LESSON 14. Computation of area and volume

The main objective of the surveying is to compute the areas and volumes.

Generally, the lands will be of irregular shaped polygons.

There are formulae readily available for regular polygons like, triangle, rectangle, square and other polygons.

But for determining the areas of irregular polygons, different methods are used.

Earthwork computation is involved in the excavation of channels, digging of trenches for laying underground pipelines, formation of bunds, earthen embankments, digging farm ponds, land levelling and smoothening. In most of the computation the cross sectional areas at different interval along the length of the channels and embankments are first calculated and the volume of the prismoids are obtained between successive cross section either by trapezoidal or prismoidal formula.

Calculation of area is carried out by any one of the following methods:

a) Mid-ordinate method

b) Average ordinate method

c) Trapezoidal rule

d) Simpson’s rule

The mid-ordinate rule

Consider figure.

Module 8 Lesson 14  fig.14.1

Let  O1, O2, O3, O4……….On= ordinates at equal intervals

l=length of base line

d= common distance between ordinates


Module 8 Lesson 14 eq..14.1

Area = common distance* sum of mid-ordinates

Average ordinate method

Let O1, O2, …..On=ordinates or offsets at regular intervals

l= length of base line

n= number of divisions

n+1= number of ordinates

Module 8 Lesson 14 eq..14.2


While applying the trapezoidal rule, boundaries between the ends of ordinates are assumed to be straight. Thus the areas enclosed between the base line and the irregular boundary line are considered as trapezoids.

Let O1, O2, …..On=ordinate at equal intervals, and   d= common distance between two ordinates

Module 8 Lesson 14 eq..14.3

Total area=d/2{ O1+2O2+2O3+…….+2On-1+On}

Module 8 Lesson 14 eq..14.4

Thus the trapezoidal rule may be stated as follows:

To the sum of the first and last ordinate, twice the sum of intermediate ordinates is added. This total sum is multiplied by the common distance. Half of this product is the required area.

Limitation: There is no limitation for this rule. This rule can be applied for any number of ordinates


In this rule, the boundaries between the ends of ordinates are assumed to form an arc of parabola. Hence simpson’s rule is some times called as parabolic rule. Refer to figure:

Module 8 Lesson 14  fig.14.2


O1, O2, O3= three consecutive ordinates

d= common distance between the ordinates

area AFeDC= area of trapezium AFDC+ area of segment FeDEF


Module 8 Lesson 14 eq..14.5

Area of segment= 2/3* area of parallelogram FfdD

                            = 2/3* eE*2d

                             = 2/3 *{ O2- O1+O/2 }*2d

So, the area between the first two divisions,

Module 8 Lesson 14 eq..14.6

     = d/3(O1+4O2+O3)

Similarly, the area of next two divisions

            ∆2 =  d/3(O1+4O2+O3) and so on

Total area = d/3[O1+On+4(O2+O4+……) + 2(O3+O5)]

Module 8 Lesson 14 eq..14.7

Thus the rule may be stated as the follows

To the sum of the first and the last ordinate, four times the sum of even ordinates and twice the sum of the remaining odd ordinates are added. This total sum is multiplied by the common distance. One third of this product is the required area.

Limitation: This rule is applicable only when the number divisions is even i.e. the number of ordinates  is odd.

The trapezoidal rule may be compared in the following manner:

 Trapezoidal rule

Simpson’s rule

  1. The boundary between the ordinates is considered to be straight


  1. There is no limitation. It can be applied for any number of ordinates



  1. It gives an approximate result

 The boundary between the ordinates is considered to be an arc of a parabola


To apply this rule, the number of ordinates must be odd



It gives a more accurate result.

Note: sometimes one or both the end of the ordinates may be zero. However they must be taken into account while applying these rules.

Worked- out problems

Problem 1: The following offsets were taken from a chain line to an irregular boundary line at an interval of 10 m:

0, 2.50, 3.50, 5.00, 4.60, 3.20, 0 m

Compute the area between the chain line, the irregular boundary line and the end of offsets by:

a) mid ordinate rule

b) the average –ordinate rule

c) the trapezoidal rule

d) Simpson’s rule

Solution: (Refer fig)

Module 8 Lesson 14  fig.14.3

Mid-ordinate rule:

Module 8 Lesson 14 eq..14.8

Required area= 10(1.25+3.00+4.25+3.90+1.60)

                         = 10*18.80=188 m2

By average-ordinate rule:

Here d=10 m and n=6(no of devices)

Base length= 10*6=60 m

Number of ordinates= 7

Required area=10((1.25+3.00+5.00+4.60+3.20+0)/7)

   Module 8 Lesson 14 eq..14.9

By trapezoidal rule:

Here d=10m

Required area=10/2{0+0+2(2.50+3.50+5.00+4.60+3.20+)}

                         = 5*37.60=188 m2

By Simpson’s rule:


required area=10/3{0+0+4(2.50+5.00+3.20)+2(3.50+4.60)}

                         = 10/3{ 42.80+16.20}=10/3*59.00

                        10/3*59= 196.66m2


Problem 2: The following offsets were taken at 15 m intervals from a survey line to an irregular boundary line

