(5) A bowl is in the shape of a zone of a sphere; the radius of the top is 16 in. and of the bottom 4 in.; the greatest depth is 10 in. Find the number of cubic feet of water that it will hold.

(6) A bowl is in the shape of a zone of a sphere; it measures 2 ft. across at the top, and 10 in. across at the bottom; and its greatest depth is 13 in. Find the number of gallons of water that it will hold, when 1 cubic foot contains 6 gallons.

ring may be defined to be a

cylinder bent into a ring.

The length of a circular ring is represented by the dotted line, and is equal to the circumference of a circle of which FH or (AB+thickness of ring) is the diameter.

The length of the ring is

also equal to half the sum of the outer and inner circumferences.

The thickness of the ring AC or BD=half the difference of the outer and inner diameters.

The outer diameter

of the ring.

inner diameter + twice the thickness

The inner diameter=outer diameter-twice the thickness of the ring.

RULES. (1) To find the volume of a circular ring.

Multiply the area of a circular section of the ring (which is found by multiplying the square of the thickness of the ring by 1) by the length of the ring.

(2) To find the area of the surface of a circular ring.

Multiply the circumference of a circular section of the ring (which is found by multiplying the thickness of the ring by 22) by the length of the ring.

Note. A circular ring may be converted into a straight rod, and its volume will then be found by the rule given for finding the volume of a cylinder.

of the outer and inner circumferences.

Example 1.-The inner diameter of a ring is 7 in., and its thickness is 1 in.; find its volume.

The outer diameter=7+2=9 in.

The area of a circular section of the ring = 12 × 1 sq. in.

The inner boundary of the ring=7 × 22=22 in.; and outer boundary=9 × 22=198 in.; therefore the length of the ring=(22+198)=254 in.

Hence, volume of ring=area of circular section x length =14×25=968=1931 cub. in.

Example 2.-The outer diameter of a ring is 10 in., and the inner diameter is 8 in.; find its volume.

10-8
2

Now, the thickness of the ring= =1 in.

The area of a circular section of the ring=12×1 =sq. in.

The length of the ring is the circumference of a circle whose diameter is FH, or AB+ thickness (see figure); that is, 9 x 22=198 in.

Then, volume of ring area of circular section x length =1×198=1089=2211 cub. in.

Example 3.-The inner diameter of a ring is 12 in., and its thickness is 2 in.; find the area of its surface.

The thickness of the ring is 2 in.; therefore the circumference of a circular section of it=2 × 23=44 in.

And the length of the ring is the circumference of a circle whose diameter is FH, or AB+thickness of ring, or 14 in.; and is therefore=14 x 22=44 in.

Hence, surface of ring=circumference of circular section x length of ring

Formulæ

=44x44-1936 sq. in.=2764 sq. in.

I. Volume area of circular section x length.

II. Surface=circumference of circular section x length.

EXAMPLES.

(1) The inner diameter of an iron ring is 12 in., and its thickness is 3 in.; find how many cubic inches of iron were used in its construction.

(2) The inner diameter of a ring is 16 in., and its thickness is 2 in.; find its volume.

(3) The inner diameter of a ring is 21 in., and its outer diameter is 24 in.; find its volume.

(4) The inner diameter of an iron ring is 14 in., and its thickness is 3 in.; find its weight, when 1 cubic inch of iron weighs 4 oz. avoirdupois.

(5) Find the cost of painting a circular ring whose outer diameter is 16 ft., and inner diameter is 14 ft., at 3d. per sq. ft.

(6) Iron weighs 4 oz. avoirdupois per cubic inch; find the weight of an iron ring 2 ft. thick, and whose outer diameter is 15 ft.

XXIX. IRREGULAR SOLIDS.

Definition.-An irregular solid is one which is irregular in shape, and is not included under any of the previous definitions—such as many blocks of stone, pieces of iron, &c.

RULES. (1) To find the volume of an irregular solid, which is heavier than water, by placing it in a vessel of known capacity.

(a) Put the solid into any vessel of convenient form, such as a rectangular cistern or cylinder. Pour into the vessel as much water as will quite cover the solid. Take out the solid, and notice how much the water sinks in consequence.

Then

the volume of the solid is equal to the volume of a column of water having for its base the base of the vessel, and for its height the distance through which the water has sunk by removing the solid.

(b) Or, fill the vessel full of water at the first; then put in the solid gently, and measure the volume of the water that runs over the sides of the vessel. (The volume of the water running over=volume of the solid.)

(2) To find the volume of an irregular solid from its weight.

Divide the weight of the solid, brought into ounces, by the weight of a cubic inch of that particular substance; and the quotient is the number of cubic inches in the solid.

(3) To find approximately the volume of an irregular solid by different measurements.

Take breadth and depth of the solid at different points equally distant from each other. Take the

average of these for the mean breadth and mean depth. Then the volume of solid mean breadth

x mean depth x length.

=

(4) To find the capacity of any vessel.

Weigh the vessel when empty, and then weigh it full of water; the difference of these two weights is the weight of the water in the vessel. Divide the weight of water, brought into ounces, by 1000 oz. ; and the quotient is the volume of the vessel in cubic feet.

EXAMPLES.

(1) The length of a cubical vessel is 3 ft.; an irregularlyshaped stone is put into it, and water poured into the vessel until the solid is quite covered: in taking out the stone, the water sinks 7 in. Find the volume of the stone.

(2) In a cylindrical vessel, whose diameter is 14 in., is placed an irregular piece of iron; water is then poured in until the solid is covered: on taking out the iron, it is found that the water in the vessel has sunk 11⁄2 in. Find the volume of the iron.

(3) An irregular piece of granite is placed in a rectangular vessel whose length is 4 ft. 2 in. and breadth 3 ft.; water is poured into the vessel until the stone is covered: when the stone is taken out, the water sinks 8 in. Find the volume of the stone.

(4) When a body entirely immersed in water is taken out of a vessel, it is found that it will require 24 gallons more to raise the water in the vessel to the same level as it was before; find the volume of the solid, when 1 cubic foot contains 6 gallons of water.

(5) Into a vessel, which is full of water, is placed a block of marble; it is found that 3 gallons of water have run over the sides of the vessel. Find the volume of the marble, when 1 cubic foot contains 6 gallons of water.