The Surveyor’s Area FormulaBart Braden

The College Mathematics Journal,September 1986, Volume 17, Number 4, pp.326–337.

Bart Braden studied mathematics and languages at Washington State University, as anundergraduate. After an encounter with abstract 20th-century mathematics in graduateschool at Yale and Oregon, which culminated in a doctoral thesis on representations ofLie algebras, he gradually returned to the more concrete parts of classical undergraduatemathematics. Currently, he is Professor of Mathematics at Northern KentuckyUniversity.

Atypical survey of a plot of land gives as data the successive displacementsrequired to traverse the boundary of a simple plane polygon. From this data, wewish to determine the area of the plot. In a simple example, such as Figure 1a,

one could break up the polygon into triangles whose areas could be laboriously foundusing trigonometric methods. A better way (Figure 1b) is to introduce rectangularcoordinates and change the displacement vectors from polar to rectangular form, so theycan be added to give the coordinates of the vertices of the polygon. Then a generalformula can be applied, which expresses the area of the polygon as a function of thecoordinates of its vertices. Such a polygonal area formula is well known to surveyorsbut, despite its elementary nature, does not appear in most precalculus or calculustextbooks.

Besides its intrinsic interest, at least two reasons can be advanced for including thissurveyor’s area formula in the calculus course, when plane vectors are introduced:

1. The derivation provides an excellent opportunity to introduce and use the geometricinterpretation of a determinant as the oriented area of a parallelogram in Thismakes it easier later on for students to understand geometric properties of the crossproduct of vectors in

2. The surveyor’s formula provides a geometric interpretation of an otherwisemysterious formula in multivariable calculus, expressing the area inside a simple closedcurve in parametric form as an integral around its boundary. An elementary derivation ofthis curvilinear area formula closely resembles that of the arc-length formula, so bothcan be derived together. An important advantage of the area formula is that the areaintegrals for many familiar curves are easily evaluated.

The Surveyor’s Formula. If the vertices of a simple polygon, listed counterclockwisearound the perimeter, are . . . , the area of the polygon is

Note that each oriented edge of the polygon corresponds to a determinant in thesurveyor’s formula.

2 3 2

A 512 5|x0

y0

x1

y1| 1 |x1

y1

x2

y2| 1 . . . 1 |xn22

yn22

xn21

yn21| 1 |xn21

yn21

x0

y0|6.

sxn21, yn21d,sx1, y1d,sx0, y0d,

R3.

R2.2 3 2

Figure 1a. Figure 1b.

Our derivation of the surveyor’s formula is based on the geometric interpretation of adeterminant as the oriented area of the parallelogram whose sides are the vectors

comprising the columns of the determinant. We formally state this and provide a proofsuitable for a calculus course.

Lemma. The absolute value of is the area of the parallelogram determined by

the vectors and the determinant is positive just if the(shorter) direction of rotation of into is counterclockwise.

To prove this,let denote the vector obtained by rotating counterclockwise by radians. If is the polar angle of the angle between thepositive x-axis and then is the polar angle of Since

where

we have

Figure 2.

N 5 k2r sinu, r cosul 5 k2v2, v1l.

r 5 |V|,

V 5 kr cosu, r sinul,

N.u 1 spy2dVd,sVupy2

V 5 kv1, v2lN

WV

W 5 kw1, w2l,V 5 kv1, v2l,|v1

v2

w1

w2|2 3 2

2

The altitude of the parallelogram with base andadjacent side is the absolute value of the componentof along that is, Therefore, thearea A of this parallelogram is but

so theabsolute value of the determinant with and ascolumn vectors. Moreover, this determinant is positivejust if the angle between and is acute; that is, ifthe angle from to measured counterclockwise,is between 0 and p.

W,V

WN

WV

A 5 |W ? N| 5 |v1w2 2 v2w1|,|N| 5 |V|,|V||W ? N|y|N|.

|W ? N|y|N|.N;W

W

V

Now let’s turn to the surveyor’s formula. The case when the polygon is atriangle, is the key to our proof.

