Theodore R. (Theodore Rudolph) Running.

Empirical formulas online

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INTERPOLATION

Let y^ = A(x-b){x-c)(x-d) . .

. {x-k)

+B(x-a)(x-c)(x-d) . .

. (x-k)

+C{x-a){x-b)(x-d) . .

. (x-k)

+D{x-a){x-b){x-c) . .

. (x-k)

+ etc. to n terms.

105

Each one of the n terms on the right-hand side of the equation
lacks one of the factors x â€” a, x â€” b, x â€” c, xâ€”d, . . . xâ€”k,
and each is affected with an arbitrary constant. The expression
on the right-hand side of the equation is a rational integral
function of x.

Letting x = a gives

and

ya = A{a â€” b)(aâ€”c)(aâ€”d) . . . aâ€”k,

ya

(a â€” b)(a â€” c){a â€” d) . . . a â€” k'
Letting x = b gives
B =

Jb

{b-a){b-c){b-d) . . . {b-k)'

Proceeding in the same way we obtain values for all of the
constants and, finally,

{x â€” b){x â€” c){x â€” d) . . . {x â€” k)

yx=ya

+yi

{a â€” b){a â€” c){a â€” d)
{x â€” a){x â€” c)(x â€” d)

+yc

{b-a){b-c){b-d) .
{x â€” a){x â€” b)(x â€” d) .

+yd

{câ€”a){câ€”b){câ€”d) .
{x â€” a){x â€” b){x â€” c) .

{d-a){d-b){d-c) .

{x â€” a)(x â€” b)(x â€” c) .
'^^\k-a){k-b){k-c) .

. {a-k)
. {x â€” k)

. (b-k)

. {x â€” k)

. ic-k)

. {x â€” k)

. {d-k)

(2)

106

EMPIRICAL FORMULAS

This is called Lagrange's theorem for interpolation.

I. Apply formula (2) to the data given under formula XIX
for finding the value of y corresponding to it: = 0.425. Select
two values on either side of the value required,

6 = .40, yi> =
X in the formula must be taken as o

>' = (-695)

+(.757)
= .744.

i(-i)(-t)

(-i)(-2)(-3)

a)(Â§)(-f)

730,

757,
780.

5-

â– f(.73o)

(2)(i)(-i)

+ (.780)

t(-|)(-f)

(l)(-l)(-2)

(3)(2)(i)

2. Required an approximate value of log 212 from the fol-
lowing data:

log 210 = 2.3222193,

log 211 = 2.3242825,

log 213 = 2.3283796,

log 2 14 = 2.3304138.

(-i)(-3)(-4)^^ ^^ ^\i)(-2)(-3)

log 212 = (2.3222193)-

+ (2.3283796)

(2)(l)(-2)

+ (2.3304138)

(2)(l)(-l )

(4)(3)(i)'

(3)(2)(-i)
= 2.326359.

This is correct to the last figure.

In case the values given are periodic it is better to use a
formula involving circular functions. In Chapter V the approxi-
mate values of the constants in formula XX were derived. This
formula could be used as an interpolation formula. But on
account of the work involved in determining the constants it is

INTERPOLATION

107

much more convenient to use an equivalent one which does not
necessitate the determination of constants.* The equivalent
formula given by Gauss is

y^=y

^-y\

+3^-

+ etc.

sin^(x â€” 6) sin |(x â€” c) . .

. sm\{x-k)

"sin J(a â€” 5) sin^(a â€” c) . .
sin |(a: â€” a) sin Â§(x â€” c) .

. sin ^{a â€” k)
. sin ^{x â€” k)

sin Â§(6 â€” a) sin \{b â€” c) .
sin \{x â€” (i) sin \{x â€” 'b) .

. sin^(x â€” ^)

" sin \{c â€” a) sin \{c â€” h) .

. . dn^{c-k)

(3)

It is evident that the value of ja is obtained from this formula
by putting x = a. The value of Jd is obtained by putting x = b,
and jc by putting x = c.

The proof that (3) is equivalent to XX need not be given
here.

Let it be required to find an approximate value of y cor-
responding to it: = 42Â° from the values given.

