Unit 04: Fundamentals of Solid Geometry  Shapes and Volumes


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1 Unit 04: Fundamentals of Solid Geometry  Shapes and Volumes Introduction. Skills you will learn: a. Classify simple 3dimensional geometrical figures. b. Calculate surface areas of simple 3dimensional figures and complex figures that can be obtained by combining such simple geometrical figures. c. Calculate volumes of simple 3dimensional figures and complex figures that can be obtained by combining such simple geometrical figures. d. Solve real world problems involving surface areas and volumes. Some Concepts. Polyhedron A polyhedron is a solid bounded by planes, for example, a cube is bounded by six planes and is a hexahedron. A tetrahedron (see figure on the right) is formed by four triangular faces. When three or more planes meet at a point, it is called a vertex. The intersections of planes are called the edges and the sections of planes are called the faces of the polyhedron. Tetrahedron We will be concerned mostly with two types of polyhedrons the prism and the pyramid. Prism Two of the prism s faces are congruent polygons contained in two parallel planes. Planes through the sides of the congruent polygons form the other faces of a prism. A rectangular prism with some relevant terms is shown below as an example. A
2 parallelepiped is a prism whose bases are parallelograms. A right circular cylinder can be viewed as a prism whose base is a circlea polygon with infinite number of sides. A Rectangular Prism. The figure shows a rectangular prism of altitude = c and base of dimensions a and b. c a b Pyramid. A pyramid is a 3dimensional figure formed by a polygon (called the base) and three or more triangular planes meeting at a common point (called the vertex). The perpendicular from the vertex to the base is called the altitude (h) of the pyramid. If the base is a regular polygon and the altitude meets the base at the center, the pyramid called regular or right. The slant height (L) of a regular pyramid is the altitude of one of the lateral faces. A right circular cone can be viewed as a pyramid whose base is a circle a polygon with infinite sides. vertex h L Base Sphere A spherical surface is a curved surface obtained by rotating a semicircle about its diameter. All points on a spherical surface are equidistant from a common point called the center. A sphere is a solid bounded by a spherical surface. Small Circle Great Circle A great circle of a sphere is any circle obtained by the intersection of the sphere by a plane containing the center of the sphere. A small circle is
3 obtained when the intersecting plane does not pass through the center.
4 Surface Areas of Solids 1. Prism The lateral area S of a prism is the sum of the areas of the lateral faces and equals the product of the base perimeter P and the altitude h: S = Ph. Thus, for a cylinder, S = 2πrh. For a rectangular prism (see figure on previous page) altitude c and base dimensions of a and b the lateral area is given by, S = 2(a+b)c. The total area A of a prism is simply the sum of the lateral area S and the two basal areas, B 1 and B 2. A = S + B 1 + B 2 For a right prism, B 1 = B 2 = B, therefore A = S+2B. For a cylinder A= 2πrh + 2πr 2 = 2πr(h + r) For a right rectangular prism of edges, a, b, and c, the total area A = 2(a+b)c + 2ab = 2ab +2bc + 2ca. 2. Pyramid Lateral area, S = ½ LP, where L is the slant height and P is the base perimeter. Thus, S = πrl for a cone of base of radius r. For a pyramid, the total surface area A = S + B, where B is the basal area of the pyramid. For a cone A = πrl + πr 2 = πr(l + r) 3. Sphere The surface area, A of a sphere of radius R is given by, A = 4 πr 2 Volumes of Solids 1. Prism The volume of a prism is the product of its basal area B and altitude h. For a cylinder V =Bh =πr 2 h. Similarly for a right rectangular prism V = abc, where a, b, and c are edge lengths of the prism. 2. Pyramid
5 For a pyramid, the volume V = (1/3)*B*h. Thus, for a cone V = πr 2 h/3. 3. Sphere Volume of a sphere,v = 4!r 3,where R is the radius of the sphere. 3
6 Solved Examples. Example 1. A right circular cone has a base of radius r = 12.0 cm and an altitude of H = 16.0cm. Determine the cone s: [a] Lateral area [b] Total area [c] Volume Solution: The cone s slant height forms the hypotenuse of the right angle triangle whose base is the radius and altitude is the altitude of the cone. Therefore, using Pythagorean theorem, the slant height, L = ( ) 1/2 = 20.0cm H r L a. The total area S = πrl = (3.14)*(12.0)*(20.0) = cm 2 (754 cm 2 rounded off to three significant figures. b. The total area = πr(l + r) = (3.14)*(12.0)*( ) = cm 2 (1206 cm 2 when approximated to the nearest whole number or to four significant digits) c. The volume, V = πr 2 h/3 = (1/3)*[(3.14)( )(16.0)] = (2412 cm 3 when approximated to the nearest whole number)
