Area&vol.

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MATH 6001

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Areas and Volumes of Solids

Learning Outcome

When you complete this module you will be able to:

Calculate the volumes of rectangular objects, cylinders, and spheres and thesurface areas of cylinders and spheres.

Learning Objectives Here is what you will be able to do when you complete each objective:

1. Convert the commonly used units of volume.

2. Calculate the volume of a rectangular prism.

3. Calculate the surface area and volume of a cylinder.

4. Calculate the surface area and volume of a sphere.

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INTRODUCTION

Figures having three dimensions such as length, width and thickness are known assolids, and the methods used for finding volume and area of selected solids arediscussed in this module.

VOLUME

Volume is a measure of the three dimensional (height, width, length) capacity of an object. In S.I., the basic unit of volume is the cubic metre, abbreviated m3.

1 m3

= 1 m x 1 m x 1 m

In practice, cubic metres are quite large, so to avoid always having to use decimalor fractional quantities of cubic metres, we employ volume units which are

derived from the S.I. submultiples of the metre. For example:

1 cm3

= 1 cm x 1 cm x 1 cm(In past years, a cm3 was also commonly referred to as a c.c. for cubic centimetre)

1 mm3

= 1 mm x 1 mm x 1 mm

Note, that in converting from one set of volume units to another, we mustmaintain the "cubic" relationships. For example, if we wish to convert 1 m

3into

cm3 we must do the following:

1 m3 = 1 m x 1 m x 1 m= 100 cm x 100 cm x 100 cm= 1 000 000 cm3

(It is a common error for students to employ the linear relationship and incorrectlyconclude that 1 m3 = 100 cm3!)

Convert 0.5 m3

into cm3.

1 m3 = 1 m x 1 m x 1 m1 m

3= 100 cm x 100 cm x 100 cm

1 m3 = 1 000 000 cm3 0.5 m3 = 0.5 x 1 000 000 cm3

= 500 000 cm3 (Ans.)

(Again, notice that 0.5 m3 ≠ 50 cm3)

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One commonly used unit of volume in S.I. is the litre. By definition 1 L = 1 dm3 (dm = decimetre or 0.1 m). You may wish to verify for yourself that thisdefinition also means that:

1 L = 1000 cm3

and 1 m

3

= 1000 L

Volumes of Rectangular Prisms

The volume of a rectangular prism equals length x width x depth, each dimensionbeing taken in the same units, that is to say, all dimensions must be in centimetresto give volume in centimetres, all in metres to give volume in metres, and so on.

Example 1:

A rectangular tank is 6 m long and 4 m wide and 2 m deep. What is its capacity

in m

3

?

Solution: Volume = length x width x depth

= 6 x 4 x 2

= 48 m3

(Ans.)

Example 2:

The base of a 27.3 kL storage tank is 4 m by 4.2 m. To what height must the tank be built?

Solution:

27.3 kL = 27.3 m3

Then divide the volume in m3

by the area of the base in m2

to get the height in m.

Height = VolumeArea

= 27.34 x 4.2

= 1.625 m (Ans.)

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SURFACE AREA OF CYLINDERS

If the shell or lateral surface of a cylinder could be unrolled and spread out asshown in Fig. 1, it would form a rectangle of length equal to the circumference

(π D) of the cylinder end and a width equal to the height (h) of the cylinder.

The total surface area of the cylinder is the lateral surface area plus the area of thetwo ends or bases.

Lateral surface area = π Dh

Area of two bases = 2 x (0.7854 D2)

Total surface area = π Dh + 2 x (0.7854 D2)

Note: 0.7854 = π /4

Figure 1

Surface Area

Example 3:

A cylindrical tank is 5 m in diameter and 22 m long. Find its lateral surface area.

Solution:

Lateral surface area = π Dh

= 3.1416 x 5 m x 22 m

= 345.6 m2 (Ans.)

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Example 4:

Calculate the surface area of a cylindrical fuel tank with flat ends if the tank is 10m long and 6 m in diameter.

Solution:

Lateral (shell) surface area ( AS):

AS

= π Dh

= 3.1416 x 6 m x 10 m

= 188.5 m²

Area of the two ends ( A E ):

A E

= 2 x (0.7854 D2)

= 2 x [0.7854 x (6 m)2]

= 56.5 m2

Total surface area ( AT ):

AT

= AS

+ A E

= 188.5 m2

+ 56.5 m2

= 245 m2 (Ans.)

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VOLUME OF CYLINDERS

The volume (V ) of a cylinder can be found by multiplying the area of the base(0.7854 D2) by the height, altitude, or length (h):

V = 0.7854 D2

h

Example 5:

Find the volume of a cylindrical tank 10 m in diameter and 30 m long.

Solution:

V = 0.7854 D2h

= 0.7854 x (10 m)2 x 30 m

= 0.7854 x 100 m2 x 30 m

= 2356.2 m3

= 2356 m3 (Ans.)

Example 6:

Calculate the volume of the fuel tank in Example 4.

Solution:

V = 0.7854 D2h

= 0.7854 x (6 m)2 x 10 m

= 282.7 m3 (Ans.)

SURFACE AREA OF SPHERES

A sphere is a solid bounded by a curved surface called the circumference, every

point of which is the same distance from a center point.

The radius (r ) of a sphere is a straight line drawn from the center point to thecircumference and is equal to one-half the diameter ( D).

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The surface area of a sphere can be found by multiplying the square of the radius

by 4π .

Surface Area ( A) = 4 πr 2

Example 7:

Find the surface area of a sphere having a diameter of 8 cm.