From (3)
>' = (io.i)

X

y

30Â°

10. 1

40"

9.8

50"

8.5

sin iÂ°sin (^4Â°)
sin(-sÂ°)sin(-ioÂ°)

f(9-8)

+ (8.5)

sin 6Â° sin ( â€” 4Â°)
sin 5Â° sin (-5Â°)
sin 6Â° sin 1Â°
sin 10Â° sin 5Â°

- rrn T^i:^II5)C^698) , . (.1 045) (.0698)

(.1045) (.01 75)

(.i736)(.o872)
= 9.618.

+ (8.5)

* Trigometrische Interpolation, Encyklopadie der Mathematischen
Wissenchaften, Vol. II, pt. I, pp. 642-693.

108 EMPIRICAL FORMULAS

A better result would have been obtained by using four sets
of values.

Differentiation of Tabulated Functions

It is frequently desirable to obtain the first and second
derivatives of a tabulated function to a closer approximation
than graphical methods will yield. For that purpose we will
derive differentiation formulas from (i) and (2). From

^ x{x-i){x-2){x-s) ^^^^^^

k

By differentiating it follows that

, ^ , 2:1:â€” 1 .0 , sx^ â€” 6x+2^^
yx=Ayo-\ â€” â€”A^yo+^ â€” I A^yo

^ 4^-i2.M-...-6 ^,^^^ (^)

Differentiating again
yJ'=A^yo+(x-i)A^yo+(ix^-x+H)A'yo+ ..... . . (s)

As an illustration let it be required to find the first and
second derivatives of the function given in the table below and
determine whether the series of observations is periodic*

The consecutive daily observations of a function being
0.099833, 0.208460, 0.314566, 0.416871, 0.514136, 0.605186,
0.688921, 0.764329, show that the function is periodic and deter-
mine its period.

* Interpolation and Numerical Integration, by David Gibb.

INTERPOLATION

109

From the given observations the following table may be
ttpn !

written:

y=f(x)

0-099833
0.208460
0.314566
0.416871
0.514136

6 0.605186

7 0.688921

8 0.764329

0.108627
0.106106
0.102305
(-)o.097265
(-)o.o9io5o
(-)o.o83735
(-)o.o754o8

From (4)

y\= .108627
.001260

. 109887
.000437

. 109450

â€” .000427
â€” .000010

-.000437

y3= .102305 -.000392
.002520 â€” .000019

.104825
.000411

. 104414

.000411

.002521
.003801
.005040
.006215

.007315

.008327

^2

A'

â€” .001280
( + ) -.001239

(+)-.ooii75
(+) â€” .001100
(+) â€” .001012

A*

.000041
.000064
.000075
.000088

.106106
.001900

. 108006
â€” .000429

.107577

.097265
.003108

.100373
â€” .000389

.099984

â€” .000413
â€” .000016

â€” .000429

.000367
.000022

.000389

For the remaining first derivatives the order must be reversed
and the resulting sign changed.

^5=- .097265 .002520 ^6= â€” .091050 .003108

,000010 .000413 â€” .000016 .000392

097275
002933

002933

094342

â€” .091066
â€˘003500

.087566

â€˘003500

110

EMPIRICAL FORMULAS

/7=". 083735
,000019

083754
004025

,003658
.000367

004025

â€” .006279
.000038

â€” .006241

y'7=-. 007315
â€” .001100

â€” .008415
.000069

â€” . 008346

y8=-.o754o8
â€” .000022

.079729

From (5)

y'i = -. 002521
.001318

.001280
.000038

.001318

.001175

.000069
.001244

y'2

â€” .001203

/';,= -. 005040
.001244

y'4

-.003796

y"r, = - . 005040
-.001239

y\

-075430

.004501
.070929

y 8

003801
001298

002503

006215
001181

005034

006215
001175

007390
000059

007331

008327
001012

009339
000081

009258

.004164
,000337

004501

.001239
,000059

001298

OOIIOO

,000081
,001181

X

y

y

y"

yi

y

I

.099833

. 109450

â€” .001203

â€” .0121

2

. 208460

.107577

â€” .002503

â€” .0120

3

.314566

. 104414

-.003796

â€” .0121

4

.416871

.099984

-.005034

â€” .0121

INTERPOLATION

111

x

y

/

/'

y

5

.514136

â€˘094342

.006241

â€” .0121

6

.605186

.087566

â€˘007331

â€” .0121

7

.688921

.079729

.008346

â€” .0121

8

.764329

.070929 â€”

