Now we step into the third dimension. 3D mensuration is about volume (how much stuff fits inside) and surface area (how much material covers the outside). The formulas might look like a lot, but once we see the pattern and practice a few problems, they become second nature. Let’s build our 3D toolkit.
The Complete 3D Formula Reference
| Shape | Volume | CSA (Curved/Lateral) | TSA (Total) |
|---|---|---|---|
| Cube (side a) | a³ | 4a² | 6a² |
| Cuboid (l,b,h) | lbh | 2h(l+b) | 2(lb+bh+hl) |
| Cylinder (r,h) | πr²h | 2πrh | 2πr(r+h) |
| Cone (r,h,l) | ⅓πr²h | πrl | πr(r+l) |
| Sphere (r) | 4/3 πr³ | 4πr² | 4πr² (same) |
| Hemisphere (r) | 2/3 πr³ | 2πr² | 3πr² |
| Frustum (R,r,h,l) | ⅓πh(R²+r²+Rr) | π(R+r)l | π(R+r)l+πR²+πr² |
Where l = slant height for cone and frustum. For a cone: l = √(r² + h²).
Key Formulas
Understanding the Relationships
Some beautiful patterns to remember:
- Cone volume = ⅓ × Cylinder volume (with same radius and height). In simple language: we need 3 cones to fill 1 cylinder.
- Hemisphere volume = ⅔ × Cylinder volume (with same radius, and cylinder height = radius)
- Sphere = 2 Hemispheres (obvious, but people forget this in TSA calculations)
- Hemisphere TSA ≠ Half of sphere’s SA. We need to add the flat circular face: TSA = 2πr² + πr² = 3πr².
Cube and Cuboid
Cube
All edges are equal (side = a).
- Volume = a³
- Longest diagonal (space diagonal) = a√3
- CSA = 4a² (4 faces, since top and bottom are excluded)
- TSA = 6a² (all 6 faces)
Cuboid
Three different dimensions (l, b, h).
- Volume = l × b × h
- Longest diagonal = √(l² + b² + h²)
- CSA = 2h(l + b) (the 4 vertical faces)
- TSA = 2(lb + bh + hl) (all 6 faces)
Cylinder
Think of it as a rectangle wrapped around to form a tube, with circles on top and bottom.
- CSA = 2πrh (the “label” around the can)
- TSA = 2πrh + 2πr² = 2πr(r + h) (label + two circular caps)
- Volume = πr²h (base area × height)
Hollow Cylinder (like a pipe)
Outer radius R, inner radius r, height h.
- Volume = πh(R² − r²) = πh(R + r)(R − r)
- CSA = 2πh(R + r) (both inner and outer curved surfaces)
Cone
A cone is essentially a cylinder that tapers to a point. That’s why its volume is exactly one-third of a cylinder.
- Slant height l = √(r² + h²) — always check if the question gives height or slant height!
- CSA = πrl (the curved surface, like an ice cream cone)
- TSA = πr(r + l) (curved surface + circular base)
- Volume = ⅓πr²h
Sphere and Hemisphere
Sphere
- SA = 4πr² (a sphere has no “curved” vs “flat” — it’s all curved)
- Volume = 4/3 πr³
Hemisphere
- CSA = 2πr² (just the dome part)
- TSA = 3πr² (dome + flat circular base)
- Volume = 2/3 πr³
Frustum (Truncated Cone)
When we cut a cone parallel to its base, the remaining piece is a frustum. Think of it like a bucket shape — wider at the top, narrower at the bottom.
- R = larger radius (top/bottom), r = smaller radius
- Slant height l = √(h² + (R − r)²)
- Volume = ⅓πh(R² + r² + Rr)
- CSA = π(R + r)l
Melting and Recasting Problems
These are the most common 3D mensuration problems in exams. The concept is simple:
When a solid is melted and recast into another shape, the volume stays the same.
So we just set Volume₁ = Volume₂ and solve for the unknown.
Common Variations:
- Sphere melted into smaller spheres → (4/3)πR³ = n × (4/3)πr³ → n = (R/r)³
- Cylinder melted into cone → πr₁²h₁ = ⅓πr₂²h₂
- Multiple small objects melted into one large object → sum of volumes = large volume
Water Flow Problems
These combine volume with rate:
- Water flowing through a pipe at speed v, with pipe cross-section area A:
- Volume per unit time = A × v
- For a circular pipe: Volume per second = πr² × v
Largest Shape Inscribed Problems
- Largest sphere in a cube (side a): diameter = a, so r = a/2
- Largest cube in a sphere (radius r): side = 2r/√3
- Largest cone in a cylinder (r, h): same r and h (volume = ⅓ of cylinder)
- Largest sphere in a cylinder (r, h): radius = min(r, h/2)
- Largest cylinder in a sphere (radius R): when h = 2R/√3 and r = R√(2/3) for maximum volume
Worked Examples
Example 1: Cube Problem
Find the volume and total surface area of a cube whose diagonal is 6√3 cm.
