88. Area of figures with curved boundaries

We have seen already in the case of the circle that the definition of area with the aid of equalities of decomposition and complementation is not successful. You cannot decompose such figures into squares or complement them into figures, bounded by lines. The method, which we have applied to the circle, suggests a solution. We only need to generalize it by including the area between two bounds. If you imagine a figure drawn on millimetre graph paper, one can count the squares and obtain a first approximation. There arises the difficulty that the boundary of the figure no longer coincides with that of the squares. But you can design two bounds, between which the sought area lies. The inner polygon, from which you obtain the area I_i by summing all squares inside the boundary, the outer polygon, from which you obtain the area I_o by including the squares intersected by the boundary, whence

I_o > I > I_i .

If you now use a denser grid of squares, you will obtain new approximate values I '_o and I '_i, so that

I_o > I '_o > I > I '_i > I_i ,

where the sequence of the I '_o values decreases and the sequence of the I '_i values increases. They have bounds and approach a definite identical value, whence we have the Definition:

The area of a figure with a curved boundary is the limit,
which is approached by the areas of the outer
and inner grids with decreasing grid size.

Calculation of area

In practical, approximate computations, you can often proceed in a more convenient manner. You select a base line AB and draw AC and BD at right angles to it. The area, you are to compute, will then appear as the difference of two areas, each bounded by a curve and three segments.

You now decompose the base line into n equal segments of length x, draw at the marked points perpendiculars y₀, y₁, etc. and the corresponding chords. Using the formula for the area of trapezoids, you find now the following approximate values, which with decreasing width of the strips approximate the required area:

Alternatively, you can proceed as follows: You select an even number of strips and draw tangents through the odd ordinates, which will then bound the area with the neighbouring ordinate strips of double width. Then

Areas of similar figures

If you want to compare the areas of two similar figures, you decompose both of them by similar grids. Then you will obtain the same number of squares for the outer and inner sums. But the squares are in the same ratio as the squares of the similarity scale a²₂ : a²₁ = m², whence I₂ : I₁ = m² : 1 and there follows the Rule:

The areas of two similar figures are in the same ratio
as the squares of any corresponding segments.

Areas of the mantles of cylinders, cones and truncated cones

If you open a cylinder along a mantle line and spread it out in a plane, you obtain a rectangle, one side of which equals the height and the other the circumference of the base circle of the cylinder, whence the mantle's area is

M = 2p rh.

For the computation of the area of the mantle of a cone, you use the formulae for the circular sector and circular arc. Since equal central angles and equal circular sectors correspond to the same arcs, you have

b : 2p r = a : 360,

whence b = ap r/180 and the area of the corresponding circular sector is

I : p r² = b : 2p r,

that is,

I = br/2.

If you now imagine the cone cut along a mantle line and unrolled in the plane, you obtain a circular sector, the radius of which is the generator s and the arc of which equals the circumference of the base of the cone b=2p r. Obviously, you have now the ratio

M : p s ² = 2p r : 2p s, that is, M = p r s.

By the Ray Theorem,

(x + s)/x = r₁/r₂,

whence

s/x = (r₁ - r₂)/r₂, that is, x = r₂ s/(r₁ - r₂)

and

M = p (r₁ + r₂)/s.

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