Surface Areas and Volumes · Lesson 3 of 4
Volume of a Combination of Solids
“Combining a cylinder and a hemisphere sounds like advanced mathematics until you realize you’re just trying to figure out how much actual ice cream fits in the cone.”
• By the end of this lesson, you should be able to. • Calculate the volume and capacity of combined 3D solids. • Understand why volumes are strictly additive when shapes are attached. • Calculate remaining volume when cavities are carved or scooped out. • Solve complex real-world application problems including industrial sheds, gulab jamuns, pen stands, and displaced water volumes.
When we work with combined three-dimensional shapes, finding surface area can sometimes be tricky because we must check which faces are actually visible and which surfaces are hidden where two solids are joined.
Volume, however, is usually more straightforward.
Volume tells us how much three-dimensional space an object occupies. It can also describe how much a hollow object can hold. For example, the volume of a water tank tells us how much water it can contain, while the volume of a solid block tells us how much space the block takes up.
When an object is made by joining two or more solids, we usually find its total volume by adding the volumes of the individual parts. If a portion has been cut out or removed, we find the remaining volume by subtracting the removed part.
So, the main idea is simple: break the object into familiar solids, calculate the volume of each part, and then add or subtract as required. This makes volume problems much easier to understand, even when the overall object looks complicated.
When two solid shapes are joined together, no space is lost or destroyed. The total volume of the combined solid is simply the sum of the individual volumes of its parts.
Type 1: Architectural Shed (Cuboid + Half Cylinder)
Problem
Shanta runs an industry in a shed shaped like a cuboid surmounted by a half-cylinder. The base is 7 m × 15 m, and height of cuboid is 8 m. Find the volume of air inside the empty shed. If machinery occupies 300 m³ and 20 workers occupy 0.08 m³ each, find the air volume when workers and machinery are present. (Take π = 22/7)
- 1.Step 1: Find Volume of cuboidal base = l × b × h = 15 × 7 × 8 = 840 m³.
- 2.Step 2: For half-cylinder: diameter = 7 m (so radius r = 3.5 m) and length/height = 15 m.
- 3.Step 3: Volume of half-cylinder = (1/2) × (πr²h) = (1/2) × (22/7) × (3.5)² × 15 = 288.75 m³.
- 4.Step 4: Total capacity of empty shed = Volume of cuboid + Volume of half-cylinder = 840 + 288.75 = 1128.75 m³.
- 5.Step 5: Calculate occupied space: Machinery = 300 m³. 20 Workers = 20 × 0.08 = 1.6 m³.
- 6.Step 6: Remaining air volume = 1128.75 - (300 + 1.6) = 1128.75 - 301.6 = 827.15 m³.
Type 2: Actual vs Apparent Capacity (Glass with Raised Base)
Some containers have internal protrusions (like a raised hemispherical bottom in a glass) that reduce the actual liquid holding capacity compared to what it visually appears to hold.
Problem
A cylindrical juice glass has inner diameter 5 cm and height 10 cm. The bottom has a raised hemispherical portion which reduces capacity. Find the apparent capacity and actual capacity of the glass. (Take π = 3.14)
- 1.Step 1: Inner radius (r) = 5 / 2 = 2.5 cm, Height (h) = 10 cm.
- 2.Step 2: Apparent Capacity = Volume of cylinder = πr²h = 3.14 × (2.5)² × 10 = 196.25 cm³.
- 3.Step 3: Volume of raised hemispherical bottom = (2/3)πr³ = (2/3) × 3.14 × (2.5)³ = 32.71 cm³.
- 4.Step 4: Actual Capacity = Apparent Capacity - Volume of Hemisphere = 196.25 - 32.71 = 163.54 cm³.
Type 3: Circumscribed Shapes and Differences in Volume
Problem
A solid toy is in the form of a hemisphere surmounted by a right circular cone of height 2 cm and base diameter 4 cm. Determine the volume of the toy. If a right circular cylinder circumscribes the toy, find the difference of volumes of the cylinder and the toy. (Take π = 3.14)
- 1.Step 1: Radius of cone and hemisphere (r) = 4 / 2 = 2 cm. Height of cone (h) = 2 cm.
