f(b)]

Which is applicable over n subintervals

3. The Attempt at a Solution [/b

My attempt was to do:

[tex]\frac{2}{6}[/tex][5.8+4[tex]\frac{5.8+6.6}{2}[/tex]+6.6] + [tex]\frac{2}{6}[/tex][6.6+4[tex]\frac{6.6+5.4}{2}[/tex]+5.4] +
...
[tex]\frac{2}{6}[/tex][4.8+4[tex]\frac{4.8+5.2}{2}[/tex]+5.2]

But that did was not right.

My gripe is that the question does not allow me to find the function of [tex]\frac{a+b}{2}[/tex]


Does anyone have any suggestions on how to tackle this problem? Thanks
 

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bns1201 said:

Homework Statement



The widths (in meters) of a kidney-shaped swimming pool were measured at 2 meter intervals as indicated in the figure. Use Simpson's Rule to estimate the area of the pool. Please round the answer to the nearest square meter.

a = 5.8
b = 6.6
c = 5.4
d = 5.8
e = 5.2
f = 4.8
g = 5.2

Homework Equations



I know that the general formula for Simpson's rule is [tex]\frac{b-a}{6}[/tex][f(a)+4f[tex]\frac{a+b}{2}[/tex]+f(b)]

Which is applicable over n subintervals No, it's over two intervals [a, (a+b)/2] and [(a+b)/2, b], in practise.[/color]

Does anyone have any suggestions on how to tackle this problem?
Denote f(x0)=a=5.8, f(x1)=b, f(x2)=c, ..., f(x6)=g.

The formula you stated (let me replace a and b with x and y, respectively, to avoid confusion with your problem) requires values of f at x, y and their midpoint (x+y)/2, so to apply it to the problem one would group the intervals as {x0, x1, x2}, {x2, x3, x4}, {x4, x5, x6}, and apply the formula to each, then sum.

Note that in doing so you'll essentially derive Simpson's rule for a function tabulated at n points.
 
i did this, but the answer is not working...i'm getting (b-a)/6 is 4/6

any other suggestions to this because I am having trouble. thanks
I don't think it makes a difference but the answer is asked for in meters squared
 
anyone have suggestions?
 
What you supposed to get (if you know)? The problem is a little ambiguous. For one, there are an odd number of points given and for another are you supposed to consider the ends of the pool where the width is zero and assume they are 2m from points a and g? One way to do it is split the seven points plus the two extra points into 3 groups of 3 points. But it's hard to know what they expect.
 
In this case the pool is 16 m wide and there are 8 subdivisions so delta x equals 2.

f(x_0)=0 and f(x_n)=0 since the edges of the pool have no height.

the resulting formula is:

(2/3)(0 + 4f(a) + 2f(b) + 4f(c) + ... + 4f(g) + 0)
 

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