Volume of Hemisphere
A hemisphere is exactly half of a sphere, obtained by cutting it along a great circle. The volume of a hemisphere is half the volume of the full sphere.
Hemispheres are common in real life — bowls, domes, igloo roofs, and half-watermelons are all hemispherical shapes. Calculating their volume is essential for capacity, storage, and material estimation problems.
This topic is part of NCERT Class 9 Mathematics, Chapter: Surface Areas and Volumes. It builds on the sphere volume formula and is frequently tested alongside sphere and cylinder problems.
What is Volume of Hemisphere?
Definition: The volume of a hemisphere is the amount of space enclosed within the hemispherical surface and its flat circular base.
For a hemisphere of radius r:
Volume = (2/3)πr³
Where:
- r = radius of the hemisphere (same as the sphere from which it was cut)
- π = 22/7 or 3.14159...
- The result is in cubic units (cm³, m³, etc.)
Important:
- Volume of hemisphere = half the volume of a sphere = (1/2) × (4/3)πr³ = (2/3)πr³.
- Volume depends on the cube of the radius — doubling the radius increases the volume 8 times.
- Do NOT confuse with surface area formulas: CSA = 2πr², TSA = 3πr².
Volume of Hemisphere Formula
Key Formulas:
1. Volume of hemisphere:
V = (2/3)πr³
2. Volume of full sphere (for comparison):
- V_sphere = (4/3)πr³ = 2 × V_hemisphere
3. In terms of diameter (d = 2r):
- V = (2/3)π(d/2)³ = πd³/12
4. Finding radius from volume:
- r³ = 3V / (2π)
- r = ³√[3V / (2π)]
5. Volume of hollow hemisphere:
- V = (2/3)π(R³ − r³), where R = outer radius, r = inner radius
6. Conversion:
- 1 litre = 1000 cm³
- 1 m³ = 1000 litres
Derivation and Proof
Derivation of Volume of Hemisphere:
Step 1: Start with the sphere volume
- Volume of a sphere of radius r = (4/3)πr³
Step 2: Cut the sphere in half
- When a sphere is divided into two equal halves by a plane through its centre, each half is a hemisphere.
- Both hemispheres are identical and have equal volumes.
Step 3: Calculate hemisphere volume
- Volume of hemisphere = (1/2) × Volume of sphere
- V = (1/2) × (4/3)πr³
- V = (4/6)πr³
- V = (2/3)πr³
Verification:
- Two hemispheres = one sphere: 2 × (2/3)πr³ = (4/3)πr³ ✓
Types and Properties
Types of hemisphere volume problems:
1. Finding volume given radius
- Direct application of V = (2/3)πr³
2. Finding volume given diameter
- First find r = d/2, then apply the formula.
3. Finding radius from volume
- Rearrange: r³ = 3V/(2π), then take the cube root.
4. Capacity problems (in litres)
- Calculate volume in cm³, then convert: litres = cm³ / 1000.
5. Hollow hemisphere
- V = (2/3)π(R³ − r³), where R and r are outer and inner radii.
- A hemisphere melted into cones, cylinders, or smaller hemispheres.
- Volume is conserved.
7. Combined solids
- Hemisphere + cylinder, hemisphere + cone, etc.
- Total volume = sum of individual volumes.
Solved Examples
Example 1: Example 1: Volume given radius
Problem: Find the volume of a hemisphere of radius 7 cm. (Use π = 22/7)
Solution:
Given:
- r = 7 cm
Using V = (2/3)πr³:
- V = (2/3) × (22/7) × 7³
- V = (2/3) × (22/7) × 343
- V = (2/3) × 22 × 49
- V = (2/3) × 1078
- V = 2156/3 = 718.67 cm³
Answer: Volume = 718.67 cm³ (approx.).
Example 2: Example 2: Volume given diameter
Problem: Find the volume of a hemisphere of diameter 42 cm. (Use π = 22/7)
Solution:
Given:
- d = 42 cm, so r = 21 cm
Using V = (2/3)πr³:
- V = (2/3) × (22/7) × 21³
- V = (2/3) × (22/7) × 9261
- V = (2/3) × 22 × 1323
- V = (2/3) × 29106
- V = 58212/3 = 19404 cm³
Answer: Volume = 19,404 cm³.