.009258

â€” .0121

Since â€” is very nearly constant and equal to â€”.0121, the

y

corresponding differential equation is

y'^-{-.oi2iy = o,
whose solution is

;y = ^ coso.iix+^sino.iix.
This shows that y is a, period function of x, and its period is

27r J

, or 57.12 days.

o.ii

Convenient formulas for the first and second derivatives may
also be obtained by differentiating Lagrange's formula for inter-
polation. Using five points the formula is

_ (x â€” b){xâ€”c)(xâ€”d)(x â€” e) (x â€” a)(xâ€”c)(xâ€”d)(xâ€”e)
^"~^" (a-b){a-c)(a-d)(a-ey^' {b-a){b-c){b-d){h-e)

{x â€” a){x â€” b){x â€” d){x â€” e) {x â€” a){x â€” b){x â€” c){x â€” e)
â– ^^^ {c-a){c-b){c-d){c-e) "^^' {d-a){d-b){d-c){d-e)

{x â€” a){x â€” b){x â€” c){x â€” d) . .

'^^' {e-a){e-b){e-c){e-d) ^^^

Selecting the points at equal intervals and letting

eâ€”d = dâ€”c = c â€” b = b â€” a = hj

and differentiation

ya=-^l-2sya+4^yb-3^yc-\-i6ya-sye],

y'h = â€” rl- 3>'a-io>+i8>'c- 63;^+ y^,
\2h

112

EMPIRICAL FORMULAS

yc = ^[ ya- ^yi> + 8>'d-3'e],

yd = ^[- >'a+ 6>-i83;c+io>'d+3>'e],

Differentiating again

y"a = j;^[35>'a - 1043^6+ 1 i4>'c - 56>'d+ 1 1}'.],

/'&= â€” jrh^^ya- 2o>'ft+ 63^0+ 4yd-ye],

12nr
y'c = ^[->'a+ l6>- 3OJC+ l63;d->'e],

yd = â€” pl-^^ 43^*+ 63'c- 20>'d+II>],

3''e = -^[ii3'a- 56>'6+ii4>'c-i04>'d+35>'J-

The results of applying these formulas to the function given
are expressed in the table below.

X

y

/

r

I

â€˘099833

.109451

.001203

2

. 208460

â€˘ 107583

002524

3

.314566

.104415 -

.003804

4

.416871

.099986 -

.005045

5

â€˘514136

â€˘094347

.006221

6

.605186

.087568 -

007322

7

.688921

â€˘079733

008334

8

.764329

^A r ^ ? 1

.070929

009258
1 1

These results agree fairly well with those previously obtained.
It is probable that the formulas derived from the interpolation
formula give the most satisfactory results.

INTERPOLATION

113

As another application let us find the maximum or minimum
value of a function having given three values near the critical
point.

Let ya, jb, and jc be three values of a function of x near
its maximum or minimum corresponding to the values of Xy
a, b, and c respectively.
From (2)

_ (x â€” b)(xâ€”c) (x â€” a)(x â€” c) (x â€” a)(x â€” b)
^'"^" (a-b){a-cy^\b-a){b-cy^' {c-a){c-b)'

Equating to zero the first derivative with respect to x

2x â€” a â€” b

2X â€” b â€” c

y'^ 7 â€” TVT â€” ^+>

2x â€” a-

^yc

{a-b){a-c) ' '" {b-a){b-c) ' â– " {c-a){c-b)
_ yajb^ - c^) +>(c^ - a^) -\-ycia^ - b^)

=0;

(6)

2[ya{b-c)-\-yb{c-a)-\-yc{a-b)]
This is equivalent to drawing the parabola

y=A+Bx-\-Cx^

through the three points and determining its maximum or
minimum.

From the table of values

6.0

6.5
7.0

10.05
10.14
10.10

the abscissa of the maximum point is found from (6).

(io.o5)(-6.75) + (io-i4)(i3) + (iaio)(-6.25) ^
2[(io.o5)(-.5) + (io.i4)(i) + (io.io)(-.5) ''â€˘^^''

3; = 10.1424.