Solution:

r = D / 2

= 8 cm / 2

= 4 cm

And:

A = 4πr 2

= 4 x 3.1416 x (4 cm)2

= 4 x 3.1416 x 16 cm2

= 201 cm2 (Ans.)

Example 8:

Find the amount of insulation required to completely cover a chlorine storagesphere having a diameter of 9 m.

Solution:

A = 4π r 2

= 4 x 3.1416 x (4.5 m)2 (since r = d /2)

= 254.5 m2 (Ans.)

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VOLUME OF SPHERES

The volume (V ) of a sphere can be found by multiplying the cube of the radius by

(4/3)π .

V = (4/3)π r 3

Example 9:

A cast iron ball has a radius of 1 m. If cast iron has a density of 7210 kg/m3, findthe mass of the ball.

Solution:

V = (4/3) πr 3

= (4/3)π x (1 m)3

= (4/3) x 3.1416 x 1 m3

= 4.189 m3

Therefore:

Mass = 4.189 m3 x 7210 kg/m3

= 30 203 kg (Ans.)

Example 10:

A boiler drum is equipped with hemispherical heads (i.e. the two heads togethermake a sphere.) The diameter of the drum is 80 cm and the drum is 5 m in length.

Calculate:(a) The area of insulation required to cover the entire drum.(b) The total volume of the drum.

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Solution:

(a) The total surface area of the boiler drum is equal to the lateral area(shell area) plus the surface area of the sphere which makes up theends.

Therefore:

A = π Dh + 4π r 2

= (3.1416 x 0.8 m x 5 m) + [4 x 3.1416 x (0.4 m)2]

= 12.57 m2 + 2.01 m2

= 14.58 m2 (Ans.)

Note that the 80 cm is changed to 0.8 m.

(b) The total volume of the drum is equal to the total volume of thecylinder plus the total volume of the sphere. Therefore the totalvolume is:

V = 0.7854 D2h + (4/3)π r 3

= (0.7854 x (0.8 m)2 x 5 m) +[(4/3) x 3.1416 x (0.8 m /2)3]

= 2.513 m3 + 0.268 m3

= 2.781 m3 (Ans.)

MORE EXAMPLES

Most problems involve a combination of two or more concepts. The followingexamples utilize information from this module as well as other modules andwhere possible, reference is made to the module containing the appropriateinformation. An effort is made to show the logical steps used to solve theproblem; but do not make the assumption that this is the only approach which can

be used.

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Example 11:

A hexagonal shaft has sides of 4 cm and an overall length of 1.3 m. The shaft ismade of mild steel, which has a density of 7860 kg/m3. Find the mass of the shaft.

*Density, relative density, and area of a hexagon are discussed in other modules.

Solution:

The first step is to find the volume of the shaft in m3; m3 is used since the densityis provided in kg/m3. The volume of the hexagonal shaft is equal to the cross-sectional area of the hexagon times the overall length of the shaft.

V = 2.6 x s2 x l

= 2.6 x (0.04 m)2 x 1.3 m

= 2.6 x 0.0016 m2 x 1.3 m

= 0.0054 m3

The mass of the shaft can then be found by multiplying the volume by the density

m = 0.0054 m3 x 7860 kg/m3

= 42.5 kg. (Ans.)

Example 12:

Calculate the stroke of a single cylinder, positive displacement pump if thecylinder is 3 cm in diameter and the pump delivers 3392.9 cm3 /min. The pump isoperating at 60 effective strokes per minute.

Solution:

The first step is to find the volume delivered for each stroke of the pump

V = 3392.9 cm3 /min

60 strokes/min

= 56.55 cm3 /stroke

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The volume of a cylinder is:

V = 0.7854 d 2h

56.55 cm3 = 0.7854 (3 cm)2 h

Transposing:

h = 56.55 cm3 0.7854 (3cm)2

= 8 cm (Ans.)

Example 13:

Calculate the cost of painting a sphere which is 12 m in diameter. A litre of paintcovers 8 m2 and costs $4.50.

Solution:

The surface area of a sphere is:

A = 4 πr 2

= 4 x 3.1416 x (6m)2

= 452.4 m2

Since each litre covers 8 m2, the total number of litres used is the total areadivided by the area covered by one litre.

# of litres = 452.4 m2 8 m 2

= 56.55 L

= 57 L

The total cost will be the number of litres times the cost per litre:

Cost = 57 L x $4.50/L

= $256.50 (Ans.)

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Self Test

After completion of the self-test, check your answers with the answer guide thatfollows.

1. What is the total surface area of a closed cylinder that is 10 cm high and 4 cmin diameter?

2. Find the lateral area (in m2) of a steam pipe 15 cm in diameter and 3 m long.

3. Find the volume in cubic centimetres of a cylinder having a radius of 3 cmand a height of 8 cm.

4. A rectangular piece of sheet iron 625 cm2

in area is rolled to form a cylinder9 cm in diameter. Find the height and the volume of this cylinder.

5. Find the surface area and the volume of spheres having the following radii:

a) 5 cm

b) 8 cm

c) 2 m

6. Find the mass of a lead ball having a diameter of 15 cm. A cubic metre of lead has a mass of 11 422 kg.

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Self Test Answers

1. 150.8 cm2

2. 1.414 m2

3. 226.2 cm3

4. 22.1 cm; 1406 cm3

5. (a) 314 cm2; 523.6 cm

3

(b) 804 cm2; 2145 cm3

(c) 50.27 m2; 33.5 m

3

6. 20.18 kg

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Notes:

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