Calculus is a fundamental tool in land surveying, enabling precise determination of lengths, areas, and volumes critical for planning and construction. In surveying, calculating perimeters of irregular plots or land parcels often involves applying calculus principles, especially when curves or complex shapes define boundaries. Mastery of perimeter calculations supports accurate mapping, boundary demarcation, and resource estimation in Kenya’s diverse terrain.

2.1 Perimeters of figures

Perimeter calculation is essential for determining the boundary length of a land parcel or feature, especially when the shape includes curved edges. Surveyors frequently encounter plots bounded by circular arcs, parabolic curves, or combinations of line segments and curves. Calculus provides the means to compute these perimeters by integrating the length of curve elements, ensuring precision beyond simple polygonal approximations.

2.1.1 Calculating perimeter of polygons using coordinate geometry

The perimeter of a polygon defined by survey points can be calculated by summing the distances between consecutive vertices. When coordinates of vertices are known, the distance formula derived from the Pythagorean theorem is applied repeatedly to find each side length.

The perimeter \(P\) of a polygon with vertices \((x_1, y_1), (x_2, y_2), \ldots, (x_n, y_n)\) is given by

$$P = \sum_{i=1}^{n} \sqrt{(x_{i+1} - x_i)^2 + (y_{i+1} - y_i)^2}$$

where \((x_{n+1}, y_{n+1}) = (x_1, y_1)\) to close the polygon.

Worked Examples

Example 1: A rectangular plot has vertices at \((0,0)\), \((50,0)\), \((50,30)\), and \((0,30)\) meters. Find the perimeter.

Given:
\((x_1,y_1) = (0,0)\), \((x_2,y_2) = (50,0)\), \((x_3,y_3) = (50,30)\), \((x_4,y_4) = (0,30)\)

$$P = \sqrt{(50-0)^2 + (0-0)^2} + \sqrt{(50-50)^2 + (30-0)^2} + \sqrt{(0-50)^2 + (30-30)^2} + \sqrt{(0-0)^2 + (0-30)^2}$$

$$P = \sqrt{50^2 + 0} + \sqrt{0 + 30^2} + \sqrt{(-50)^2 + 0} + \sqrt{0 + (-30)^2}$$

$$P = 50 + 30 + 50 + 30$$

Answer: 160 meters

Example 2: A triangular land parcel has vertices at \((100, 50)\), \((130, 80)\), and \((120, 20)\) meters. Calculate the perimeter.

Given:
\((x_1,y_1) = (100,50)\), \((x_2,y_2) = (130,80)\), \((x_3,y_3) = (120,20)\)

$$P = \sqrt{(130-100)^2 + (80-50)^2} + \sqrt{(120-130)^2 + (20-80)^2} + \sqrt{(100-120)^2 + (50-20)^2}$$

$$P = \sqrt{30^2 + 30^2} + \sqrt{(-10)^2 + (-60)^2} + \sqrt{(-20)^2 + 30^2}$$

$$P = \sqrt{900 + 900} + \sqrt{100 + 3600} + \sqrt{400 + 900}$$

$$P = \sqrt{1800} + \sqrt{3700} + \sqrt{1300}$$

$$P = 42.43 + 60.83 + 36.06$$

Answer: 139.32 meters

Example 3: A pentagonal parcel has vertices: \((0,0)\), \((40,0)\), \((50,30)\), \((20,50)\), \((0,30)\). Find the perimeter.

Given:
\((x_1,y_1) = (0,0)\), \((x_2,y_2) = (40,0)\), \((x_3,y_3) = (50,30)\), \((x_4,y_4) = (20,50)\), \((x_5,y_5) = (0,30)\)

$$P = \sqrt{(40-0)^2 + (0-0)^2} + \sqrt{(50-40)^2 + (30-0)^2} + \sqrt{(20-50)^2 + (50-30)^2} + \sqrt{(0-20)^2 + (30-50)^2} + \sqrt{(0-0)^2 + (0-30)^2}$$

$$P = \sqrt{40^2 + 0} + \sqrt{10^2 + 30^2} + \sqrt{(-30)^2 + 20^2} + \sqrt{(-20)^2 + (-20)^2} + \sqrt{0 + (-30)^2}$$

$$P = 40 + \sqrt{100 + 900} + \sqrt{900 + 400} + \sqrt{400 + 400} + 30$$

$$P = 40 + \sqrt{1000} + \sqrt{1300} + \sqrt{800} + 30$$

$$P = 40 + 31.62 + 36.06 + 28.28 + 30$$

Answer: 165.96 meters

2.1.2 Perimeter of curves using calculus: Arc length formula

For curves defined by a function \(y = f(x)\), the perimeter or length of the curve between two points is found by integrating the infinitesimal arc lengths. This is critical when boundaries are curved, such as riverbanks or circular plots.