CHAPTER VIII

NUMERICAL INTEGRATION

Areas

An area bounded by the curve, y=J{x), the axis of x, and
two given ordinates is represented by the definite integral

r

ydx,

where the ordinates are taken at :r = a and x = n. It may be
said that the definite integral represents the area under the
curve, or that the area under the curve represents the value of
the definite integral.

If a function is given by its graph, it is possible, by means
of the planimeter, to find roughly the area bounded by the curve,
two given ordinates and the x â€” axis, or, what amounts to the
same thing, the area enclosed by a curve. This method is used
in finding the area of the indicator diagrams of steam, gas or
oil engines, and various other diagrams. The approximations
in these cases are close enough to satisfy the requirements.

If, however, considerable accuracy is sought, or whenever
the function is defined by a table of numerical values another
method must be employed.

Mechanical Quadrature or Numerical Integration is the method
of evaluating the definite integral of a function when the func-
tion is given by a series of numerical values. Even when the
function is defined by an analytical expression but which can-
not be integrated in terms of known functions by the method
of the integral calculus, numerical integration must be resorted
to for its evaluation.

The formulas employed in numerical integration are derived
from those established for interpolation.

114

NUMERICAL INTEGRATION 115

In interpolation it was found that the order of differences
which must be taken into account depends upon the rapidity
with which the differences decrease as the order increases.
This is also true of numerical integration. It is the same as
saying that if the series employed does not converge the process
will give unsatisfactory results. An illustration will be given
later.

Formulas for numerical integration will be derived from (i)
of Chapter VII.

In this formula it was assumed that the ordinates are given
at equal intervals.

, ^ .x(xâ€”l).o , x{xâ€”l){x â€” 2) .o

yx=yo-{-xAyo+^ -A^yo+ â€” , ^A^yo

11 11

^ x(x-i){x-2)(x-s) ^. ^ x{x-i)(x-2)(x-s){x-4) ^.^

^ x(x-i){x-2){x-s){x-4){x-s) ^, J ^j>j

[6

Integrating the right-hand member,
\ ydx=yQ I dx-\-Ayo\ xdx-\ â€” p- | x{x â€” i)dx

Jo Jo \2_J

+^Vx{x-l){x-2)dx

b Jo

^^ C\{^^-^){0c-2){x-^)dx

k Jo

H-^ \x{x-i){x-2){x-i){x-4)dx
15 Jo

\ x{x-i){x-2){x-^){x-4){x-s)dx-\- . . .

=nyo-\ â€” Ayo+ -r^+ n^-^n^ ) - -

2 \3 2/ |2 V4 / |3

â– "|6

\5 2 3 /

A^o

k

116 EMPIRICAL FORMULAS

\6 4 3 / |5

\7 2 ' 4 3 / P

The data given in any particular problem will enable us
to compute the successive differences of ^'o up to A">'o. On the
assumption that all succeeding differences are so small as to
be negligible the above formula gives an approximate value
of the integral. It is only necessary to assign particular values
tow.

Let w = 2, then

I yxdx = 2yo+2Ayo+iA^yoj

Ayo=yi-yo,

A^yo = Ayi - Ayo = y2-yi-yi +yo,
=y2-2yi+yo.

Substituting these values in the above integral it becomes

I y2dx = 2yo-]-2yi-2yo+iy2-iyi+^yo,
_yo+4yi+y2

This is equivalent to assuming that the curve coincides with
a parabola of the second degree.

If the common distance between the ordinates is h, the
value becomes

'2h

ydx=U(yo+4yi+y2) (7)

X

IfÂ»=3

ydx = syo+^Ayo+iA^yo+i^^yoj

Ayo=yi-yo,

A^yo = Ayi-Ayo=y2- 2yi +yoy

i

NUMERICAL INTEGRATION 117

=yz-2>y2+2>yi-yo'

Substituting these values in the equations,

1 ydx = 7^yo-\-^yi-^yQ+ly2-hi+ho-\-h^-h2+hi-hoj

= l>'0 + ljl+f3'2+|>'3,

= l(3'o+33'i+3>'2+3'3).