7 Example 2 Consider a right triangular prism of basal edges 6.0 cm, 8.0 cm, and 10.0 cm and an altitude of 15.0 cm. Note that the base of the prism is a right angle triangle. Calculate the prism s: a. Lateral area b. Total area c. Volume 6 cm 10 cm 8 cm Solution a. The perimeter of the base of the prism is 15 cm P = = 24.0 cm. Therefore the lateral area S = Ph = (24.0)(15.0) = 360 cm 2. b. The total area, A = S + area of the two triangular bases = *(1/2)(base)(altitude) = *(1/2)*(6.0)(8.0) = = 408 cm 2. Example 3 Find the mass of a spherical wooden shell whose inner and outer radii are, respectively, 10.0 cm and 12.0 cm. The mass density ρ of the wood is 20.0 kg/m 3 (Note: the mass density is defined as mass per unit volume ρ = M/V ). Express your answer in kg. Solution. The spherical shell can be viewed as a sphere of radius 12.0 cm from which a concentric sphere of radius 10.0 cm has been removed. [Onehalf of a spherical shell is depicted in the diagram on the right] The volume of the spherical shell : V = (4/3) πr 3 (4/3) πr 3 = (4/3) π(r 3 r 3 ) = (4/3)(3.14)[(12.0) 3 (10.0) 3 } 12 cm 10 cm
8 = cm 3 = x 10 6 m 3 = 3.05 x 10 3 m 3 The mass M = ρv = (20.0 kg/m 3 )(3.05 x 10 3 m 3 ) = kg. Example 4 The mass of the earth is given to be 6.0 x kg. Determine the average mass density ρ of the earth assuming it to be a perfect sphere of radius R = 6.4x10 3 km. (Express your answer both in kg/m 3 and gm/cm 3 units). Solution. M = 6.0 x kg. ; V = (4π/3)(6.4 x 10 6 ) 3 Substitute these values of M and V in ρ = M/V to get, R = 5.5 x 10 3 kg/m 3 = 5.5 gm/cm 3. Note: we have used the following conversion factors: 1 kg = 10 3 gm, and 1m 3 = 10 6 cm 3 Example 5 The magnitude of the acceleration due to gravity, g at the surface of a planet of mass M and radius R is given by: g = GM, where G = 6.67x1011 Nm 2 /kg 2 is the R 2 gravitational constant. Calculate the value of g at the surface of the Earth. Solution: g = GM 6.67!10"11 = = 9.8m / R 2 (6.4!10 "11 s2 2 ) Example 6 The acceleration due to gravity at the surface of the Moon is g m = 1.6 m/s 2. Assuming the Moon has the same average mass density as the Earth, calculate the radius of the Moon. Solution g m = GM m R m 2 R m = = 4! R 3 mg" = 4! R mg" 2 3R m 3 3g m 4!G" = 3#1.6 4 # 3.14 # 6.67 #10 $11 # 5.5 #10 3 = 1.05 #106 m
9 (Something to think about: the mean radius of the Moon is known to be 1.74x10 6 m. From your calculations, what can you say about the density of the Moon?) Example 7 You have a 25.0kg chunk of copper. You would like to melt it and draw it into a wire of circular crosssection of radius r. Calculate the length of the wire when [a] r = 0.30cm, and [b] r = 0.15cm. The mass density of copper is 9000 kg/m 3. Solution: M =!V =!"r 2 L or the length of the wire, L = [a] For r = 0.30 cm L = ! 3.14! (3.0!10 "3 ) 2 = 98.3m [b] For r = 0.15 cm 25.0 L = 9000! 3.14! (1.5!10 "3 ) = 393m 2 M!"r 2. Example 8 Water flows through a cylindrical pipe of radius R = 1.25cm at the rate of 20.0 liters/sec. What is the speed of water in the pipe? a b A D x Solution: Suppose the water flows inside the cylinder from a to b in time Δt. Then, the water that flowed through the crosssection A at a is contained in the cylindrical volume ΔV = A Δx. Therefore, the flow rate (Vol./time) is:!v!t = A!x!x = As, where s =!t!t! s = 1 $ " # A% & 'V 't =! 1 $ " # ((1.25 )10 *2 ) 2 % & 20.0 )10*3 m 3 s *1 = 40.7ms *1 is the speed of the water flowing through the pipe.
10 Example 9 A hypodermic syringe contains a medicine with the density of water. The syringe plunger (P) is of radius R = 0.5cm, and the syringe needle (N) is of inner radius r = 0.2mm. The medicine is delivered by moving the plunger 2.4 cm in 2.0 s at a constant rate. What is the speed of the medicine at the needle s tip? Plunger Needle Solution: Volume ejected from the needle in one second: V = A P s P = A N s N, where A P and A N are, respectively, the crosssectional areas of the plunger and the needle, and s P and s N are the speeds in the barrel and in the needle, respectively. s N = (A P / A N ) s P = (R/r) 2 s P = (0.5/0.02) 2 *(2.4/2.0) = 750 cm/s = 7.50m/s.
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