The arc length \(L\) from \(x=a\) to \(x=b\) is given by

$$L = \int_a^b \sqrt{1 + (\frac{dy}{dx})^2} \, dx$$

where \(\frac{dy}{dx}\) is the derivative of \(y\) with respect to \(x\).

Worked Examples

Example 1: Find the length of the curve \(y = 3x\) from \(x=0\) to \(x=4\).

Given:
\(y = 3x\), \(a=0\), \(b=4\)

Calculate derivative:

$$\frac{dy}{dx} = 3$$

Arc length formula:

$$L = \int_0^4 \sqrt{1 + (3)^2} \, dx = \int_0^4 \sqrt{1 + 9} \, dx = \int_0^4 \sqrt{10} \, dx$$

Since \(\sqrt{10}\) is constant:

$$L = \sqrt{10} \times (4-0) = 4\sqrt{10}$$

Calculate numerical value:

$$4 \times 3.1623 = 12.65 \text{ meters}$$

Answer: 12.65 meters

Example 2: Calculate the length of curve \(y = x^2\) from \(x=0\) to \(x=2\).

Given:
\(y = x^2\), \(a=0\), \(b=2\)

Calculate derivative:

$$\frac{dy}{dx} = 2x$$

Arc length:

$$L = \int_0^2 \sqrt{1 + (2x)^2} \, dx = \int_0^2 \sqrt{1 + 4x^2} \, dx$$

This integral requires a substitution:

Let \(x = \frac{1}{2} \sinh t\), so \(dx = \frac{1}{2} \cosh t \, dt\)

Then:

$$\sqrt{1 + 4x^2} = \sqrt{1 + 4 \times \frac{1}{4} \sinh^2 t} = \sqrt{1 + \sinh^2 t} = \cosh t$$

Change limits: when \(x=0\), \(t=0\); when \(x=2\), \(2 = \frac{1}{2} \sinh t \Rightarrow \sinh t = 4 \Rightarrow t = \sinh^{-1} 4\)

Integral becomes:

$$L = \int_0^{\sinh^{-1} 4} \cosh t \times \frac{1}{2} \cosh t \, dt = \frac{1}{2} \int_0^{\sinh^{-1} 4} \cosh^2 t \, dt$$

Use identity: \(\cosh^2 t = \frac{1 + \cosh 2t}{2}\)

$$L = \frac{1}{2} \int_0^{\sinh^{-1} 4} \frac{1 + \cosh 2t}{2} \, dt = \frac{1}{4} \int_0^{\sinh^{-1} 4} (1 + \cosh 2t) \, dt$$

Integrate:

$$L = \frac{1}{4} [ t + \frac{\sinh 2t}{2} ]_0^{\sinh^{-1} 4}$$

Calculate:

\(t = \sinh^{-1} 4 = \ln (4 + \sqrt{16 + 1}) = \ln (4 + 4.1231) = \ln (8.1231) = 2.095\)

\(\sinh 2t = 2 \sinh t \cosh t\)

\(\sinh t = 4\), \(\cosh t = \sqrt{1 + \sinh^2 t} = \sqrt{1 + 16} = \sqrt{17} = 4.1231\)

Calculate \(\sinh 2t\):

$$\sinh 2t = 2 \times 4 \times 4.1231 = 33.0$$

Plug values:

$$L = \frac{1}{4} ( 2.095 + \frac{33.0}{2} ) = \frac{1}{4} (2.095 + 16.5) = \frac{1}{4} (18.595) = 4.649 \text{ meters}$$

Answer: 4.65 meters

Example 3: Find the length of the semicircular arc defined by \(y = \sqrt{16 - x^2}\) from \(x=-4\) to \(x=4\).