If the common distance between the ordinates is h the
formula becomes

â€˘3ft

X

= iKyQ-\-^yi-{-2>y2+y^) (8)

This is equivalent to assuming that the curve coincides with
a parabola of the third degree.

If there are five equidistant ordinates, h representing the
distance between successive ordinates

/â€˘4ft

Jo ^

yd^ - 14(3^0+^4) +64(3^1 +^3) +24y2 ^^^ ... (9)

b 45

If the area is divided into six parts bounded by seven equi-
distant ordinates the integral becomes

i

6

ydx = 6>'o + 1 ^^yo + 2 7 A^^yo + 24 A^^yo + ^-L^yo
+ffA5>'o+AVA63;o.

Sine 3 the last coefficient, 1^, differs but slightly from t^
and by the assumption that A^^'o is small the error will be slight
if the last coefficient is replaced by tq-

Doing this and replacing

^yohyyi-yo,
^^yohyy2-2yi+yQ,
A^yo by yd-;^y2+^yi -yo,

118

EMPIRICAL FORMULAS

A*yo by yA-4ys+6y2 -4yi+yo,
A^yo by y5-5>'4H-ioy3-io>'2+53'i-3'o,
A^^'o by / - 6>'5 -f- 1 5>'4 - 2oy3 + 1 53'2 - 6yi +yo,
gives the formula

X

ydx = TV/b'o+>'2+>'4+3'6+5(>'i +>'5) +6>'3]. . (lo)

The application of these
formulas is illustrated by
finding the area in Fig. 27.

Fig. 27.

By (7)

A=ih(yo+4yi + 2y2-\-4y3 + 2y4:+4y5+2ye+4y7+2y8+4y9

-\-2yio+4yn+yi2)'

By (8)

^=l^(yo+3>'i+3>'2 + 2y3+3>'4+3>'5+2y6+3>'7+33'8+2y9

+3>'io+33'ii+>'i2).

By (9)

A=-hhli4(yo+2y^-[-2y8+yi2)+64(yi-\-y3+y5+y7+y9+yii)

+ 24(y2+yQ-\-yio)].
By (10)

A=^hlyo+y2+y4.+2y6-\-y8+yio-\-yi2+5(yi+y5+y7+yii)
+6(^3 +^9)].. .

I. A rough comparison of the approximations by the use
of these formulas will be obtained by finding the value of

'ax

Jn
â€” . The value of this definite integral is log 13 = 2.565. It
I X

is also equal to the area under the curve

I

y=-

X

NUMERICAL INTEGRATION 119

from x = i to a: -=13. Dividing the area up into 12 strips of
unit width by 13 ordinates the corresponding values of x and
y are

X

I

2

3

4

5

6

7

8

9

10

II

12

13

y

I

1

2

i

i

i

i

1

T

1

i

A

tV

A

tV

By (7)

^ =Mi+2+i+i+f +1 +f +Ht+f +A+i+Ty

= 2.578, error .5%;

By (8), ^=2.585, error .8%;

By (9), ^ = 2.573, error .3%;

By (10), A = 2.572, error .3%.

2. The accuracy of the approximation is much increased
by taking the ordinates nearer together, as is shown by the
following evaluation of

^^ dx^

h 1+^

X'

The value of this integral is equal to the area under the
curve

I

i-\-x

from x=o tG x = i. Dividing the area into twelve parts by

dx

thirteen equidistant ordinates the value of | is found to be

Jo I ' "

By (7), 0.69314866, error 0.00000148;

By (8), 0.69315046, error 0.00000328;

By (9), 0.69314725, error 0.00000007;

By (10), 0.69314722, error 0.00000004.

The correct value is, of course, loge 2, which is 0.69314718.
Formulas (7) and (8) are Simpson's Rules, (10) is Weddle's
Rule.

120

EMPIRICAL FORMULAS

3. Apply the above formulas to the area of that part of the
semi-ellipse included between the two perpendiculars erected at
the middle points of the semi-major axes. Let this area be
divided into twelve parts by equidistant ordinates.

Since the equation of the ellipse is

these ordinates are

Ws b, ^V7^ b, iVs b, iVYs b, iVJ~s b, tVVm3 b, b,
t^jVh3 ^, WYs b, \y/Vs b, JVS 6, ^V^ b, iV3 b.

