EXAM. 2. To find the length of a parabola. EXAM. 3. To find the length of the semicubical parabola, whose equation is ar2=y3. EXAM. 4. To find the length of an elliptical curve: EXAM. 5. To find the length of an hyperbolic curve. OF QUADRATURES; OR, FINDING THE AREAS of CURVES. 62. The Quadrature of Curves, is the measuring their areas, or finding a square, or other right-lined space, equal to a proposed curvilineal one. By art. 9 it appears, that any flowing quantity being drawn into the fluxion of the line along which it flows, or in the direction of its motion, there is produced the fluxion of the quantity generated by the flowing. That is, Dd x DE or yx is the fluxion of the area ADE. Hence this rule. Y 2 A RULE. RULE. 63. From the given equation of the curve, find the value either of or of y; which value substitute instead of it in the expression ; then the fluent of that expression, being taken, will be the area of the curve sought. EXAMPLES. EXAM. 1. To find the area of the common parabola. The equation of the parabola being ax = y; where a is the parameter, x the absciss AD, or part of the axis, and y the ordinate DE. From the equation of the curve is found yax. This substituted in the general fluxion of the area yx gives xax or axx the fluxion of the parabolic area; and the fluent of this, or ax = Zx √ ax = 3xy, is the area of the parabola ADE, and which is therefore equal to of its circum433 scribing rectangle. - I EXAM. 2. To square the circle, or find its area. The equation of the circle being y2 = ax-x2, or y = √ ax , where a is the diameter; by substitution, the general fluxion of the area yx, becomes /ax - x2, for the fluxion of the circular area. But as the fluent of this cannot be found in finite terms, the quantity Wax - x2 is thrown into a series, by extracting the root, and then the fluxion of the area becomes and then the fluent of every term being taken, it gives 2 1.2 1.2.2 1.3.x3 1.3.5.x4 xaxx ( for the general expression of the semisegment ADE. &c); &c); And when the point D arrives at the extremity of the diameter, then the space becomes a semicircle, and x = a; and then the series above becomes barely for the area of the semicircle whose diameter is a. EXAM. S. EXAM. 3. To find the area of any parabola, whose tion is az EXAM. 4. To find the area of an ellipse. EXAM. 5. To find the area of an hyperbola. equa EXAM. 6. To find the area between the curve and asymptote of an hyperbola. EXAM. 7. To find the like area in any other hyperbola whose general equation is "y" = a”+”. TO FIND THE SURFACES OF SOLIDS. 64. In the solid formed by the rotation of any curve about its axis, the surface may be considered as generated by the circumference of an expanding circle, moving perpendicularly along the axis, but the expanding circumference moving along the arc or curve of the solid. Therefore, as the fluxion of any generated quantity, is produced by drawing the generating quantity into the fluxion of the line or direction in which it moves, the fluxion of the surface will be found by drawing the circumference of the generating circle into the fluxion of the curve. That is, the fluxion of the surface BAE, is equal to AE drawn into the circumference BCEF, whose radius is the ordinate DE. C 65. But, if c be 3.1416, the circumference of a circle whose diameter is 1, r = AD the absciss, y = DE the ordinate, and z = AE the curve; then 2y = the diameter BE, and 2cy the circumference BCEF; also, AE = = +: therefore 2cyż or 2cy is the fluxion of the surface. And consequently if, from the given equation of the curve, the value of x or ý be found, and substituted in this expression 2cy√x2+j2, the fluent of the expression, being then taken, will be the surface of the solid required. EXAMPLES. EXAM. 1. To find the surface of a sphere, or of any seg ment. In this case, AE is a circular arc, whose equation is This value of 2, the fluxion of a circular arc, may be found more easily thus: In the fig. to art. 60, the two triangles EDC, Eae are equiangular, being each of them equiangular to the triangle ETC: conseq. ED EC: Ea; Ee, that is, The value of being found, by substitution is obtained Beyżack for the fluxion of the spherical surface, generated by the circular arc in revolving about the diameter AD. And the fluent of this gives acx for the said surface of the spherical segment BAE. But ac is equal to the whole circumference of the generating circle; and therefore it follows, that the surface of any spherical segment, is equal to the same circumference of the generating circle, drawn into a or AD, the height of the segment. Also when .r or AD becomes equal to the whole diameter, the expression acx becomes aca or ca2, or 4 times the area of the generating circle, for the surface of the whole sphere, And these agree with the rules before found in Mensuration of Solids. EXAM. 2. To find the surface of a spheroid. TO FIND THE CONTENTS OF SOLIDS. 66. ANY solid which is formed by the revolution of a curve about its axis (see last fig.), may also be conceived to be generated by the motion of the plane of an expanding circle, moving perpendicularly along the axis. And there fore fore the area of that circle being drawn into the fluxion of the axis, will produce the fluxion of the solid. That is, AD X area of the circle BCF, whose radius is DE, or diameter BE, is the fluxion of the solid, by art. 9. 67. Hence, if ADX, DEy, c = 3.1416; because cy is equal to the area of the circle BCF; therefore sy is the fluxion of the solid. Consequently if, from the given equation of the curve, the value of either y2 or x be found, and that value substituted for it in the expression cy2x, the fluent of the resulting quantity, being taken, will be the solidity of the figure proposed." EXAMPLES. EXAM. 1. To find the solidity of a sphere, or any segment. The equation to the generating circle being y2= ax — x2, where a denotes the diameter, by substitution, the general fluxion of the solid cyx, becomes caxxcxx, the fluent of which gives cax2- cx3, or cx (3a -2x), for the solid content of the spherical segment BAE, whose height AD İs x. When the segment becomes equal to the whole sphere, then a = a, and the above expression for the solidity, becomes ca3 for the solid content of the whole sphere. And these deductions agree with the rules before given and demonstrated in the Mensuration of Solids. EXAM. 2. To find the solidity of a spheroid. TO FIND LOGARITHMS. 68. IT has been proved, art. 23, that the fluxion of the hyperbolic logarithm of a quantity, is equal to the fluxion of the quantity divided by the same quantity. Therefore, when any quantity is proposed, to find its logarithm; take the fluxion of that quantity, and divide it by the same quantity; then take the fluent of the quotient, either in a series or otherwise, and it will be the logarithm sought; when corrected as usual, if need be; that is, the hyperbolic logarithm. 69. But, for any other logarithm, multiply the hyperbolic logarithm, above found, by the modulus of the system, for the logarithm sought. Note |