Figure 3.

From the Lemma,we know that the area of a triangle having vertices listed counterclockwise (so that the shorter direction of rotation of

into is counterclockwise),is

Now let D be the determinant whose rows are Comparing the two expansions of D,

and

we obtain the surveyor’s formula

for the area of a triangle. Note that the determinants appearing in the formulacorrespond to the three orientededges of the triangle.

To establish the surveyor’s formula for a polygon with sides,we use the fact (see[4], p. 286) that any oriented simple polygon can be triangulated; that is, we can add

auxiliary diagonals through the interior to decompose the polygon into triangles,each diagonal being an edge of two adjacent triangles but inheriting oppositeorientations form them (Figure 4). Since the vertices of our polygon are listedcounterclockwise, all the triangles inherit this positive orientation; so the oriented areaof the polygon is the sum of the areas of the triangles.

n 2 2n 2 3

n > 3

A 5125|x0

y0

x1

y1| 1 |x1

y1

x2

y2| 1 |x2

y2

x0

y0|6 5 |x0

y0

x1

y1| 1 |x1

y1

x2

y2| 1 |x2

y2

x0

y0|, D 5 |x1

y1

x2

y2| 2 |x0

y0

x2

y2| 1 |x0

y0

x1

y1|

D 5 | 1x0

y0

0 x1 2 x0

y1 2 y0

0 x2 2 x0

y2 2 y0| 5 |x1 2 x0

y1 2 y0

x2 2 x0

y2 2 y0| 5 2A

r3 5 sy0, y1, y2d.r2 5 sx0, x1, x2d,r1 5 s1, 1, 1d,3 3 3

A 512|x1 2 x0

y1 2 y0

x2 2 x0

y2 2 y0|.W 5 kx2 2 x0, y2 2 y0lV 5 kx1 2 x0, y1 2 y0l

sx2, y2dsx1, y1d,sx0, y0d,

n 5 3,

3

Figure 4.

Applying the surveyor’s formula to each triangle and summing gives

with one determinant for each oriented edge in our collection of triangles. Since eachdiagonal occurs twice (as the common edge of adjacent triangles) and with oppositeorientations,the two determinants corresponding to each diagonal cancel out

because and we’re left with the sum of the determinants

corresponding to the oriented edges of the original polygon. This completes the proof.

Exercises base on a sketch like Figure 1,or on survey data (e.g. [7]) will reinforce thestudent’s grasp of the surveyor’s formula.

The Ar ea Inside a Simple Closed Curve. In a typical calculus course, parametricequations of curves are introduced in the 2nd semester and the arc-length formula for acurve in parametric form is derived, but no mention is made of the problem of findingthe area inside a simple closed curve. Except for curves in polar coordinates,the studentcan find area only by breaking up the region into pieces bounded by graphs of functionsand lines parallel to the coordinate axes.

Late in the 3rd semester, or perhaps in a multivariable calculus course, the formula

appears as a trivial consequence of Green’s theorem. Often no geometric explanation isgiven,and the formula makes little impression. The surveyor’s formula leads naturally tothis integral formula, if one thinks of a curve as the limit of inscribed polygons. Andsince the area formula in polar coordinates is an easy consequence of the generalintegral formula for parametric curves,class time spent developing the surveyor’sformula can be partially regained. The result is an elementary treatment of the calculusof curves,which makes clear the fundamental role of the parametric form.

A 512

rx dy 2 y dx

|xi

yi

xj

yj| 1 |xj

yj

xi

yi| 5 02,1

A 512o|xi

yi

xj

yj|,

4

Example. Let the circle C of radius r, with center at the origin, be given by theparametric equations and For any naturalnumber n, the points form a regular partition of the parameterinterval and the corresponding points are the verticesof a positively oriented, regular n-sided polygon inscribed in our circle. (See Figure 5a.)Note that

Figure 5a.