Given:
\(y = \sqrt{16 - x^2}\), \(a=-4\), \(b=4\)

Calculate derivative:

$$\frac{dy}{dx} = \frac{-x}{\sqrt{16 - x^2}}$$

Arc length:

$$L = \int_{-4}^4 \sqrt{1 + (\frac{-x}{\sqrt{16 - x^2}})^2} \, dx = \int_{-4}^4 \sqrt{1 + \frac{x^2}{16 - x^2}} \, dx$$

Simplify inside the square root:

$$\sqrt{\frac{16 - x^2 + x^2}{16 - x^2}} = \sqrt{\frac{16}{16 - x^2}} = \frac{4}{\sqrt{16 - x^2}}$$

So:

$$L = \int_{-4}^4 \frac{4}{\sqrt{16 - x^2}} \, dx$$

This integral equals the length of a semicircle of radius 4:

$$L = 4 \times \int_{-4}^4 \frac{1}{\sqrt{16 - x^2}} \, dx = 4 \times \pi = 4\pi$$

Calculate numerical value:

$$4 \times 3.1416 = 12.566 \text{ meters}$$

Answer: 12.57 meters

2.1.3 Perimeter of parametric curves

Survey boundaries may be described parametrically as \(x = x(t)\), \(y = y(t)\) for \(t\) in \([a,b]\). The perimeter or curve length is computed by integrating the speed over the parameter interval.

The formula for arc length \(L\) is

$$L = \int_a^b \sqrt{(\frac{dx}{dt})^2 + (\frac{dy}{dt})^2} \, dt$$

This method applies to curves such as ellipses or more complex paths common in land parcel boundaries.

Worked Examples

Example 1: Find the length of the curve defined by \(x = t^2\), \(y = t^3\) for \(t\) from 0 to 2.

Given:
\(x = t^2\), \(y = t^3\), \(a=0\), \(b=2\)

Calculate derivatives:

$$\frac{dx}{dt} = 2t$$

$$\frac{dy}{dt} = 3t^2$$

Arc length:

$$L = \int_0^2 \sqrt{(2t)^2 + (3t^2)^2} \, dt = \int_0^2 \sqrt{4t^2 + 9t^4} \, dt = \int_0^2 \sqrt{t^2(4 + 9t^2)} \, dt$$

Simplify:

$$L = \int_0^2 t \sqrt{4 + 9t^2} \, dt$$

Use substitution:

Let \(u = 4 + 9t^2\), then \(du = 18t \, dt\), or \(t \, dt = \frac{du}{18}\)

Change limits: when \(t=0\), \(u=4\); when \(t=2\), \(u=4 + 9 \times 4 = 40\)

Integral becomes:

$$L = \int_{u=4}^{40} \sqrt{u} \times \frac{du}{18} = \frac{1}{18} \int_4^{40} u^{1/2} \, du$$

Integrate:

$$\frac{1}{18} [ \frac{2}{3} u^{3/2} ]_4^{40} = \frac{1}{27} (40^{3/2} - 4^{3/2})$$

Calculate powers:

$$40^{3/2} = 40 \times \sqrt{40} = 40 \times 6.3246 = 252.98$$

$$4^{3/2} = 4 \times \sqrt{4} = 4 \times 2 = 8$$

Calculate difference:

$$252.98 - 8 = 244.98$$

Calculate final length:

$$L = \frac{244.98}{27} = 9.07 \text{ meters}$$

Answer: 9.07 meters

Example 2: Find the length of the ellipse parameterized by \(x = 5 \cos t\), \(y = 3 \sin t\) for \(t\) from 0 to \(\pi/2\).

Given:
\(x = 5 \cos t\), \(y = 3 \sin t\), \(a=0\), \(b=\pi/2\)

Calculate derivatives:

$$\frac{dx}{dt} = -5 \sin t$$

$$\frac{dy}{dt} = 3 \cos t$$

Arc length:

$$L = \int_0^{\pi/2} \sqrt{(-5 \sin t)^2 + (3 \cos t)^2} \, dt = \int_0^{\pi/2} \sqrt{25 \sin^2 t + 9 \cos^2 t} \, dt$$

Rewrite:

$$L = \int_0^{\pi/2} \sqrt{9 + 16 \sin^2 t} \, dt$$

This integral is elliptic and cannot be expressed in elementary functions. Approximate numerically.

