Hence the value of the curve, from the fluent of each of these, gives the four following forms, in series, viz. putting d = 2r the diameter, the curve is Now, it is evident that the simplest of these series, is the third in order, or that which is expressed in terms of the tangent. That form will therefore be the fittest to calculate an example by in numbers. And for this purpose it will be convenient to assume some arc whose tangent, or at least the square of it, is known to be some small simple number. Now, the arc of 45 degrees, it is known, has its tangent equal to the radius; and therefore, taking the radius. r = 1, and consequently the tangent of 45°, or t, = 1 also, in this case the arc of 45° to the radius 1, or the arc of the quadrant to the diameter 1, will be equal to the infinite se ries 1 But as this series converges very slowly, it will be proper to take some smaller arc, that the series may converge faster; such as the arc of 30 degrees, the tangent of which is = √, or its square t2 = : which being substituted in the series, the length of the arc of 30° comes out 1 I (1-3.3 pute these terms in decimal numbers, after the first, the suc ceeding terms will be found by dividing, always by 3, an these quotients again by the absolute numbers 3, 5, 7, 9, &c and lastly, adding every other term together, into two sum. the one the sum of the positive terms, and the other the sur of the negative ones; then lastly, the one sum taken fro the other, leaves the length of the arc of 30 degrees; whic. being the 12th part of the whole circumference when th radius is 1, or the 6th part when the diameter is 1, cons quently 6 times that arc will be the length of the whole ci cumference to the diameter 1. Therefore, multiplying t. first term by 6, the product is 12 = 3.4641016; a hence the operation will be conveniently made as follows + Ter 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 = y2; where a is the parameter, & the absciss AD, or part of the axis, and y the ordinate DE. From the equation of the curve is found y = ax. This substituted in the general fluxion of the area yx gives ax or ax*x the fluxion of the parabolic area; and the fluent of this, ora x = x√ ax = xy, is the area of the parabola ADE, and which is therefore equal to of its circumscribing rectangle. I EXAM. 2. To square the circle, or find its area. The equation of the circle being y = ax - x2, or y = Vax - x2, 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 Vax - 2 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. 1.2 1.3.3 1.3.5.x4 xaxx (3-5 4.7a2 4.6.943 4.6.8.1104 &c); - &c); for the general expression of the semisegment ADE. 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 a2 2 1 5 1 4.7 &c) 4.6.9 4.6.8.11 for the area of the semicircle whose diameter is a. EXAM. 3. EXAM. 3. To find the area of any parabola, whose equa tion is az = y+". EXAM. 4. To find the area of an ellipse. EXAM. 5. To find the area of an hyperbola. EXAM. 6. To find the area between the curve and asymp tote of an hyperbola. EXAM. 7. To find the like area in any other hyperbola whose general equation is rty = amt. 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 C B E D Y 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, 1 whose radius is the ordinate DE. 1. 65. But, if cbe = 3·1416, the circumference of a circle whose diameter is 1, x = AD the absciss, y = DE the ordinate, and z = AE the curve; then 2y = the diameter BE, and 2cy = the circumference BCEF; also, AE == √x2 + j2: therefore 2cyż or 2cyx2 + y2 is the fluxion of the surface. And consequently if, from the given equation of the curve, the value of & or jy be found, and substituted in this expression 2cyx2 + j2, the fluent of the expression, being then taken, will be the surface of the solid required. 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 ż, 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, ax 2y y: a ::: = the same as before. The value of & being found, by substitution is obtained 2cyx = acx 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 ar or AD becomes equal to the whole diameter a, 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 therefore |