Applying the surveyor’s formula to each triangle and summing gives

Thus,the area of this circle is Note also that

To show that the limit of the areas of inscribed polygons is given by the integralfor any simple, closed, rectifiable curve C, we modify our notation

slightly, writing the surveyor’s formula ass1y2deC x dy 2 y dx

12E

C

x dy 2 y dx 512E2p

0

fr cos tsr cot td 2 r sin ts2r sin tdg dt 5 pr2.

A 5 limnAn 5 pr2.

5r2

2? n sin

2p

n5 pr23sin

2p

n2p

n4.

512

r25sin 2p

n1 sin

2p

n1 . . . 1 sin

2p

n 6

. . . 1 | 2sn 2 1dpr cos

n 2sn 2 1dp

r sin n

r cos 2p

r sin 2p

|6An 5

125|r cos 0

r sin 0

2pr cos

n 2p

r sin n | 1 | 2pr cos

n 2p

r sin n

4pr cos

n 4p

r sin n | 1 . . .

sx0, y0d 5 sxn, ynd.

sxk, ykd 5 sr cos tk, r sin tkdf0, 2pg,s0 ≤ k ≤ ndtk 5 s2kpdyn

ystd 5 r sin t s0 ≤ t ≤ 2pd.xstd 5 r cos t

5

(1)

with the convention Setting and andusing the identity

we obtain

(2)

Now suppose for is a simple, closed, smooth,planecurve traversed in the positive (counterclockwise) direction; so that the interior of thecurve lies to the left of the moving point Each partition of the parameter interval determines the vertices of a positively orientedpolygon ‘inscribed’ in the curve. (See Figure 5b).

Figure 5b.

Using the Mean Value theorem,the length of the inscribed polygon

can be expressed as

where the derivatives are evaluated at points in Similarly, the surveyor’sformula (2) for the area of the polygon,

can be expressed as

12 o

n

i51

Hxsti21dy9svid 2 ysti21dx9suidJsti 2 ti21d.

12 o

n

i51

Hxsti21dfystid 2 ysti21dg 2 ysti21dfxstid 2 xsti21dgJ,

sti21, tid.viui,

on

i51!Hx9suidJ2 1 Hy9svidJ2sti 2 ti21d,

on

i51|Cstid 2 Csti21d| 5 o

n

i51

Hfxstid 2 xsti21dg2 1 fystid 2 ysti21dg2J1y2

Cstidfa, bga 5 t0 < t1 < . . . < tn 5 bCstd.

t [ fa, bgC: t → Cstd 5 sxstd, ystdd

A 512

on

i51 |xi21

yi21

Dxi

Dyi|.|xi21

yi21

xi

yi| 5 |xi21

yi21

Dxi

Dyi|,Dyi 5 yi 2 yi21,Dxi 5 xi 2 xi21sxn, ynd 5 sx0, y0d.

A 512o

n

i51 |xi21

yi21

xi

yi|,

6

As the mesh of the partition of tends to zero, these ‘generalized Riemann sums’converge (see [2],p. 133) to the integrals

and

(3)

That the integral L does in fact give the arc-length of C is shown convincingly in [6].See Remark 2 below for a proof that A is indeed the area inside C.

The connection between the area formula (3) and the surveyor’s formula (2) can bedisplayed most clearly by using the notation for the integral of a differential form over acurve. In this notation ([5], p. 290),the integral (3) over is the integral of the form

over the curve C, denoted

or

One might simply say that as and become infinitesimals dx and dy, thesummation in the surveyor’s formula (2) is transformed into the integral (3).

Remarks. 1. By including in each partition of all t-values at which either iszero or fails to be differentiable, the arc-length and area formulas are seen to bevalid for piecewise smooth curves,a large enough class to include all curves studied inelementary calculus.

2. For certain pathological curves,the inscribed polygons corresponding to partitions ofthe parameter interval might cross themselves,even for arbitrarily fine partitions.This unfortunate complication could be circumvented either by simply excluding suchcurves from consideration ([3], p. 187),or better, by broadening the discussion to coverpolygons and curves with self-intersections ([2],p. 311).

