Volume and Capacity
Interactive curriculum lessons, worked examples, and geometric problem-solving techniques designed to help Kenyan students master CBC, KPSEA, KCSE, and IGCSE mathematics.
First Principles
Objective: Master the Geometry and Practical Metrics of Volume and Capacity
Volume measures the three-dimensional space enclosed by a solid boundary, whereas Capacity measures the volume of fluid (liquid or gas) or granular material that a container can hold.
The Foundation of Extrusion: For any uniform prism or cylinder, volume is simply the product of its constant cross-sectional base area (\(A\)) and its height or length (\(h\)):
\[ \text{Volume} = \text{Base Area} \times \text{Height} = A \times h \]
Key Conversions to Memorise:
- \(1\text{ cm}^3 = 1\text{ mL}\)
- \(1\text{ litre} (\text{L}) = 1\,000\text{ cm}^3 = 1\,000\text{ mL}\)
- \(1\text{ m}^3 = 1\,000\text{ L} = 1\,000\,000\text{ cm}^3\)
Interactive Solid Visualiser & Tank Capacity
Key Formulas
Core Geometric & Capacity Formulas
1. Right Rectangular Prism (Cuboid)
\[ V = l \times w \times h \]- \(l\) = length, \(w\) = width, \(h\) = height
- Total Base Area \(A = l \times w\)
2. Right Circular Cylinder
\[ V = \pi r^2 h = \frac{\pi d^2 h}{4} \]- \(r\) = base radius, \(d = 2r\) = diameter, \(h\) = perpendicular height
- Circular Base Area \(A = \pi r^2\)
3. Right Triangular Prism
\[ V = \left( \frac{1}{2} b h_{\text{triangle}} \right) \times L \]- \(b\) = base of triangular cross-section
- \(h_{\text{triangle}}\) = perpendicular height of triangle
- \(L\) = length/depth of the prism
4. Unit Conversion Bridge
\[ 1\text{ m}^3 = 1\,000\text{ dm}^3 = 1\,000\text{ Litres} \]\[ 1\text{ Litre} = 1\,000\text{ cm}^3 = 1\,000\text{ mL} \]\[ 1\text{ m}^3 = 1\,000\,000\text{ cm}^3 \]Worked Examples
Example 1 (Easy): Rectangular Rainwater Tank
A rectangular underground water tank at a homestead in Machakos has dimensions \(3\text{ m}\) long, \(2\text{ m}\) wide, and \(1.5\text{ m}\) deep. Calculate its total capacity in litres.
Step-by-step Solution:
- Calculate Volume (\(\text{m}^3\)):\[ V = l \times w \times h = 3\text{ m} \times 2\text{ m} \times 1.5\text{ m} = 9.0\text{ m}^3 \]
- Convert to Litres: Since \(1\text{ m}^3 = 1\,000\text{ L}\),\[ \text{Capacity} = 9.0 \times 1\,000 = 9\,000\text{ Litres} \]
Example 2 (Medium): Cylindrical Drum from Diameter
A cylindrical storage drum used by a dairy cooperative in Eldoret has a diameter of \(70\text{ cm}\) and a height of \(120\text{ cm}\). Taking \(\pi = \frac{22}{7}\), find the maximum volume of milk it can hold in litres.
Step-by-step Solution:
- Find radius \(r\):\[ r = \frac{d}{2} = \frac{70}{2} = 35\text{ cm} \]
- Compute Volume in \(\text{cm}^3\):\[ V = \pi r^2 h = \frac{22}{7} \times 35^2 \times 120 = \frac{22}{7} \times 1225 \times 120 = 22 \times 175 \times 120 = 462\,000\text{ cm}^3 \]
- Convert \(\text{cm}^3\) to Litres:\[ \text{Capacity} = \frac{462\,000}{1\,000} = 462\text{ Litres} \]
Example 3 (Hard): Rate of Pumping into a Cylindrical Silo
A community posho mill grain silo is cylindrical with an internal diameter of \(4\text{ m}\) and a height of \(6\text{ m}\). A grain elevator pours maize into the silo at a rate of \(1.5\text{ m}^3/\text{min}\). Taking \(\pi = 3.142\), how many minutes will it take to fill the empty silo to \(75\%\) of its capacity? (Give your answer to 1 decimal place).
Step-by-step Solution:
- Find radius and total silo volume:\[ r = \frac{4}{2} = 2\text{ m} \]\[ V_{\text{total}} = \pi r^2 h = 3.142 \times 2^2 \times 6 = 3.142 \times 4 \times 6 = 75.408\text{ m}^3 \]
- Calculate target volume (75%):\[ V_{\text{target}} = 0.75 \times 75.408 = 56.556\text{ m}^3 \]
- Calculate time required:\[ \text{Time} = \frac{\text{Volume}}{\text{Rate}} = \frac{56.556\text{ m}^3}{1.5\text{ m}^3/\text{min}} = 37.704\text{ minutes} \approx 37.7\text{ min} \]
Common Mistakes
1. Confusing Diameter with Radius in Cylinder Formulas
The Mistake: Plugging the diameter directly into \(V = \pi r^2 h\) instead of first dividing by 2.
Why it feels right: The problem statement provides the diameter (e.g. "diameter of 14 cm"), and students quickly substitute 14 into \(r^2\), which overestimates the volume by a factor of 4 (since \(d^2 = 4r^2\)).
The Correction: Always write down \(r = \frac{d}{2}\) before calculating.
2. Linear Conversion Applied to Cubic Units
The Mistake: Assuming \(1\text{ m}^3 = 100\text{ cm}^3\) because \(1\text{ m} = 100\text{ cm}\).
Why it feels right: We intuitively carry over linear conversion factors into 3D space.
The Correction: Three dimensions must all scale: \(1\text{ m}^3 = 100\text{ cm} \times 100\text{ cm} \times 100\text{ cm} = 1\,000\,000\text{ cm}^3\).
3. Mismatching Unit Dimensions
The Mistake: Calculating volume when the radius is in metres and height is in centimetres without converting to the same unit first (e.g., \(r = 0.5\text{ m}\), \(h = 80\text{ cm} \rightarrow V = \pi \times 0.5^2 \times 80\)).
The Correction: Convert all dimensions to the same unit before performing multiplications.
Real World
Practical Applications Across Kenya and East Africa
Rainwater Harvesting in Schools
Schools across arid and semi-arid regions install corrugated plastic cylindrical tanks (e.g., Roto or Kentank). Calculating tank capacity lets headteachers estimate how many days a 10,000-litre tank will serve 400 students assuming an allocation of 5 litres per student per day.
Commercial Milk & Water Bowsers
Dairy processors in Nyandarua and water vendors in Nairobi use cylindrical truck tanks. Knowing the precise cross-section allows installation of dipsticks graduated directly in litres for trade calibration.
Grain Storage in Posho Mills
National Cereals and Produce Board (NCPB) silos and local cooperative granaries use cylindrical silos. Converting volume to tonnes of maize (where \(1\text{ m}^3 \approx 0.75\text{ tonnes}\)) helps farmers plan post-harvest logistics.
Practice