Surface Area of Combined Solids
In real life, most objects are not simple geometric solids. A toy rocket may be a cylinder topped with a cone. An ice cream may be a hemisphere on a cone. A capsule is two hemispheres joined by a cylinder.
To find the surface area of such combined solids, we identify the individual shapes, calculate their visible surface areas, and add them — excluding the parts that are joined (hidden).
This topic requires a strong understanding of surface area formulas for individual solids (cylinder, cone, sphere, hemisphere, cuboid).
What is Surface Area of Combined Solids?
Key principle:
TSA of combined solid = Sum of visible surface areas of each part
Important: When two solids are joined, the joining surface is hidden and must be excluded from the total surface area calculation.
Surface Area of Combined Solids Formula
Surface area formulas (recap):
- Cylinder CSA: 2πrh
- Cylinder TSA: 2πr(r + h)
- Cone CSA: πrl (where l = slant height = √(r² + h²))
- Sphere SA: 4πr²
- Hemisphere CSA: 2πr²
- Hemisphere TSA: 3πr² (curved + flat base)
- Cuboid TSA: 2(lb + bh + hl)
Types and Properties
Common combinations:
- Cylinder + Cone: Rocket, tent. SA = CSA of cylinder + CSA of cone + base circle of cylinder (if visible).
- Cylinder + Hemisphere: Capsule end, storage tank. SA = CSA of cylinder + CSA of hemisphere(s).
- Cone + Hemisphere: Ice cream, spinning top. SA = CSA of cone + CSA of hemisphere.
- Cuboid + Cylinder: Toy house. SA = TSA of cuboid − top area + CSA of cylinder + top circle.
- Cylinder + Cone + Hemisphere: Toy rocket (three parts joined).
Solved Examples
Example 1: Example 1: Cylinder topped with a cone
Problem: A solid is made of a cylinder of radius 7 cm and height 10 cm, topped with a cone of same radius and height 5 cm. Find the total surface area.
Solution:
Given: r = 7 cm, cylinder h = 10 cm, cone h = 5 cm.
- Slant height of cone: l = √(7² + 5²) = √(49 + 25) = √74 ≈ 8.6 cm
- CSA of cylinder = 2πrh = 2 × 22/7 × 7 × 10 = 440 cm²
- CSA of cone = πrl = 22/7 × 7 × 8.6 = 189.2 cm²
- Base of cylinder = πr² = 22/7 × 49 = 154 cm²
- TSA = CSA of cylinder + CSA of cone + base = 440 + 189.2 + 154 = 783.2 cm²
Answer: TSA ≈ 783.2 cm².
Example 2: Example 2: Hemisphere on a cylinder
Problem: A medicine capsule is a cylinder of diameter 1 cm and length 1.4 cm with hemispheres at both ends. Find its surface area.
Solution:
Given: r = 0.5 cm, cylinder length = 1.4 cm.
- CSA of cylinder = 2πrh = 2 × π × 0.5 × 1.4 = 1.4π cm²
- CSA of 2 hemispheres = 2 × 2πr² = 4π × 0.25 = π cm²
- TSA = 1.4π + π = 2.4π ≈ 7.54 cm²
Answer: TSA ≈ 7.54 cm².
Example 3: Example 3: Cone topped with hemisphere
Problem: An ice cream is shaped as a cone of height 12 cm and radius 5 cm with a hemisphere of ice cream on top. Find the outer surface area.
Solution:
- Slant height of cone: l = √(5² + 12²) = √(25 + 144) = √169 = 13 cm
- CSA of cone = πrl = π × 5 × 13 = 65π cm²
- CSA of hemisphere = 2πr² = 2π × 25 = 50π cm²
- TSA = 65π + 50π = 115π ≈ 361.43 cm²
Answer: TSA ≈ 361.43 cm².
Example 4: Example 4: Toy — cylinder between two cones
Problem: A toy is in the form of a cylinder of radius 3 cm and height 8 cm with a cone at each end, each of height 4 cm. Find the total surface area.
Solution:
- Slant height of each cone: l = √(3² + 4²) = √(9 + 16) = 5 cm
- CSA of cylinder = 2πrh = 2π × 3 × 8 = 48π cm²
- CSA of 2 cones = 2 × πrl = 2 × π × 3 × 5 = 30π cm²
- TSA = 48π + 30π = 78π ≈ 245.14 cm²
(No flat surfaces are visible — all are hidden at the joints.)
Answer: TSA ≈ 245.14 cm².
Example 5: Example 5: Tent — cylinder base with conical top
Problem: A tent is cylindrical up to a height of 3 m with a conical roof of height 2.8 m. The radius is 4 m. Find the canvas required (no floor).
Solution:
- Slant height of cone: l = √(4² + 2.8²) = √(16 + 7.84) = √23.84 ≈ 4.88 m
- CSA of cylinder = 2πrh = 2 × 22/7 × 4 × 3 = 75.43 m²
- CSA of cone = πrl = 22/7 × 4 × 4.88 = 61.33 m²
- Canvas needed = 75.43 + 61.33 = 136.76 m²
Answer: Canvas required ≈ 136.76 m².
Example 6: Example 6: Cuboid with hemispherical depression
Problem: A cuboid of dimensions 10 × 10 × 5 cm has a hemisphere of radius 3 cm scooped out from one face. Find the surface area.
Solution:
- TSA of cuboid = 2(10×10 + 10×5 + 10×5) = 2(100 + 50 + 50) = 400 cm²
- Area of circular opening removed from cuboid face = πr² = 9π cm²
- CSA of hemisphere added = 2πr² = 18π cm²
- TSA = 400 − 9π + 18π = 400 + 9π ≈ 400 + 28.27 = 428.27 cm²
Answer: TSA ≈ 428.27 cm².