3.50,4.30, 6.75, 5.25, 7.50, 8.80, 7.90, 6.40, 4.40, 3.25 m

Calculate the area enclosed between the survey line, the irregular boundary line, and the offsets, by:

a) the trapezoidal rule

b) simpson’s rule


Module 8 Lesson 14  fig.14.4

a) the trapezoidal rule

required area=15/2{3.50+3.25+2(4.30+6.75+5.25+7.50+8.80+7.90+6.40+4.40)}

                        = 15/2{6.75+102.60} = 820.125 m2

c) simpson’s rule

if this rule is to be applied, the number of ordinates must be odd. But here the number of ordinates must be odd. But here the number of  ordinate is even(ten).

So, simpson’s rule is applied from O1 to O9  and the area between O9 and O10  is found out by the trapezoidal rule.

A1= 15/3{ 3.50+4.40+4( 4.30+5.25+8.80+6.40)}+2(6.75+7.50+7.90)

    = 15/3( 7.90+99.00+44.30)= 756.00 m2

A2= 15/2(4.40+3.25)=  57.38 m2

Total area= A1+ A2 =756.00+57.38 = 813.38  m2

Problem 3: the following offsets are taken from a survey line to a curves boundary line, and the first and the last offsets by:

a) the trapezoidal rule

b) simpson’s rule


Module 8 Lesson 14  fig.14.5

here the intervals between the offsets are not reglar through out the length.

So, the section is divided into three compartments


 ∆I= area of the first section

 ∆II= area of 2nd section

III= area of 3rd section


d1= 5 m

d2=10 m

d3=20 m

a) by trapezoidal rule

I= 5/2{2.50+6.10+2(3.80+4.60+5.20)} = 89.50 m2

II= 10/2{6.10+5.80+2(4.70)} =106.50 m2

III= 20/2{5.80+2.20+2(3.90)} = 158.00 m2

Total area = 89.50+106.50+158.00 = 354.00 m2

b) by simpson’s rule

I= 5/3{2.50+6.10+4(3.8+5.20) + 2(4.60)} = 89.66 m2

II= 10/3{6.10+5.80+4(4.70)} =102.33 m2

III= 20/3{5.80+2.20+4(3.90)} = 157.33 m2

Total area= 89.66+102.33+157.33 = 349.32 m2



Module 8 Lesson 14  fig.14.6

D= common distance between the sections

A. trapezoidal rule

volume (cutting or filling), V=D/2(A1+An+2(A2+A3+….+An-1))

Module 8 Lesson 14 eq..14.10

  1. Prismoidal formula

Volume( cutting or filling), V= D/3{A1+ An +4(A2+ A4+ An-1)+ 2(A3+ A5+….+ Ann-1)}


i.e. V=common distance  {area of 1st section+ area of last section+ 4(sum of areas of even sections)

                       3                                                 +2(sum of areas of odd sections)


Note: the prismoidal formula is applicable whrn there is an odd number of sections. If the number of sections is even, the end strip is treated separately and the area is calculated according to the trapezoidal rule. The volume of the remaining strips is calculated in the usual manner by the prismoidal formula. Then both the results are added to obtain the total volume.

Works out problems

Problem 1: an embankment of width 10 m and side slopes 1 ½:1 is required to be made on a ground which is level in a direction transverseto the centre line. The central heights at 40 m intervals are as follows:

0.90,1.25,2.15,2.50,1.85,1.35, and 0.85

Calculate the volume of earth work according to

i) Trapezoidal formula

ii) Prismoidal formula

Solution:  the c/s areas are calculated by

∆= (b+sh)*h

1= (10+1.5*0.90)*0.90 = 10.22 m2

2= (10+1.5*1.25)*0.90 = 14.84 m2

3= (10+1.5*1.25)*2.15 = 28.43 m2

4= (10+1.5*2.50)*2.50 = 34.38 m2

5= (10+1.5*1.85)*1.85 = 23.63 m2

6=(10+1.5*1.35)*1.35 = 16.23 m2

7=(10+1.5*0.85)*0.85= 9.58 m2

(a) Volume according to trapezoidal formula

V= 40/2{10.22+ 9.58+2(14.84+28.43+34.38+23.63+16.23)}

  = 20{19.80+235.02} = 5096.4 m2

(b) Volume calculated in prismoidal formula:

V = 40/3 {10.22+9.58+4(14.84+34.38+16.23)+2(28.43+23.63)}

    = 40/3 (19.80+ 261.80+104.12) = 5142.9 m2

Problem the areas enclosed by the contours in the lake are as follows:

Contour (m)






Area (m2)








Calculate the volume of water between the contours 270 m and 290 m by:

i) Trapezoidal formula

ii) Prismoidal formula

Volume according to trapezoidal formula:


=330,250 m3

Last modified: Friday, 1 November 2013, 4:21 AM