76. Spherical Surface.-Let this result be applied to the calculation of the area of the earth's surface, assuming it to be spherical. The whole surface may be divided up into narrow rings of uniform width lying between parallels of latitude. Thus, if the difference. of latitude be taken to be 1°, the uniform width of each ring will be about 17 miles. The meridian arc throughout this length may be considered straight without appreciable error. The ring at the equator forms practically a cylindrical ring of radius equal to that of the earth, R. A ring taken at latitude à has a mean radius R cos; and the circumferential length of its centre line is therefore 2TR cos λ. Naming the difference of latitude for one ring SA, the width of the ring is R.SA, and its area therefore 2πR cos A.R.Sλ = 2πR2 cos A.SA. The factor 27R2 being the same for all the rings, we may first sum up all the products cos A.SA, and afterwards multiply this sum by the common factor 2R2. If we perform this integration from the equator to the north pole, that is, between the limits λ=0° and λ = 90°, we obtain the surface of the hemisphere. The integral of cos A.A from 0° to 90° is 1; and therefore the hemispherical surface is 2R2, and the whole spherical surface 4πR2. We used this result in § 70, p. 35.

n

N

n

77. Spherical Surface integrated otherwise.-The above total is 2TR x 2R. Here 2πR is the circumference of a cylinder touching the sphere, and 2R is the diameter of the sphere; so that the whole spherical surface equals that of a touching cylindrical surface whose length equals the diameter (or length) of the sphere.

In fig. 14 this circumscribing cylinder is represented by its axial section nn, ss. For each strip of spherical surface of radius 7 bounded by parallels of latitude AA, AA, there corresponds a strip of cylindric surface ll, ll of radius R, which latter is, in fact, the radial projection on the cylindric surface of the spherical strip. It is easy to prove that the arc

R

is greater than its projection l in the same ratio that R is greater than r. Hence the areas of the two differential strips are equal; and, therefore, the integral areas from end to end are also equal. This proof is more elementary than that given in the previous paragraph.

Here is a constant multiplier, and, if any other constant multiplier k were used, the result would be

Written in the inverse manner, the same result appears as

62. Rectangular Area.-A third simple example is shown in fig. 11. Here a vertical line of length k is moved horizontally to the right from an initial position 1. Its lower end moves along the axis of x. Its upper end moves along a horizontal straight

2

18x

FIG. 11.

line at the height k above this axis.

The vertical sweeps out a

rectangular area equal to k multiplied by the length of horizontal movement. At any stage x of the movement the area swept out

* See Classified List, III. A. 2.

Call its increment Sx. is k(x-x1). Here the variable is x. While the vertical moves horizontally dx, the area swept out is a narrow rectangular strip equal to kox. This is the increment of the integral area. Thus the x-gradient of the area is k. Calling the constant-kx1 by the letter C, this result may be written in the two forms

This result is identical in form with that of § 60, fig. 9, except that in the latter the integration constant is zero and does not appear, and the variable is named p, while x is the name adopted in fig. 11. The choosing of one or other name to indicate the variable is, of course, of no consequence; but taken as two illustrations of purely geometric integration, the two results are of different kinds, although taken as formulas simply they are identical.

63. Triangular Area. In fig. 12 we have an area swept out by a vertical line as it moves from left to right, and increases in height at a uniform rate during the motion. Its lower end lies always in the axis of x. Its upper

ཀ

m

1

FIG. 12.

end lies in a straight line whose inclination to the x-axis is called m.

During the movement dx, from (x - 18x) to (x+18), the mean height of the small strip of area swept out is (k+mx), and its area, therefore, is

(k+mx)dx.

The whole area swept out during the motion from the lower limit x1 to any upper limit x may be divided into two parts; one rectangular with the base (x-2) and the height (k+mx); the other triangular with the same base and the height m(x-x1).

* See Classified List, III. A. 1.

each equal to X'dx projected by horizontal lines on the final vertical X2, where it is perhaps a little more plainly seen that they all sum up to X2-X, than without this projection, as in our previous example of climbing a hill step by step as in fig. 1.