- 2.Step 2: Volume of toy = Volume of hemisphere + Volume of cone = (2/3)πr³ + (1/3)πr²h.
- 3.Step 3: Volume of toy = (1/3)πr²(2r + h) = (1/3) × 3.14 × (2)² × (2×2 + 2) = (1/3) × 3.14 × 4 × 6 = 25.12 cm³.
- 4.Step 4: For circumscribing cylinder: Radius = 2 cm. Height = height of cone + radius of hemisphere = 2 + 2 = 4 cm.
- 5.Step 5: Volume of cylinder = πr²H = 3.14 × (2)² × 4 = 50.24 cm³.
- 6.Step 6: Difference in volumes = Volume of cylinder - Volume of toy = 50.24 - 25.12 = 25.12 cm³.
Type 4: Displacement Principle (Archimedes' Principle)
When solid objects are dropped into a vessel filled with liquid, the volume of water overflowed or displaced equals the total volume of the submerged solid objects.
Problem
A vessel is in the form of an inverted cone of height 8 cm and top radius 5 cm filled with water to the brim. When spherical lead shots of radius 0.5 cm are dropped into it, 1/4th of the water flows out. Find the number of lead shots dropped.
- 1.Step 1: Volume of cone = (1/3)πr²h = (1/3) × π × 5² × 8 = 200π / 3 cm³.
- 2.Step 2: Volume of water overflowed = (1/4) × (200π / 3) = 50π / 3 cm³.
- 3.Step 3: Volume of 1 spherical lead shot = (4/3)πr³ = (4/3) × π × (0.5)³ = (4/3) × π × (1/8) = π / 6 cm³.
- 4.Step 4: Number of lead shots (n) = Total Displaced Volume / Volume of 1 Shot = (50π / 3) / (π / 6) = (50π / 3) × (6 / π) = 100 lead shots.
Quiz
Which rule is used to find the volume of a solid formed by attaching two basic solids?
A shed consists of a cuboid measuring 15 m × 7 m × 8 m, surmounted by a half-cylinder of radius 3.5 m and length 15 m. What is its total capacity? Use π = 22/7.
A cylindrical glass has an apparent capacity of 196.25 cm³. A raised hemispherical base occupies 32.71 cm³. What is the actual capacity of the glass?
A toy consists of a hemisphere surmounted by a cone. Its volume is 25.12 cm³, while the volume of the cylinder enclosing it is 50.24 cm³. What is the difference between their volumes?
Spherical lead shots are completely submerged in a vessel filled with water. Which statement correctly describes the water displaced?
Practice Problems
- A solid is in the shape of a cone standing on a hemisphere with both radii equal to 1 cm and height of cone equal to 1 cm. Find the volume of the solid in terms of π.
- Rachel made a model shaped like a cylinder with two cones attached at its two ends. Diameter is 3 cm and total length is 12 cm. If each cone has height 2 cm, find the volume of air inside the model.
- A gulab jamun contains sugar syrup up to about 30% of its volume. Find how much syrup would be found in 45 gulab jamuns, each shaped like a cylinder with two hemispherical ends with length 5 cm and diameter 2.8 cm.
- A wooden pen stand is a cuboid (15 cm × 10 cm × 3.5 cm) with four conical depressions (radius 0.5 cm, depth 1.4 cm). Find the volume of wood in the entire stand.
- A solid iron pole consists of a cylinder (height 220 cm, diameter 24 cm) surmounted by another cylinder (height 60 cm, radius 8 cm). Find the mass of the pole if 1 cm³ of iron has 8 g mass. (Use π = 3.14).
- A spherical glass vessel has a cylindrical neck 8 cm long, 2 cm in diameter; diameter of spherical part is 8.5 cm. Find its water capacity. (Use π = 3.14).
Key Takeaways
• Combined Volume = Direct sum of volumes of individual attached components. • For scooped-out cavities: Remaining Volume = Original Volume - Volume of Cavity. • Volume of displaced liquid = Total volume of submerged solids dropped into the liquid. • Apparent capacity is based on outer/gross dimensions, while actual capacity excludes internal raised bases or wall thickness.