Example 7: Example 7: Storage tank — cylinder with hemispherical ends
Problem: A storage tank is a cylinder of length 6 m and radius 2 m with hemispherical ends. Find the outer surface area.
Solution:
- CSA of cylinder = 2πrh = 2π × 2 × 6 = 24π m²
- SA of 2 hemispheres = 2 × 2πr² = 4π × 4 = 16π m²
- (Two hemispheres = one full sphere SA = 4πr²)
- TSA = 24π + 16π = 40π ≈ 125.71 m²
Answer: TSA ≈ 125.71 m².
Example 8: Example 8: Solid with three parts
Problem: A solid consists of a cone (h = 4 cm, r = 3 cm) on top of a cylinder (h = 5 cm, r = 3 cm) on top of a hemisphere (r = 3 cm). Find the TSA.
Solution:
- Slant height of cone: l = √(3² + 4²) = 5 cm
- CSA of cone = πrl = π × 3 × 5 = 15π
- CSA of cylinder = 2πrh = 2π × 3 × 5 = 30π
- CSA of hemisphere = 2πr² = 2π × 9 = 18π
- TSA = 15π + 30π + 18π = 63π ≈ 197.92 cm²
Answer: TSA ≈ 197.92 cm².
Example 9: Example 9: Pen stand — cylinder with conical cavity
Problem: A wooden pen stand is a cylinder of radius 5 cm and height 12 cm with a conical cavity at the top (depth 4 cm, radius 5 cm). Find the total surface area of the wood.
Solution:
- Slant height of cone cavity: l = √(5² + 4²) = √41 ≈ 6.4 cm
- CSA of cylinder = 2πrh = 2π × 5 × 12 = 120π
- Base of cylinder = πr² = 25π
- CSA of cone cavity = πrl = π × 5 × 6.4 = 32π
- TSA = CSA cylinder + base + cone cavity CSA = 120π + 25π + 32π = 177π ≈ 556.0 cm²
(The top flat ring area cancels: full circle − cone opening + cone CSA; cone cavity replaces the flat top.)
Answer: TSA ≈ 556 cm².
Example 10: Example 10: Gulab jamun shaped solid
Problem: A gulab jamun is modelled as a cylinder of length 3 cm and diameter 2.8 cm, with hemispheres at both ends. Find the surface area.
Solution:
Given: r = 1.4 cm, cylinder length = 3 cm.
- CSA of cylinder = 2πrh = 2 × 22/7 × 1.4 × 3 = 26.4 cm²
- CSA of 2 hemispheres = 2 × 2πr² = 4 × 22/7 × 1.96 = 24.64 cm²
- TSA = 26.4 + 24.64 = 51.04 cm²
Answer: TSA ≈ 51.04 cm².
Real-World Applications
Real-life applications:
- Manufacturing: Calculating material needed for capsules, tanks, bottles.
- Construction: Canvas for tents, paint for structures.
- Packaging: Surface area determines packaging material needed.
- Architecture: Domes (hemispheres) on cylindrical buildings.
- Toys: Designing toys that combine different shapes.
Key Points to Remember
- TSA of combined solid = Sum of visible surface areas only.
- Exclude hidden joining surfaces — the circular faces where solids meet.
- Always identify each component solid and its dimensions first.
- Use the correct formula for CSA (not TSA) when a base is hidden.
- For slant height of cone: l = √(r² + h²).
- Two hemispheres of same radius = one sphere (SA = 4πr²).
- Check whether the base of the combined solid is visible (e.g., tent has no floor, but a pen stand does).
- Use π = 22/7 or 3.14 as specified in the problem.
Practice Problems
- A solid toy is a hemisphere of radius 4.2 cm mounted on a cone of same radius and height 7 cm. Find TSA.
- A storage silo is a cylinder of radius 3 m and height 8 m with a hemispherical top. Find the outer surface area.
- A tent is a cone of height 6 m on a cylinder of height 3 m, both of radius 5 m. Find the canvas needed.
- A capsule is a 2 cm long cylinder (diameter 0.8 cm) with hemispherical ends. Find its surface area.
- A solid is formed by joining a cone and cylinder of same radius 7 cm. Cone height = 24 cm, cylinder height = 10 cm. Find TSA.
- A building is a cuboid 10 m × 8 m × 6 m topped by a half-cylinder of radius 4 m along its length. Find the total outer surface area.
Frequently Asked Questions
Q1. Why do we exclude the joining surface?
The joining surface is internal (hidden). It is not exposed to the outside, so it is not part of the total surface area.
Q2. When do we use CSA vs TSA?
Use CSA for surfaces that are visible (curved part only). Use TSA only when all faces of a solid are exposed. For combined solids, most components contribute only their CSA.
Q3. How do you find slant height of a cone?
l = √(r² + h²), where r is the radius and h is the vertical height.
Q4. What about a hollow combined solid?
For hollow solids, you must include both inner and outer curved surfaces, plus any exposed ring-shaped areas.
Q5. Do all parts need to have the same radius?
Usually yes for smooth joining. If radii differ, there will be a step or ring that adds to the surface area.
Q6. What is the surface area of a capsule?
A capsule = cylinder + 2 hemispheres. SA = 2πrh + 4πr² (since 2 hemispheres = 1 sphere).
Q7. How do you handle a cone with hemisphere on top?
TSA = CSA of cone + CSA of hemisphere = πrl + 2πr². The flat faces of both are hidden at the joint.
Q8. Can you have more than two solids combined?
Yes. A rocket shape might have cone + cylinder + hemisphere. Just add all visible surface areas and exclude all hidden joining areas.
