Page images
PDF
EPUB

2. The diameters of two concentric circles being 20 and 10; required the area of the ring between their circumferences. Ans. 235.62.

3. What is the area of the ring, the diameters of whose bounding circles are 6 and 4?

Ans. 15.708.

PROBLEM XVI.

To measure long irregular figures.

RULE.

Take the breadth in several places at equal distances. Add all the breadths together, and divide the sum by the number of them, for the mean breadth, which multiply by the length, for the area.

[ocr errors]

EXAMPLES.

1. The breadths of an irregular figure, at five equidis tant places being AD 81, mp 74, ng 9°2, or 10°1, BC 86; and the length AB 39; required the area.

[ocr errors][merged small][merged small][ocr errors][merged small][merged small]

COR. If DI be a perpendicular at the point D, then will the area of the ring be equal to that of a circle, whose radius is DI.

[ocr errors]

RULE 2.

Multiply half the sum of the circumferences by half the difference of the diameters, and the product will be the area.

8.1

74

9'2

10 I

8.6

5)43°4

8.68

39

7812

2604

338.52 answer.

2. The length of an irregular figure being 84, and the breadths at 6 places 17'4, 2016, 14°2, 16′5, 20°1, 24°3; what is the area? Ans. 1583'4.

PROBLEM XVII.

To find the circumference of an ellipse.*

RULE.t

Add the two axes together, and multiply the sum by 15708, for the circumference nearly.

EXAMPLES.

* For definitions of ellipse, parabola, and hyperbola, see CONIC SECTIONS.

† It will be evident, that this rule is very near the truth, if it be considered, that this arithmetical mean between the axes exceeds their geometrical mean; and, that the geometrical mean is the diameter of a circle, equal in area to the ellipse; which circle is of less ambit than the ellipse, or any other figure, of the

same area.

EXAMPLES.

1. Required the circumference of an ellipse, whose two axes are 70 and 50.

[blocks in formation]

2. What is the periphery of an ellipse, whose two axes

are 24 and 20?

Ans. 69°1152

PROBLEM XVIH.

To find the area of an ellipse.

RULE.

Multiply the transverse by the conjugate, and the product, multiplied by 7854, will be the area.

Or, Multiply 7854 by one axe, and the product by the other.

EXAMPLES.

1. To find the area of an ellipse, whose two axes are 70 and 50.

*7854'

* The demonstration of this rule is contained in that of the next problem.

7854

50

39'2700

70

2748 9000 answer.

2. What is the area of an ellipse, whose two axes are 24 and 18 ?

Ans. 339 2928.

PROBLEM XIX.

To find the area of an elliptic segment.

RULE.*

Divide the height of the segment by the axis of the ellipse, of which it is a part; and find, in the table of cir

cular

* DEMONSTRATION. Let the transverse diameter AB-a, the conjugate CD=c, AG=x, and EG=y; then, by the property of the curve, we shall have y —✔ax-x1, and the flux

ion of the area EAF=(yx)= -xx √ ax-x2. But x1ax—x2 —;

a

is known to express the fluxion of the corresponding circular segment, whose versed sine is x, and diameter a. Let the fluent of this expression, therefore, be denoted by A, and then the fluent of

[merged small][merged small][ocr errors][merged small][merged small][merged small]

COR. mean proportional between the two axes, and hence the rule is formed for the whole ellipse.

The ellipse is equal to a circle, whose diameter is a

cular segments, a circular segment having the same versed sine as this quotient. Then multiply continually together this segment and the two axes, for the area required.

1. What is the area

EXAMPLES.

of

an elliptic segment EAF, whose height AP is 20; the transverse AB being fo, and the conjugate CD 50?

[merged small][merged small][merged small][ocr errors][merged small]

2. What is the area of an elliptic segment, cut off parallel to the shorter axis, the height being io, and the axes 25 and 35? Ans. 162 0210.

3. What is the area of an elliptic segment, cut off parallel to the longer axis, the height being 5, and the axes 25 and 35 ? Ans. 97.8458.

PROBLEM XX.

To find the length of a parabolic curve.

RULE.*

square

of the ordinate add of the

square

of

To the "the absciss, extract the square root of the sum, and double

-it

* DEMONSTRATION. Let x any abscissa, y its ordinate, the parameter of the axe, and g. Then it is shewn

a

by

« PreviousContinue »