Diagonal = a√3 = 6√3, so a = 6 cm Volume = 6³ = 216 cm³ TSA = 6 × 6² = 216 cm²
(Fun coincidence — the numbers match for a cube with side 6!)
Example 2: Cylinder
A cylindrical tank has a radius of 7 m and height 14 m. Find the volume of water it can hold and the cost of painting its outer surface at Rs. 10/m². (Use π = 22/7)
Volume = πr²h = 22/7 × 49 × 14 = 22 × 7 × 14 = 2156 m³
For painting the outer surface (CSA + top, assuming open-top tank): CSA = 2πrh = 2 × 22/7 × 7 × 14 = 616 m² Top = πr² = 22/7 × 49 = 154 m²
If we paint only the curved surface: Cost = 616 × 10 = Rs. 6160 If we include the top: Cost = (616 + 154) × 10 = Rs. 7700
Example 3: Melting/Recasting
A metallic sphere of radius 6 cm is melted and recast into 27 smaller spheres. Find the radius of each small sphere.
Volume of large sphere = Volume of 27 small spheres
4/3 × π × 6³ = 27 × 4/3 × π × r³
6³ = 27 × r³
216 = 27r³
r³ = 8
r = 2 cm
Shortcut: n = (R/r)³, so 27 = (6/r)³, meaning 6/r = 3, so r = 2. Much faster!
Example 4: Water Flow
Water flows through a cylindrical pipe of diameter 14 cm at the rate of 5 m per second. How long will it take to fill a rectangular tank 50 m × 44 m × 7 m?
Pipe radius = 7 cm = 0.07 m Volume flow per second = πr²v = 22/7 × 0.07² × 5 = 22/7 × 0.0049 × 5 = 0.077 m³/s
Tank volume = 50 × 44 × 7 = 15400 m³
Time = 15400 / 0.077 = 200000 seconds = 55 hours 33 minutes 20 seconds
Example 5: Frustum
A bucket is in the shape of a frustum with top radius 20 cm, bottom radius 12 cm, and height 15 cm. Find its volume.
Volume = ⅓πh(R² + r² + Rr) = ⅓ × 22/7 × 15 × (400 + 144 + 240) = ⅓ × 22/7 × 15 × 784 = 22/7 × 5 × 784 = 22 × 5 × 112 = 12320 cm³ ≈ 12.32 liters
Common Exam Variations
- Melting a cylinder into cones (or vice versa) — equate volumes
- Water rise problems — object immersed in water raises the level (volume of object = l × b × rise in water level)
- Painting/wrapping problems — need surface area, not volume
- Hollow shapes — subtract inner volume from outer volume
- Compound shapes — hemisphere on top of cylinder, cone on top of cylinder, etc. Add volumes and adjust surface areas (remove the common circular face)
Practice Problems
Q1: A cone has a radius of 7 cm and height of 24 cm. Find its slant height, volume, and curved surface area. (Use π = 22/7)
Q2: A solid metallic cylinder of radius 3 cm and height 12 cm is melted and recast into a cone of radius 6 cm. Find the height of the cone.
Q3: A hemispherical bowl has an inner radius of 10 cm. Find the volume of milk it can hold and the cost of painting the inner surface at Rs. 2 per cm². (Use π = 3.14)
Answers
A1: Slant height l = √(7² + 24²) = √(49 + 576) = √625 = 25 cm (7-24-25 triplet!). Volume = ⅓ × 22/7 × 49 × 24 = 22 × 7 × 24/3 = 1232 cm³. CSA = 22/7 × 7 × 25 = 550 cm².
A2: Volume of cylinder = Volume of cone. π × 9 × 12 = ⅓ × π × 36 × h. 108 = 12h. h = 9 cm.
A3: Volume = 2/3 × 3.14 × 1000 = 2093.33 cm³ ≈ 2.09 liters. Inner surface (CSA of hemisphere) = 2 × 3.14 × 100 = 628 cm². Cost = 628 × 2 = Rs. 1256.