Approximate by Simpson’s rule or use approximate formula for quarter ellipse perimeter:

Approximate quarter perimeter:

$$L \approx \frac{\pi}{2} ( \frac{3(a + b) - \sqrt{(3a + b)(a + 3b)}}{2} )$$

where \(a=5\), \(b=3\).

Calculate:

$$3(a + b) = 3(5 + 3) = 3 \times 8 = 24$$

$$(3a + b)(a + 3b) = (15 + 3)(5 + 9) = 18 \times 14 = 252$$

$$\sqrt{252} = 15.87$$

Calculate expression inside parentheses:

$$\frac{24 - 15.87}{2} = \frac{8.13}{2} = 4.065$$

Multiply by \(\pi/2\):

$$\frac{\pi}{2} \times 4.065 = 1.5708 \times 4.065 = 6.38 \text{ meters}$$

Answer: Approximately 6.38 meters

Example 3: A parametric curve is defined by \(x = 3t - t^2\), \(y = 2t + t^3\) for \(t\) from 0 to 1. Find the curve length.

Given:
\(x = 3t - t^2\), \(y = 2t + t^3\), \(a=0\), \(b=1\)

Calculate derivatives:

$$\frac{dx}{dt} = 3 - 2t$$

$$\frac{dy}{dt} = 2 + 3t^2$$

Arc length:

$$L = \int_0^1 \sqrt{(3 - 2t)^2 + (2 + 3t^2)^2} \, dt$$

Expand squares:

$$(3 - 2t)^2 = 9 - 12t + 4t^2$$

$$(2 + 3t^2)^2 = 4 + 12t^2 + 9t^4$$

Sum:

$$9 - 12t + 4t^2 + 4 + 12t^2 + 9t^4 = 13 - 12t + 16t^2 + 9t^4$$

Integral:

$$L = \int_0^1 \sqrt{13 - 12t + 16t^2 + 9t^4} \, dt$$

Numerical integration is required. Approximate using Simpson’s rule with 4 intervals:

Evaluate integrand at \(t=0, 0.25, 0.5, 0.75, 1\):

• \(t=0\): \(\sqrt{13} = 3.6056\)

• \(t=0.25\):
  Calculate inside sqrt:
  \(13 - 12(0.25) + 16(0.25)^2 + 9(0.25)^4 = 13 - 3 + 1 + 0.088 = 11.088\)
  \(\sqrt{11.088} = 3.33\)

• \(t=0.5\):
  \(13 - 6 + 4 + 0.5625 = 11.5625\)
  \(\sqrt{11.5625} = 3.4\)

• \(t=0.75\):
  \(13 - 9 + 9 + 2.85 = 15.85\)
  \(\sqrt{15.85} = 3.98\)

• \(t=1\):
  \(13 - 12 + 16 + 9 = 26\)
  \(\sqrt{26} = 5.10\)

Apply Simpson’s rule:

$$L \approx \frac{h}{3} [f_0 + 4f_1 + 2f_2 + 4f_3 + f_4]$$

where \(h = \frac{1-0}{4} = 0.25\)

Calculate:

$$L = \frac{0.25}{3} [3.6056 + 4(3.33) + 2(3.4) + 4(3.98) + 5.10]$$

$$= 0.0833 [3.6056 + 13.32 + 6.8 + 15.92 + 5.10]$$

$$= 0.0833 \times 44.7456 = 3.73 \text{ meters}$$

Answer: 3.73 meters

2.1.4 Perimeter of composite figures combining straight lines and curves

Surveyed land parcels often have boundaries composed of straight segments and curved arcs. The total perimeter is the sum of all straight-line distances and arc lengths. Calculus is applied to curves, while coordinate geometry handles straight segments.

The total perimeter \(P\) is

$$P = \sum \text{length of straight segments} + \sum \text{arc lengths of curves}$$

Each curved segment length is calculated using the formulas from previous subtopics.

Worked Examples

Example 1: A land parcel has three straight sides measuring 40 m, 30 m, and 50 m, and a curved boundary which is a quarter circle of radius 20 m. Find the total perimeter.

