Lesson Notes By Weeks and Term v3 - Senior Secondary 3
Surface Area and volume of sphere
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Surface Area and volume of sphere
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Subject: General Mathematics
Class: Senior Secondary 3
Term: 1st Term
Week: 3
Theme: Geometry
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Find the surface area ofa sphere Find the volume of as phere
V = (4/3) × (22/7) × (21 × 21 × 21) V = (4/3) × (22/7) × 9261 V = (4 × 22 × 9261) / (3 × 7) V = (88 × 9261) / 21 V = 815088 / 21 V = 38800.86 cm3 (approximately) Alternatively, simplify before multiplying: V = (4/3) × (22/7) × 21 × 21 × 21 V = 4 × (22/7) × 7 × 21 × 21 (after dividing one 21 by 3) V = 4 × 22 × 21 × 21 (after dividing 7 by 7) V = 88 × 441 V = 38808 cm3 Therefore, the volume of the spherical storage container is 38808 cubic centimeters.
Example 4: Finding the Radius from Volume The volume of a spherical buoy used for navigation in the Lagos lagoon is 33.51 m
3. Find its radius. (Use π = 3.14).
Solution:
1. Identify the given values: Volume (V) = 33.51 m3.
2. Recall the formula: V = (4/3)πr3.
3. Substitute the values: 33.51 = (4/3) × 3.14 × r3 33.51 = (12.56/3) × r3 33.51 × 3 = 12.56 × r3 100.53 = 12.56 × r3
4. Solve for r3: r3 = 100.53 / 12.56 r3 ≈ 8.0039...
5. Find r by taking the cube root: r = 3√8 r = 2 m Thus, the radius of the spherical buoy is approximately 2 meters. A sphere is a perfectly round geometrical object in three-dimensional space that is the surface of a ball. It is defined as the set of all points that are equidistant from a given point, which is the centre of the sphere. The constant distance from the centre to any point on the surface of the sphere is called the radius (r). The diameter (d) is the distance between any two points on the sphere passing through its centre (d = 2r). 2.1 Surface Area of a Sphere The surface area of a sphere is the total area of its outer surface. Imagine peeling an orange in one continuous strip and trying to cover a flat surface – this relates to its surface area. Mathematically, the surface area of a sphere is equivalent to the area of four great circles with the same radius as the sphere. A great circle is a circle on the sphere whose plane passes through the centre of the sphere. The formula for the surface area of a sphere is: A = 4πr2 Where: A = Surface Area π (pi) ≈ 3.142 or 22/7 (depending on the precision required or specified in the question) r = Radius of the sphere Worked
Examples: Example 1: Calculating Surface Area of a Spherical Water Tank A community in Kaduna plans to construct a spherical water tank with a radius of 3.5 meters. Calculate the surface area of the tank that needs to be painted. (Use π = 22/7).
Solution:
1. Identify the given values: Radius (r) = 3.5 m.
2. Recall the formula for surface area of a sphere: A = 4πr2.
3. Substitute the values into the formula: A = 4 × (22/7) × (3.5)2 A = 4 × (22/7) × (3.5 × 3.5) A = 4 × (22/7) × 12.25 A = 4 × 22 × (12.25 / 7) A = 88 × 1.75 A = 154 m2 Therefore, the surface area of the water tank that needs to be painted is 154 square meters.
Example 2: Finding the Radius from Surface Area The surface area of a spherical football is 1256 cm
2. Find the radius of the football. (Use π = 3.14).
Solution:
1. Identify the given values: Surface Area (A) = 1256 cm2.
2. Recall the formula: A = 4πr2.
3. Substitute the values: 1256 = 4 × 3.14 × r2 1256 = 12.56 × r2
4. Solve for r2: r2 = 1256 / 12.56 r2 = 100
5. Find r by taking the square root: r = √100 r = 10 cm Thus, the radius of the football is 10 cm. 2.2 Volume of a Sphere The volume of a sphere is the amount of space it occupies. Imagine filling a spherical container with water or sand – the amount of water/sand represents its volume. The formula for the volume of a sphere can be conceptually related to stacking infinitesimally thin discs or using calculus principles.