Better still,by an application of Green’s theorem,

which is valid for an arbitrary Jordan region R with boundary curve C ([1], p. 289),the difficulty with pathological curves vanishes. Choosing and

we get which (by definition) is the area ofthe region R. Thus,our formula (3) is valid for any simple, closed, rectifiable, orientedcurve. The only point in mentioning the inscribed polygons is to provide a geometricmotivation for the line integral formula,and for this purpose we may restrict attention towell-behaved curves.

Variants of the Ar ea Formulas. The formula for the arc-length of a polar curveis usually derived a s a special case of the arc-length formula for

curves in parametric form. using the parametrization by polar angle:and ysud 5 fsud sin u.

xsud 5 fsud cos ur 5 fsudsa ≤ u ≤ bd

s12deCx dy 2 y dx 5 eeR1 dA,gsx, yd 5 s1

2dx,

fsx, yd 5 2s12dy

EC

fsx, yd dx 1 gsx, yd dy 5 EER1­g

­x2

­f­y2 dA,

fa, bg

CstdC9stdfa, bg

DyiDxi

12E

C|xy

dxdy|.1

2EC

x dy 2 y dx,

v 5 s12dsx dy 2 y dxd

fa, bg

A 512Eb

a

fxstdy9std 2 ystdx9stdg dt 512Eb

a |xstdystd

x9stdy9std| dt.

L 5 Eb

a

!Hx9stdJ2 1 Hy9stdJ2 dt

fa, bg

7

Since we have

Similarly, and this yields the area formula

for a simple, closed, polar curve.

The usual geometric argument,based on the formula for the area of acircular sector, shows that the polar area formula applies not only to simple closedcurves but more generally gives the area of the ‘sector’bounded by therays and and the curve (See Figure 6.)

Figure 6.

This more general result can be derived by observing that is identically 0 along

any ray through the origin, since along such a ray the tangent vector is a multiple of the radius vector So the integral along the segmentsfrom the pole to the point with polar coordinates and from back to the pole will both vanish,leaving as the only nonvanishing part of the integral of around the boundary of the “sector”.

In this connection,another view of the surveyor’s area formula should be mentioned.Observe that

is the area of the triangle having vertices and with theorientation determined by the orientation of the ith edge in the original polygon. Thus,the surveyor’s formula can be viewed as expressing the area of a polygon as the sum ofthe oriented areas of the triangles formed by joining successive pairs of vertices to theorigin. (See Figure 7.)

s0, 0d,sxi, yi dsxi21, yi21d,

12|xi21

yi21

xi

yi|

s12dsx dy 2 y dxd

s12d e

baHfsudJ2 du

sfsbd, bdsfsad, ads1

2dex dy 2 y dxkx, yl.kx9, y9l

|xy x9

y9|

r 5 fsud.u 5 b,u 5 as1

2d ebaHfsudJ2 du

A 5 s12dr2u

A 512Eb

a

HfsudJ2 du

xsudy9sud 2 ysudx9sud 5 HfsudJ2,

L 5 Eb

a

!HfsudJ2 1 Hf9sudJ2 du.

Hx9sudJ2 1 Hy9sudJ2 5 HfsudJ2 1 Hf9sudJ2,

8

Figure 7. the sum of the oriented areas of thetriangles subtended by the edges of the polygon.

This view leads to considering the differential form as the ‘radial

area element’in rectangular coordinates,just as one speaks of as the areaelement in polar coordinates. As the parameter t runs from a to b, the ray from the originto the point sweeps out an area which is computed by integration of over the curve C. Note that this interpretation of provides a simple explanation ofKepler’s law of equal areas for motion in a central force field. In such a field, isa multiple of so

This means is constant; that is, the area is swept out at a constant rate.

We round out this discussion by briefly considering two useful variants of the integralformula for the area inside a simple, closed curve:

(4)

and

(5)A 5 2Eb

a

ystdx9std dt.