By (7), .4=0.9566099^6;

By (8), ^=0.956608006;

By (9), ^=0.9566x1406;

By (10), A =0.9566x1406.

The correct value to seven
places is 0.9566x1506.

In the application of these
formulas it is highly desirable
to avoid large dififerences among
the ordinates. For that reason
the formulas do not give so
good results when applied to

4. The area under the curve,
Fig. 28, determined by the
following sets of values:

yi
3.0

2.5
2.0
1.5

y

â€” ,

>v

/

Y

N

^

/

/

0.5

A

X I

A .6 .8 1.0 1.2 Â»

Fig. 28.
) .2 .4

x.o

1.2

1.5

2.2

2.7

2.6

2.3

y \ i-o

is by (7)

^ =i - 5-(i-o+6.o+4.4+xo.8+5. 2+9.2 + 2. x) = 2.58,
and by (8),

.4 =|-i(x.o+4.5+6.6+5.4+7.8+6.9+2.x) = 2.5725.

2.1

NUMERICAL INTEGRATION

121

X1.2
ydx.

The area found is therefore the approximate value of this
integral

5. Find the area under the curve determined by the points

2.8 3.2 3.6 4.0 4.6 4.8

XI 1.5 1.9 2.3

5-0

y\o .40 1.08 1.82 2.06 2.20 2.30 2.25 2.00 1.80 1.5

The points located by the above sets of values are plotted
in Fig. 29 and a smooth curve drawn through them. The area

y

^'T~^=T^

2 â–  - Jr\

L ~^

"2

^

/

A

V

J

-Ai

~l

1

* f

t

/

/
^

y"

â€” 1 â€” 1 â€” 1 1 X

12345

Fig. 29.

is divided into strips each having a width of .4. Rectangles
are formed with the same area as the corresponding strips.
The eye is a very good judge of the position of the upper bound-
ary of each rectangle. Adding the lengths of these rectangles
and multiplying the sum by .4 the area is found to be 6.644.
By Simpson's Rule, formula (7), are found

for h = .2,

^=6.639,
^=6.645.

The graphical determination of areas can be made to yield
a close approximation by taking narrow strips, and where the
points are given at irregular intervals the area can be found
more rapidly than by the application of Simpson's Rules.

122 EMPIRICAL FORMULAS

6. A gas expands from volume 2 to volume 10, so that its
pressure p and volume v satisfy the equation pv = ioo. Find
the average pressure between v = 2 and 7; = 10.

The average pressure is equal to the work done divided by

8. The work is equal to the area under the curve p = ^^ from

V

v = 2 to z; = 10, which is

n^ 100 r 1 ^Â°

I â€” dv = 100 log V = 160.044.

J2 V L J2

That this area represents the work done in expanding the
volume from 2 to 10 becomes evident in the following way.
Let s represent the surface inclosing the gas, ps will then be
the total pressure on that surface. The element of work will
then be

dW = psdn,

when dn lepresents the element along the normal.

PF = j psdn.

But

sdn = dv,
and

W^j'pdv.

This is the equation above. The average pressure over the
change of volume from 2 to 10 is

160.944 -^ 8 = 20. 1x8.

7. Find the mean value of sin^ x from x = o to ii; = 27r. Plot
the curve y = sin^ x by the following values of x and y :

TT
12

TT

6

TT

4

TT

3

12

TT
2

.0670

.2500

.5000

.7500

â€˘9330

I. 0000

11

12

27r
3

31
4

6

IITT
12

9330 -7500 .5000 .2500 .0670

NUMERICAL INTEGRATION

123

X

IT

12

77r
6

4

47r

3

12

31

2

y

o

.0670

.2500

.5000

.7500

â€˘9330

I. 0000

X

IQTT
12

Si
3

Ttt
4

IITT

6

2_3Z
12

27r

â€˘9330 -7500 .5000 .2500 .0670 o

Applying Simpson's Rule, formula (7), the area is found to
be TT. The mean value is the area divided by 27r or .5.

8. A body weighing 100 lb. moves along a straight line
without rotating, so that its velocity v at time t is given by the
following table:

/sec

V ft./sec

1.47 1.58 . 1.67 1.76 1.86

Find the mean value of its kinetic energy from t = i to / = 9.

t

I

3

5

7

9

v^

2.1609

2.4964

2.7889

3.0976

3-4596

Kinetic energy .

