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ExAM. 3. To find the area of any parabola, whose equation 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 asymptote of an hyperbola.

Exam. 7. To find the like area in any other hyperbola whose general equation is rhyn = "+".

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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:

65. But, if c be = 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,

= the circumference BCEF; also, AE = = Vä? + j* : therefore 2cyż or 2cyd ** + ja is the fluxion of the surface. And consequently if, from the given equation of the curve, the value of 3 or j be found, and substituted in this expression 2cy v šo? + j, the fluent of the expression, being then taken, will be the surface of the solid required.

and 2cy


Exam. 1. To find the surface of a sphere, or of any segment.


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In this case, AE is a circular arc, whose equation is y* = ax – , or y = Vax -- .*.*.

a = 2.2

23 The fuxion of this gives j =

2 var 22 2y

4ax + 4.2.2 a? hence j2 =


**; consequently

442 a'%?

az * + tje = and z = N2 + j' =

" 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 :: EQ : Ee, that is,

y:a::* := the same as before.

The value of į being found, by substitution is obtained 2cyž = aci 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 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 x or AD, the height of the segment,

Also when r or AD becomes equal to the whole diameter the expression acx becomes aca or ca“, 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.
EXAM. 3. To find the surface of a paraboloid.
Exam. 4. To find the surface of an hyperboloid.

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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 the area of that circle being drawn into the fuxion of the axis, will produce the flution 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 ad = x, DE = y,c= 3:1416; because cy? is equal to the area of the circle BCF; therefore cy** is the fluxion of the solid. Consequently if, from the given equation of the curve, the value of either y or x be found, and that value substituted for it in the expression cy's, the fluent of the resulting quantity, being taken, will be the solidity of the figure proposed.


Exam. 1. To find the solidity of a sphere, or any segment. The equation to the generating circle being y' = ax – x?, where a denotes the diameter, by substitution, the general fluxion of the solid cy**, becomes caxi - cx'x, the fluent of which gives cax? - {cx}, or {cx? (30 – 2x), for the solid content of the spherical segment BAE, whose height ad is x..

When the segment becomes equal to the whole sphere, then x = a, and the above expression for the solidity, becomes fca' 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.
Exam. 3. To find the solidity of a paraboloid.
ExAM. 4. To find the solidity of an hyperboloid.


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.




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Note. The modulus of the hyperbolic logarithms, is 1; and the modulus of the common logarithms, is .43429448190 &c; and, in general, the modulus of any system, is equal to the logarithm of 10 in that system divided by the number 2.3025850929940&c, which is the hyp. log. of 10. Also, the hyp. log. of any number, is in proportion to the com. log. of the same number, as unity or i is to .43429&c, or as the number 2:302585&c, is to l; and therefore, if the common log. of any number be multiplied by 2.302585&c, it will give the hyp. log. of the same number; or if the hyp. log. be divided by 2.302585&c, or multiplied by 43429&c, it will give the common logarithm.

atx Exam. 1. To find the log. of

Denoting any proposed number %, whose logarithm is required to be found, by the compound expression a + x

* the fluxion of the number z, is- and the fluxion

2 a

Xü x2x x3x

+ &c.
a + x

a? a3 Then the fluent of these terms give the logarithm of a + x

x2 x3 or logarithm of


&c. 2a? 3a3

x3 Writing – x for x, gives log. .


2a2 323' 4a4 Div. these numb. and a + x 2.7

2.73 2.xs subtr. their logs. gives ) log.

+ &c. 303

bas afx

afx Also, because =)

or log.

=0– log. 3 atx


x 2013 24 therefore log. of is

+ &c,

2a2 303 4a

23 and the log. of is +



a> x2

+ + + &c. x2

q2 Now, for an example in numbers, suppose it were required to compute the common logarithm of the number 2. This will be best done by the series, a + x

x3 x5 x7 log. of 2m x 6 + + +

&c. 3a3





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ats Making

= ; .

= ,

a? , which is the constant factor for every succeeding term; also, 2m = 2 X •43429448190 = •868588964; therefore the calculation will be conveniently made, by first dividing this number by 3, then the quotients successively by 9, and lastly these quotients in order by the respective numbers 1, 3, 5, 7, 9, &c, and after that, adding all the terms together, as follows:

3 ) •868588964 9) 289529654 1) 289529654 (-280529654 9) 32169962 3) 32169962 ( 10723321 9 ) 3574440 5). 3574440

714888 9 ) 397160 7

397160 ( 56737 9 ) 44129 9 ) 44129

4903 9 4903 ! 11 )

4903 (

446 9) 545 13 )

545 (

42 9) 61

61 (


15 )

Sum of the terms gives log. 2 = '301029995

Exam. 2. To find the log. of **

b Exam. 3. To find the log. of a – X. Exam. 4. To find the log. of 3. EXAM. 5. To find the log. of 5. EXAM. 6. To find the log, of 11.



T Inflexion in a curve, is that point of it which

E separates the concare from the convex part, lying between thetwo; or where the curve

A changes from concave to convex, or from convex to concave, on the same side of the curve. Such as the point e in the annexed figures, where the former of the two is concave


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