A 5 Eb

a

xstdy9std dt

xy9 2 yx9

ddt|xy x9

y9| 5 |xy x0

y0| 5 0.

kx, yl,kx0, y0 l

vvCstd 5 sxstd, ystdd

s12dr2 du

v 5 s12dsx dy 2 y dxd

A 5 |A1| 1 |A2| 2 |A3| 1 |A4| 2 |A5|,

9

Adding these together and dividing by 2 gives the symmetric formula (3). How are (4)and (5) related to the surveyor’s formula for the area of a polygon? Defining

to be and collecting the coefficients of each in the surveyor’sformula,one easily verif ies the identity

Writing the right side as

it follows (as noted earlier) that as the mesh of the partition of approaches 0,thisconverges to

A similar argument gives formula (5). As a matter of fact,the surveyor’s formula isoften expressed in the corresponding forms

or

in surveying books. (See [7],p. 202,or [8], p. 483.)

Some Exercises. A drawback of the formula for classroom purposes,is that this integral is non-elementary for most curves. The area integrals (3),(4), or (5) for many familiar curves,however, are easilyevaluated. Indeed, we invite readers to verify the area formulas for the following curves given in parametric form.

Ellipse: Astroid:

Area Area538pa35 pab

x 5 a cos3 ty 5 a sin3 t

s0 ≤ t ≤ 2pdx 5 b cos ty 5 a sin t

s0 ≤ t ≤ 2pd

L 5 eba!Hx9stdJ2 1 Hy9stdJ2 dt,

A 512oyisxi21 2 xi11dA 5

12oxisyi11 2 yi21d

12Eb

a

xstdy9std dt 112Eb

a

xstdy9std dt 5 Eb

a

xstdy9std dt.

fa, bg

12 o

n

i51xisyi11 2 yid 1

12 o

n

i51xisyi 2 yi21d,

12 o

n

i51|xi21

yi21

xi

yi| 512 o

n

i51xisyi11 2 yi21d.

xisx1, y1dsxn11, yn11d

10

Cardioid: Deltoid:

Area Area

Trisectrix:inner loop:

Strophoid: loop:

Area Area5 a2s4 2 pdy25 a21p 23!3

2 2

s2py3 ≤ t ≤ 4py3dx 5 a cos t 1 a cos 2ty 5 a sin t 1 a sin 2t

x 5 ast2 2 1dyst2 1 1d

5 2pa25 6pa2

x 5 2a cos t 1 a cos 2ty 5 2a sin t 2 a sin 2t

s0 ≤ t ≤ 2pdx 5 2a cos t 2 a cos 2ty 5 2a sin t 2 a sin 2t

s0 ≤ t ≤ 2pd

11

y 5 atst2 2 1dyst2 1 1ds21 ≤ t ≤ 1d

Hourglass: One loop: Teardrop:

Area Area

Acknowledgement.I thank the Editor for his many suggestions for improving this article.

REFERENCES

1. Tom M. Apostol,Mathematical Analysis, Addison-Wesley, Reading, MA,. 1957.

2. Richard Courant,Differential and Integral Calculus, Vol. 1, Interscience, New York,1936.

3. Edouard Goursat, A Course of Mathematical Analysis, Vol. 1, tr. E. R. Hedrick, 1905,Dover, New York, 1959.

4. Einar Hille, Analytic Function Theory, Vol. 1,Ginn,Boston,1959.

5. Jerrold E. Marsden and Anthony J. Tromba,Vector Calculus, Freeman,San Francisco,1976.

6. Warren Page, “The Formula for Arc Length DoesMeasure Arc Length,” Two-YearCollege Mathematics Readings, ed. W. Page, The Mathematical Association ofAmerica,1981,pp. 111–114.

7. William H. Rayner and Milton O. Schmidt,Fundamentals of Surveying, 5th Ed.,Van Nostand, New York, 1969.

8. Harold S. Rice and Raymond M. Knight,Technical Mathematics, 3rd Ed., McGraw-Hill, New York, 1973.

5 2pab534ab

x 5 2a cos t 2 a sin 2ty 5 b sin t

s0 ≤ t ≤ 2pdx 5 a sin 2ty 5 b sin t

12

s0 ≤ t ≤ pd