3-355

3.876

4-331

4.810

5-372

Plotting kinetic energy to /, the area under the curve is
34.755. This divided by 8 gives the mean kinetic energy as
4-357-

Volumes Ai/

Fig. 30 explains the ap-
plication of the formulas
to the problem of finding
the approximate volume of
an irregular figure. Tiie area
of the sections at right angles
to the axis of x are :

Ai=\k{yi-\-^y5-^y^),

^2=i^(3'6 + 4>'9+3'8),

Az=\k{y2+Ay7+y^)' Fig. 30.

124 EMPIRICAL FORMULAS

If the areas of these sections be looked upon as ordinates,
h being the distance between two adjacent ones, it is evident
that the volume may be represented by the area under the
curve drawn through the extremities of these ordinates.

Substituting the values of A\, A2, and A3 in this equation,
the volume becomes

V = ih[ik{yi-{-4y5+y4)+ik(y6-^4yc)-\-y8)+lHy2-\-4y7-\-y3)]
= iM[yi +>'2 +>'3 +3'4 +4(75 +>'G+>'7 +y8) + 16^9]

In order to apply formulas (8), (9) and (10), the solid would
have to be divided differently, but the method of application
is at once evident from the above and needs no further discussion.

1. The following are values of the area in square feet of the
cross-section of a railway cutting taken at intervals of 6 ft.
How many cubic feet of earth must be removed in making the
cutting between the two end sections given?

91, 95, 100, 102, 98, 90, 79.

These cross-section -areas were obtained by the application
of Simpson's Rules.
By (7),

F = | â€˘ 6(91 -1-380 -f 200 -f 408 + 196 H-36o-f 79) =3428;

By (8),

7 = 1-6(91 + 285+300 -F204-f 294 -f-27o-f79) =3426.8.

2. A is the area of the surface of the water in a reservoir
when full to a depth h.

h ft I 30 25 20 15 10 5 o

^ sq.ft. . 1 26,700 22,400 19,000 16,500 14,000 10,000 5,000

NUMERICAL INTEGRATION 125

- Find (a) the volume of water in the reservoir, (b) the work
done in pumping water out of the reservoir to a height of loo ft.
above the bottom until the remaining water has a depth of
lO ft.

F = f (26, 700 +89, 600 +38,000 +66,000 +28,000+40,000+5,000)
= 488,833 cu. ft.

Work = 'ZÂŁ; ( A(ioo â€” h)dh, where ze; = weight of i cu.ft. of water

Jio
= 62.3 lb. The value of this integral will be approximately the
area under the curve determined by the points

30 25 20 15 10

^(100 â€” /?) -i 1,869,000 1,680,000 1,520,000 1,402,500 1,260,000

multiplied by 62.3.
This area is equal to

1(1,869,000+6,920,000+3,040,000+5,610,000 + 1,260,000)
= 31,165,000.

Multiplying this by 62.3 gives the work equal to 1,941,579,500
ft.-lb.

3. When the curve in Fig. 29 revolves about the ::i;-axis,
find the volume generated.

The areas of the cross-sections corresponding to the given
values of x are given in the following table:

I .0 1.2

TT 2.257r 4.847r 7.29^ 6.76^ 5 â€˘ 29^ 4.41^

By (7) F = 5.8627r = 18.416.

By (8) F = 5.8o37r = 18.231.

4. When the curve in Fig. 30 revolves about the x-axis,
find the volume generated from x=i tox = 4.2. From the curve
the following sets of values are obtained :

126 EMPIRICAL FORMULAS

X

i.o 1.2 1.4 1.6

1.8

2.0

2.2

2.4

2.6

y

.11 .29 .53

.87

1.37

1. 71

1.90

2.01

X

.012 .084 .281
2.8 3.0 3.2

â€˘757
3-4

1.877
3.6

2.924
3.8

3.610
4.0

4.040
4.2

y_

2 . 06 2.12 2.2

2.27

2.30

2.28

2.25

2.20

f 4.2444.4944.84 5.153 5.290 5.198 5.062 4.84

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