Example 3: Example 3: Finding radius from volume
Problem: The volume of a hemisphere is 2425.5 cm³. Find the radius. (Use π = 22/7)
Solution:
Given:
- V = 2425.5 cm³
Using V = (2/3)πr³:
- 2425.5 = (2/3) × (22/7) × r³
- 2425.5 = (44/21) × r³
- r³ = 2425.5 × 21/44 = 50935.5/44 = 1157.625
- r = ³√1157.625 = 10.5 cm
Answer: Radius = 10.5 cm.
Example 4: Example 4: Capacity of a hemispherical bowl
Problem: A hemispherical bowl has an inner radius of 10.5 cm. Find the volume of milk it can hold in litres. (Use π = 22/7)
Solution:
Given:
- r = 10.5 cm
Volume:
- V = (2/3) × (22/7) × (10.5)³
- V = (2/3) × (22/7) × 1157.625
- V = (2/3) × 3637.5
- V = 2425 cm³
Converting to litres:
- V = 2425/1000 = 2.425 litres
Answer: The bowl can hold 2.425 litres of milk.
Example 5: Example 5: Hollow hemisphere
Problem: A hollow hemispherical vessel has an outer radius of 14 cm and inner radius of 10.5 cm. Find the volume of material used. (Use π = 22/7)
Solution:
Given:
- R = 14 cm, r = 10.5 cm
Using V = (2/3)π(R³ − r³):
- R³ = 2744, r³ = 1157.625
- R³ − r³ = 2744 − 1157.625 = 1586.375
- V = (2/3) × (22/7) × 1586.375
- V = (2/3) × 4985.75
- V = 9971.5/3 ≈ 3323.83 cm³
Answer: Volume of material ≈ 3323.83 cm³.
Example 6: Example 6: Hemisphere melted into cone
Problem: A hemisphere of radius 9 cm is melted and recast into a cone of radius 6 cm. Find the height of the cone.
Solution:
Volume of hemisphere:
- V = (2/3)π(9)³ = (2/3)π(729) = 486π cm³
Volume of cone:
- V = (1/3)π(6)²h = (1/3)π(36)h = 12πh
Equating volumes:
- 12πh = 486π
- h = 486/12 = 40.5 cm
Answer: Height of the cone = 40.5 cm.
Example 7: Example 7: Comparing hemisphere and sphere volumes
Problem: Show that the volume of a hemisphere is exactly half the volume of the sphere with the same radius.
Solution:
Volume of sphere:
- V_sphere = (4/3)πr³
Volume of hemisphere:
- V_hemi = (2/3)πr³
Ratio:
- V_hemi / V_sphere = [(2/3)πr³] / [(4/3)πr³] = 2/4 = 1/2
Verified: Volume of hemisphere = (1/2) × Volume of sphere.
Example 8: Example 8: Hemisphere on top of cylinder
Problem: A solid consists of a cylinder of radius 7 cm and height 10 cm, with a hemisphere of the same radius mounted on top. Find the total volume. (Use π = 22/7)
Solution:
Volume of cylinder:
- V_cyl = πr²h = (22/7) × 49 × 10 = 1540 cm³
Volume of hemisphere:
- V_hemi = (2/3) × (22/7) × 343 = (2/3) × 1078 = 718.67 cm³
Total volume:
- V = 1540 + 718.67 = 2258.67 cm³
Answer: Total volume ≈ 2258.67 cm³.
Example 9: Example 9: Effect of tripling the radius
Problem: If the radius of a hemisphere is tripled, by what factor does the volume increase?
Solution:
Original:
- V = (2/3)πr³
New (radius = 3r):
- V' = (2/3)π(3r)³ = (2/3)π(27r³) = 27 × (2/3)πr³
- V' = 27V
Answer: The volume increases 27 times.
Example 10: Example 10: Water tank problem
Problem: A hemispherical water tank has an internal diameter of 2.8 m. Find the capacity of the tank in litres. (Use π = 22/7)
Solution:
Given:
- d = 2.8 m, so r = 1.4 m
Volume:
- V = (2/3) × (22/7) × (1.4)³
- V = (2/3) × (22/7) × 2.744
- V = (2/3) × 8.624
- V = 17.248/3 ≈ 5.749 m³
Converting to litres:
- 1 m³ = 1000 litres
- V = 5.749 × 1000 = 5749 litres
Answer: The tank can hold approximately 5749 litres.