2

47. The Increment deduced from the Average Gradient.If this graphic method be actually employed for practical integration, then care must be taken to let the short stretch drawn at each slope X' lie partly before and partly beyond the vertical drawn at the corresponding distance x. This is necessary, because in actual graphic construction on paper we cannot make the increments dx immeasurably small. It is best to let each short curve length lied behind and 8x beyond x.

48. Area Graphic Diagram of Integration.-In fig. 5 the "x-gradient," or "differential coefficient with respect to x," or X', is represented by the tangent of an angle on the paper.

In fig. 6 the x's are again represented by horizontal lines, and the differential coefficients are also represented by vertical lines, while the integral is represented by an area. The known values of X' for successive values of a are plotted upwards co-ordinate with the x's, and a curve is thus obtained. The area of the vertical strip, bounded at the top by the curve, at its sides by two vertical straight lines dx apart, and at its foot by the axis of x, equals the product X'dx, where X' is the mean value throughout the length 8. The successive strips thus formed are contiguous to each other, so that the sum of them, that is, the integral X'dx, is the total area underneath the curve and above the axis of x included between the two verticals at x1 and x2, the limiting values of x.

Here, again, an error would be made if, instead of taking the mean value of X' for each length dx, there were taken either its value at the beginning of Sx or that at the end of dx. If for each Sx the value of X' at its beginning were used, the total area, or integral, obtained would be too small by the sum of the small triangular areas rafined as in the lower parts of the rectangles in fig. 6. If, on the other hand, the value at the end of each 8x were used, the calculated integral would be greater than the true integral by the sum of the small triangular areas rafined as in the upper parts of the rectangles in fig. 6. This is clearly true whether the dx's be uniform in size or of differing sizes.

49. Diminishing Error.-Not only does the size of each such error for each individual de diminish indefinitely as Sx is made smaller, but their sum also diminishes indefinitely in spite of their number being indefinitely increased. Although this is not really a

GENERAL IDEAS AND PRINCIPLES.

proposition of any special theoretical importance, since it is always the average, and not either the initial or final values of X' that are to be used, still it is interesting and customary to prove it to be true. The neatest proof is that given by Newton, and is explained by fig. 6. For the purposes of the proof all the Sa's are assumed to be of one uniform size, and, therefore, each of them equal to the last before X2 is reached. This being so, all the above-mentioned triangular area-errors are transferred horizontally so as to lie between the vertical X2 and (x2-x). The sum of each set of errors is now easily seen to be equal to about (X'2-X'1)dx, the sum of both sets taken together being exactly equal to this last strip. Now the height (X2-X') of this strip has a definite finite known value; and, therefore, the area (X'2-X'1) diminishes always in size as Sx is made smaller.

2

It must be clearly understood that no such error, large or small, occurs if the X's employed are the true average heights of the narrow strips. The average height of the strip is found at the middle of its breadth when the upper boundary is straight, and deviates from the middle by only a minute distance, if the upper boundary be very nearly straight, as it does become when its length is made very short.

dh

dl

50. Integration through Infinite Gradient.-A simple consideration of the vertical cliff below J, in fig. 1, where the gradient is infinite, will show that, while the process of integration written dl is possible up to the base-point of the cliff, and also possible from the top edge of the same onwards, it is impossible throughout the height of the cliff, simply because it has no meaning. Here any existence. neither the gradient nor the increment 87 (or dx) has Mathematicians so constantly speak of the infinitely great and the infinitely small, or infinity and zero, that probably many novices in mathematical reasoning fall into the habit of imagining that there are really such things. But neither has any sort of existence except as symbols, and in this case the symbols correspond to no physical fact, and to no mental conception. Throughout the vertical cliff there is no horizontal distance, i.e., there is no §; and there is no gradient, because a gradient is the ratio between the vertical and horizontal projections of a sloping length, and since there is here no horizontal projection, there can be no ratio between

the two.

The phrases "infinity" and "zero" are merely word-symbols to indicate the directions in which very large things and very small things vanish beyond the range of our perceptions, as they grow larger and smaller.