The formula for the volume of a sphere is: V = (4/3)πr3 Where: V = Volume π (pi) ≈ 3.142 or 22/7 r = Radius of the sphere Worked
Examples: Example 3: Calculating Volume of a Spherical Storage Container A food processing company uses spherical containers to store a new liquid product. If a container has a radius of 21 cm, calculate its volume. (Use π = 22/7).
Solution:
1. Identify the given values: Radius (r) = 21 cm.
2. Recall the formula for volume of a sphere: V = (4/3)πr3.
3. Substitute the values: V = (4/3) × (22/7) × (21)3 V = (4/3) × (22/7) × (21 × 21 × 21) V = (4/3) × (22/7) × 9261 V = (4 × 22 × 9261) / (3 × 7) V = (88 × 9261) / 21 V = 815088 / 21 V = 38800.86 cm3 (approximately) Alternatively, simplify before multiplying: V = (4/3) × (22/7) × 21 × 21 × 21 V = 4 × (22/7) × 7 × 21 × 21 (after dividing one 21 by 3) V = 4 × 22 × 21 × 21 (after dividing 7 by 7)
Materials: Spherical objects of various sizes (e.g., football, orange, globe, marble). Measuring tape or string, ruler. Measuring cylinder or beaker, water, sand (for practical volume demonstration). Chart paper or whiteboard. Markers. Calculators (optional, but encouraged for complex calculations).
Introduction (10 minutes): Teacher Activity: Begin by displaying various spherical objects. Ask students to identify them and discuss where they encounter such shapes in daily life (e.g., 'bola' (ball), fruits, cooking pots).
Introduce the topic: "Today, we will learn how to measure the surface area and volume of these spherical objects." Student Activity: Observe the objects, participate in the discussion, and brainstorm real-life examples of spheres.
Development (30-40 minutes): Activity 1: Understanding and Calculating Surface Area (15-20 minutes)
Teacher Activity: Define a sphere, its radius, and diameter. Introduce the formula for the surface area of a sphere: A = 4πr
2. Explain the components of the formula. Demonstrate Example 1 (Spherical Water Tank) step-by-step on the board, emphasizing correct substitution and units. Guide students through Example 2 (Football Surface Area) by asking questions and allowing them to suggest steps.
Student Activity: Take notes on the definition and formula. Actively follow the worked examples, ask clarifying questions. Attempt to solve Example 2 as guided by the teacher.
Activity 2: Practical Demonstration of Surface Area (Optional, 5 minutes)
Teacher Activity: Take an orange. Cut it in half. Trace the circular cross-section on a paper to get the area of a great circle (πr2). Explain that the surface area of the whole orange is approximately 4 times this area. This provides a visual intuition for A=4πr
2. Student Activity: Observe the demonstration and relate it to the formula.
Activity 3: Understanding and Calculating Volume (15-20 minutes)
Teacher Activity: Introduce the formula for the volume of a sphere: V = (4/3)πr
3. Explain the components and contrast it with the surface area formula. Demonstrate Example 3 (Spherical Storage Container) step-by-step, paying attention to cubed radius and fractional multiplication. Guide students through Example 4 (Spherical Buoy Volume).
Student Activity: Take notes on the volume formula. Follow the worked examples. Attempt to solve Example 4 as guided.
Activity 4: Practical Demonstration for Volume (15-20 minutes) - Addresses Evaluation Guide Point 2 & 4 Teacher Activity: Prepare a spherical object (e.g., a hard plastic ball or an orange that can be used for displacement), a measuring cylinder, and water.
Method 1 (Displacement): Show how to measure the volume of a solid sphere using water displacement. Measure the radius (r) of the spherical object using a string and ruler. Pour a known volume of water into the measuring cylinder (V1). Carefully submerge the spherical object into the water. Read the new volume (V2). The volume of the sphere (V_practical) = V2 - V
1. Using the measured radius (r), calculate the volume of the sphere using the formula V = (4/3)πr
3. Compare V_practical with V_formula. Discuss possible reasons for differences (measurement errors, object not perfectly spherical, etc.).
Method 2 (Filling a Spherical Container): If a hollow spherical container is available. Measure its internal radius. Fill it completely with sand or water. Carefully pour the sand/water into a measuring cylinder to read the practical volume. Compare with the calculated volume using the formula.