Real-World Applications
Applications of Volume of Hemisphere:
- Water storage: Calculating the capacity of hemispherical tanks and reservoirs.
- Cooking: Determining how much food a hemispherical bowl or wok can hold.
- Architecture: Estimating the internal volume of domes for ventilation and air conditioning design.
- Manufacturing: Calculating the amount of material for hemispherical moulds, bowls, and containers.
- Ice cream: Determining the volume of a hemispherical scoop of ice cream.
- Science: Estimating volumes of hemispherical lenses, reflectors, and laboratory vessels.
Key Points to Remember
- Volume of hemisphere = (2/3)πr³, where r is the radius.
- Volume of hemisphere = half the volume of a sphere.
- In terms of diameter: V = πd³/12.
- Volume depends on the cube of the radius — doubling the radius makes volume 8 times.
- For hollow hemispheres: V = (2/3)π(R³ − r³).
- To convert cm³ to litres, divide by 1000.
- To convert m³ to litres, multiply by 1000.
- In combined solids (hemisphere + cylinder), add the individual volumes.
- In melting problems, volume is conserved: volume of original = volume of new shape.
- Do NOT confuse volume formula (2/3)πr³ with surface area formulas (2πr² for CSA, 3πr² for TSA).
Practice Problems
- Find the volume of a hemisphere of radius 3.5 cm. (Use π = 22/7)
- A hemisphere has diameter 28 cm. Find its volume.
- The volume of a hemisphere is 19404 cm³. Find the radius.
- A hemispherical bowl of internal radius 7 cm is full of water. If the water is poured into a cylindrical vessel of radius 7 cm, find the height of water in the cylinder.
- A hemisphere of radius 12 cm is melted and recast into small spheres of radius 3 cm. How many spheres are formed?
- A solid is made of a cone of height 10 cm mounted on a hemisphere of radius 7 cm. Find the total volume.
- A hollow hemisphere has outer radius 10 cm and thickness 2 cm. Find the volume of the material.
- The radius of a hemisphere is increased by 25%. Find the percentage increase in volume.
Frequently Asked Questions
Q1. What is the formula for the volume of a hemisphere?
Volume of hemisphere = (2/3)πr³, where r is the radius. This is exactly half the volume of a sphere with the same radius.
Q2. How is the volume of a hemisphere derived?
Volume of sphere = (4/3)πr³. A hemisphere is half a sphere, so volume of hemisphere = (1/2) × (4/3)πr³ = (2/3)πr³.
Q3. What is the difference between volume and surface area of a hemisphere?
Volume measures the space inside (in cubic units): V = (2/3)πr³. Surface area measures the outer covering (in square units): CSA = 2πr², TSA = 3πr².
Q4. How do you convert volume from cm³ to litres?
Divide by 1000. Since 1 litre = 1000 cm³, a volume of 2425 cm³ = 2.425 litres.
Q5. What is a hollow hemisphere?
A hollow hemisphere has an outer radius R and inner radius r. Volume of material = (2/3)π(R³ − r³).
Q6. How does doubling the radius affect hemisphere volume?
Volume increases 8 times. Since V depends on r³: V_new = (2/3)π(2r)³ = 8 × (2/3)πr³ = 8V.
Q7. Is volume of hemisphere in the CBSE Class 9 syllabus?
Yes. Volume of hemisphere is part of CBSE Class 9 Mathematics, Chapter: Surface Areas and Volumes.
Q8. What is the volume of a hemisphere in terms of diameter?
If the diameter is d, then r = d/2. Substituting: V = (2/3)π(d/2)³ = πd³/12.
Related Topics
- Volume of Sphere
- Surface Area of Hemisphere
- Volume of Cone
- Volume of Cylinder
- Surface Area of Cone
- Surface Area of Sphere
- Slant Height of Cone
- Combination of Solids
- Conversion of Solids
- Frustum of a Cone
- Surface Area of Combined Solids
- Volume Word Problems
- Volume of Frustum of Cone
- Curved Surface Area of Frustum