Given:
Straight sides: 40 m, 30 m, 50 m
Curved side: quarter circle, radius \(r=20\) m

Calculate arc length of quarter circle:

$$L = \frac{1}{4} \times 2 \pi r = \frac{\pi r}{2} = \frac{3.1416 \times 20}{2} = 31.416 \text{ m}$$

Sum straight sides:

$$40 + 30 + 50 = 120 \text{ m}$$

Total perimeter:

$$P = 120 + 31.416 = 151.416 \text{ m}$$

Answer: 151.42 meters

Example 2: A plot boundary consists of two straight sides: 100 m and 80 m, and a curved boundary defined by \(y = \sqrt{100 - x^2}\) from \(x=0\) to \(x=6\). Calculate the total perimeter.

Given:
Straight sides: 100 m and 80 m
Curve: \(y = \sqrt{100 - x^2}\), \(x=0\) to \(x=6\)

Calculate arc length of curve:

Calculate derivative:

$$\frac{dy}{dx} = \frac{-x}{\sqrt{100 - x^2}}$$

Arc length:

$$L = \int_0^6 \sqrt{1 + (\frac{-x}{\sqrt{100 - x^2}})^2} \, dx = \int_0^6 \frac{10}{\sqrt{100 - x^2}} \, dx$$

Simplify inside integral:

$$\sqrt{1 + \frac{x^2}{100 - x^2}} = \frac{10}{\sqrt{100 - x^2}}$$

Integral:

$$L = 10 \int_0^6 \frac{1}{\sqrt{100 - x^2}} \, dx = 10 \sin^{-1} (\frac{x}{10}) \Big|_0^6 = 10 (\sin^{-1}(0.6) - \sin^{-1}(0))$$

Calculate:

\(\sin^{-1}(0.6) = 0.6435\) radians

So:

$$L = 10 \times 0.6435 = 6.435 \text{ m}$$

Sum straight sides:

$$100 + 80 = 180 \text{ m}$$

Total perimeter:

$$P = 180 + 6.435 = 186.435 \text{ m}$$

Answer: 186.44 meters

Example 3: A land parcel is bounded by two straight sides of 60 m and 40 m and a curve defined parametrically by \(x = 10 \cos t\), \(y = 10 \sin t\) for \(t\) from 0 to \(\pi/3\). Find the total perimeter.

Given:
Straight sides: 60 m, 40 m
Curve: parametric circle arc radius 10 m, \(t=0\) to \(\pi/3\)

Calculate curve length:

Derivatives:

$$\frac{dx}{dt} = -10 \sin t$$

$$\frac{dy}{dt} = 10 \cos t$$

Arc length:

$$L = \int_0^{\pi/3} \sqrt{(-10 \sin t)^2 + (10 \cos t)^2} \, dt = \int_0^{\pi/3} \sqrt{100 \sin^2 t + 100 \cos^2 t} \, dt$$

Simplify:

$$L = \int_0^{\pi/3} \sqrt{100 (\sin^2 t + \cos^2 t)} \, dt = \int_0^{\pi/3} 10 \, dt = 10 \times \frac{\pi}{3} = \frac{10\pi}{3}$$

Calculate numerical:

$$\frac{10 \times 3.1416}{3} = 10.472 \text{ m}$$

Sum straight sides:

$$60 + 40 = 100 \text{ m}$$

Total perimeter:

$$P = 100 + 10.472 = 110.472 \text{ m}$$

Answer: 110.47 meters

Chapter Summary

This chapter focused on applying calculus to solve practical problems in engineering by calculating perimeters, areas, volumes, and surface areas of various figures and solids. It began with techniques for finding the perimeter of complex plane figures using integration to sum infinitesimal lengths along curves. Next, the chapter detailed methods to determine areas enclosed by curves by integrating functions over specified intervals, providing a foundation for handling irregular shapes. The discussion then extended to three-dimensional solids, explaining how to compute volumes through integration of cross-sectional areas and revolving curves around axes. Surface area calculations were also covered by integrating the lengths of curve elements rotated around an axis. Throughout, the chapter emphasized step-by-step problem solving with real-world civil engineering applications. This comprehensive approach enables precise measurement of geometric properties essential for design and construction tasks.