Student Activity: Work in small groups (if feasible) to assist with measuring and recording. Observe the practical demonstration intently. Record the practical volume. Calculate the theoretical volume using the formula and compare the results. Discuss the discrepancies with the teacher.
Consolidation (5-10 minutes): Teacher Activity: Summarize the key formulas for surface area and volume of a sphere. Address any lingering questions. Distribute guided practice questions.
Student Activity: Ask final questions, prepare for guided practice. Students are encouraged to work through these problems, showing all steps.
Question 1: A spherical orange from a farm in Benue State has a radius of 4.2 cm. Calculate its surface area. (Use π = 22/7).
Solution 1: Given: r = 4.2 cm Formula: A = 4πr2 Substitution: A = 4 × (22/7) × (4.2)2 A = 4 × (22/7) × (4.2 × 4.2) A = 4 × (22/7) × 17.64 A = 88 × (17.64 / 7) A = 88 × 2.52 A = 221.76 cm2
Commentary: This question assesses the direct application of the surface area formula. It's straightforward and checks for correct substitution and calculation.
Question 2: A spherical gas tank in a factory in Port Harcourt has a diameter of 10 meters. What is the volume of gas it can hold? (Use π = 3.142).
Solution 2: Given: Diameter (d) = 10 m.
Calculate radius: r = d/2 = 10/2 = 5 m.
Formula: V = (4/3)πr3 Substitution: V = (4/3) × 3.142 × (5)3 V = (4/3) × 3.142 × 125 V = (4 × 3.142 × 125) / 3 V = 1571 / 3 V = 523.67 m3 (to 2 decimal places)
Commentary: This question requires an initial step of converting diameter to radius, a common potential error point for students. It then tests the volume formula application.
Question 3: A spherical ball used in a traditional Nigerian game has a volume of 904.32 cm
3. Find the radius of the ball. (Use π = 3.14).
Solution 3: Given: V = 904.32 cm3 Formula: V = (4/3)πr3 Substitution: 904.32 = (4/3) × 3.14 × r3 904.32 = (12.56 / 3) × r3 Isolate r3: r3 = (904.32 × 3) / 12.56 r3 = 2712.96 / 12.56 r3 = 216 Find r: r = 3√216 r = 6 cm
Commentary: This question requires algebraic manipulation to find the radius when the volume is given, a good test of inverse operations.
Example 1: Calculating Surface Area of a Spherical Water Tank
A community in Kaduna plans to construct a spherical water tank with a radius of 3.5 meters. Calculate the surface area of the tank that needs to be painted. (Use π = 22/7).
Solution:
Identify the given values: Radius (r) = 3.5 m.
Recall the formula for surface area of a sphere: A = 4πr².
Substitute the values into the formula:
A = 4 × (22/7) × (3.5)²
A = 4 × (22/7) × (3.5 × 3.5)
A = 4 × (22/7) × 12.25
A = 4 × 22 × (12.25 / 7)
A = 88 × 1.75
A = 154 m²
Therefore, the surface area of the water tank that needs to be painted is 154 square meters.
Example 2: Finding the Radius from Surface Area
The surface area of a spherical football is 1256 cm². Find the radius of the football. (Use π = 3.14).
Water and Fuel Storage: In many Nigerian communities and industries, water and fuel are stored in large spherical tanks. Understanding the surface area helps engineers calculate the amount of material needed for construction, the cost of painting, or the heat transfer characteristics. The volume calculation determines the storage capacity, which is vital for planning water supply or fuel reserves. For example, a village water project might use spherical tanks, and calculating their volume ensures adequate supply for the population.
Food Processing and Agriculture: Nigerian farmers and food processors deal with spherical produce like oranges, tomatoes, and certain varieties of yams. Knowledge of volume helps in estimating yield, designing efficient packaging, and calculating storage requirements in silos or warehouses. For instance, knowing the average volume of an orange allows a farmer to estimate the total quantity of juice that can be extracted from a harvest.
Sports and Manufacturing: The manufacturing of sports equipment like footballs, basketballs, and shot put balls in Nigeria relies on precise calculations of surface area and volume. Surface area determines the amount of leather or synthetic material needed, while volume influences the ball's weight, air pressure, and performance characteristics according to international standards (e.g., FIFA regulations for footballs).