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(2.) To determine the Area of a Circle.
If n is the number of sides of a regular polygon inscribed in a circle whose radius is r, then, as we have already seen (Trig: Art: 45),
Now if we increase the number of sides of the polygon, it becomes more and more nearly equal to the circle, and in the limiting case when the
is infinitely great, it becomes the circle;
but when n
= ∞ ..
number of sides
:0 and (Trig:
* Compare with the statement in the text what is said on page 160.
(11.) To find the space between two Concentric Circles, viz. ABCEDF.
.. TBP2: = π (1,2 — 7.2)
also since OPB is a right angle, BP touches the circle, or the area of the space in question is equal to the area of a circle whose radius is the length of the line drawn from any point
in the exterior circle to touch the interior circle.
The above cases of areas of plane figures are the chief of those which belong to Elementary Mathematics. The determination of areas bounded by curved lines belongs, of right, to the Integral Calculus; the following propositions, however, are best given here, though the reasoning is not of a strictly elementary character.
and hence, if we suppose a series of parallelograms to be described in circle and ellipse, the same proportion will hold good between each of these, and therefore
Sum of parallelograms in semicircle: sum of parallelograms in semi-ellipse::a: b. and this being true, however great the number may be, is true in the limit. Now the semi-ellipse is the limit of the parallelograms inscribed in it, and the semicircle is the limit of parallelograms inscribed in it.
*Let APQ be an area included by two straight lines AB. AP, and the curve PQ; divide AQ into equal parts Aa, ab, bc, cd, dQ, and on these lines draw the rectangles Ap, aq, br, cs within the
curved areas, and complete the parallelograms Ap', aq', br', cs', dt. Then it is plain that the difference between the interior parallelograms (Ap, aq, br, cs) and the exterior parallelograms (Ap', aq', br', cs', dt) will equal Ap'. Now, by making the number of parallelograms very large, Aa, and .. Ap' will become very small, and may be made less than any magnitude that may be assigned. Now the curvilinear area is clearly greater than the interior and less than the exterior parallelograms, and therefore differs from the interior parallelogram by a quantity less than Ap'; i. e., a quantity that can be made less than any that can be assigned: and therefore the curvilinear area is the limit to which the sum of the interior parallelograms continually approaches when their number is increased.
(13.) Let q PQ be portion of a Parabola. It is required to find its Arca. Bisect Qq in V, draw the diameter PV. Through P draw » PR parallel to QV q, and draw QR and qr parallel to PV. Then the area q PQ is two thirds of qr RQ.
Take p and p1, two points in PQ, and through p, draw pi mi, pi ni parallel to PV and QV, and through p draw pm pn also parallel to PV, QV, and produce them to meet pi niin qı and pi mi in q respectively. Then
Parallelogram pm, parallelogram pn¡ :: pn × nn ̧: pm × mm2•
:: pn × (Pn, — Pn) : Pn × (P1n2 - pn)
.. parallelogram pm: parallelogram pn1 : : pn : P1n1 + pn.
Now in the limit pin differs from pn by a quantity less than any that can be assigned.
.. In the limit pn: P1n + pn :: 1:2
.. In the limit parallelogram pm1 : parallelogram pn, : : 1 : 2
which proportion is true for cach pair of parallelograms inscribed in PQR and PQV, and.. is true of all.
. In the limit sum of parallelograms in PQR: sum of parallelograms in PQV::1:2 But the area PQR is the limit of the sum of the parallelograms inscribed in it; and PQV is the limit of the sum of the parallelograms inscribed in it.
.. PQR PQV :: 1:2
.. PQVR: PQV :: 3 : 2
Now PQVR is the half of qr QR, and PQV is the half of PQq.
(14). To find approximately the Area of a Plane Figure bounded by a Curve. Let A BPQ be the figure; divide A B into equal parts, AN, N, N2, N, N3, &c., and draw ordinates P, N,,P, NP, N3, &c. parallel to AP or BQ, and perpendicular to AB.
(1). As a first approximation the curved area may be considered as identical with the polygonal area inclosed by PABQ and the chords PP, P, P2, P2Pg, &c. In this case, the area required = AN,P.P + N, N,P,P, + N, N, P,P2, &c.
PAN, +P2N, AP + P,N, =AN, X
+ N,N, 2
P.N, +P,Na, &c.
+ ... + D)
=, + PLN, + P2N, + P,
2 Or the area= (the distance between any two consecutive ordinates) x (half the sum of the extreme ordinates, together with the sum of all the intermediate ordinates) Or if AN=1
P2N, =Yz &c.
+ Yu + y2 + yg +. (2.) The above is a good approximation, but a much better may be found in the following manner. Consider the portion of the curve between any three consecutive Ps, such as P, P, P4, through P, draw a line n P, m touching the curve, and let it meet N, P2, N. P. produced in n and m. Draw P, P perpendicular to P NĄ, join P, P. meeting P, N, in
Now we may consider P, Pg Pa a portion of a parabola, * and
2 :. the curvilinear area P, P, PA of n P, Pam.
3 m PA x P2P.
* It appears from Newton, (Lemma XI. Lect. 1,) that in any curve of finite curvature, P3 q is in the limit as the square of q P4. Now it is the characteristic property of the parabola that P3 , is as the square of q P 4; hence every curve of finite curvature tends to a parabola as its limit, and so if we take a small arc P, P3 P4 we may consider it to be a parabolic arc, without making any appreciable error,