Surface Area of Cylinder
A cylinder is a 3D solid with two parallel circular bases connected by a curved surface. Common examples include pipes, cans, drums, pillars, and pencils.
The surface area of a cylinder is the total area of all its surfaces. It consists of two parts: the curved surface area (CSA) — the area of the curved part that wraps around — and the area of the two circular bases.
Understanding surface area is essential in many practical situations. When you want to paint a cylindrical pillar, you need the CSA. When you want to make a tin can with a lid, you need the TSA. When a label wraps around a bottle, its area equals the CSA of the bottle.
The cylinder is a right circular cylinder — "right" because the curved surface is perpendicular to the base, and "circular" because the base is a circle. The two bases are congruent (identical) circles, and the line joining their centres is the axis of the cylinder.
In this topic, you will learn the formulas for curved surface area (CSA = 2πrh) and total surface area (TSA = 2πr(r + h)), understand how these formulas are derived from the net of a cylinder, solve problems involving open and closed cylinders, and apply these formulas to real-world contexts like painting, labelling, and manufacturing.
What is Surface Area of Cylinder?
Definition: A cylinder is a 3D shape with two congruent, parallel circular bases joined by a curved surface that is perpendicular to the bases.
Curved Surface Area (CSA) or Lateral Surface Area:
CSA = 2πrh
Total Surface Area (TSA):
TSA = 2πr(r + h)
Where:
- r = radius of the circular base
- h = height of the cylinder (perpendicular distance between the two bases)
- π = 22/7 or 3.14159...
Breakdown of TSA:
- TSA = CSA + 2 × (Area of circular base)
- TSA = 2πrh + 2πr²
- TSA = 2πr(h + r)
Methods
Derivation of CSA:
If you cut a cylinder along its height and unroll the curved surface, you get a rectangle.
- The length of this rectangle = circumference of the base = 2πr.
- The width of this rectangle = height of the cylinder = h.
- Area of rectangle = length × width = 2πr × h = 2πrh.
Therefore, CSA = 2πrh.
Derivation of TSA:
- The cylinder has a curved surface + two circular bases.
- CSA = 2πrh
- Area of one circular base = πr²
- Area of two bases = 2πr²
- TSA = 2πrh + 2πr² = 2πr(h + r)
For a hollow cylinder (pipe):
- Let R = outer radius, r = inner radius, h = height.
- CSA = 2π(R + r)h (outer + inner curved surfaces)
- TSA = 2π(R + r)h + 2π(R² − r²) (curved surfaces + two ring-shaped ends)
For a cylinder open at one end:
- Surface area = CSA + πr² = 2πrh + πr²
Solved Examples
Example 1: Example 1: Finding CSA
Problem: Find the curved surface area of a cylinder with radius 7 cm and height 10 cm.
Solution:
Given:
- r = 7 cm, h = 10 cm
Using CSA = 2πrh:
- CSA = 2 × (22/7) × 7 × 10
- = 2 × 22 × 10
- = 440 cm²
Answer: CSA = 440 cm².
Example 2: Example 2: Finding TSA
Problem: Find the total surface area of a cylinder with radius 3.5 cm and height 12 cm.
Solution:
Given:
- r = 3.5 cm, h = 12 cm
Using TSA = 2πr(r + h):
- TSA = 2 × (22/7) × 3.5 × (3.5 + 12)
- = 2 × (22/7) × 3.5 × 15.5
- = 2 × 22 × 0.5 × 15.5
- = 22 × 15.5
- = 341 cm²
Answer: TSA = 341 cm².
Example 3: Example 3: Given diameter
Problem: The diameter of a cylindrical pillar is 28 cm and height is 3 m. Find the cost of painting the CSA at Rs 5 per cm².
Solution:
Given:
- d = 28 cm, so r = 14 cm
- h = 3 m = 300 cm
CSA = 2πrh:
- = 2 × (22/7) × 14 × 300
- = 2 × 22 × 2 × 300
- = 26,400 cm²
Cost of painting:
- = 26,400 × 5 = Rs 1,32,000
Answer: CSA = 26,400 cm². Cost = Rs 1,32,000.
Example 4: Example 4: Finding height from CSA
Problem: The CSA of a cylinder is 1320 cm² and radius is 10.5 cm. Find the height.
Solution:
Given:
- CSA = 1320 cm², r = 10.5 cm
Using CSA = 2πrh:
- 1320 = 2 × (22/7) × 10.5 × h
- 1320 = 2 × 22 × 1.5 × h
- 1320 = 66h
- h = 1320/66 = 20 cm
Answer: Height = 20 cm.
Example 5: Example 5: Finding radius from TSA
Problem: The TSA of a cylinder is 616 cm² and height is 7 cm. Find the radius.
Solution:
Given:
- TSA = 616 cm², h = 7 cm
Using TSA = 2πr(r + h):
- 616 = 2 × (22/7) × r × (r + 7)
- 616 = (44/7) × r × (r + 7)
- 616 × 7/44 = r(r + 7)
- 98 = r² + 7r
- r² + 7r − 98 = 0
- (r + 14)(r − 7) = 0
- r = 7 (rejecting r = −14)
Answer: Radius = 7 cm.
Example 6: Example 6: Open cylinder
Problem: A cylindrical container is open at the top. Its radius is 7 cm and height is 14 cm. Find the total surface area.
Solution:
Given:
- r = 7 cm, h = 14 cm, open at one end
Surface area = CSA + one base:
- CSA = 2 × (22/7) × 7 × 14 = 616 cm²
- One base = πr² = (22/7) × 7² = 154 cm²
- Total = 616 + 154 = 770 cm²
Answer: Surface area = 770 cm².
Example 7: Example 7: Comparing CSA and TSA
Problem: A cylinder has r = 5 cm and h = 10 cm. Find both CSA and TSA.
Solution:
CSA:
- CSA = 2πrh = 2 × 3.14 × 5 × 10 = 314 cm²
TSA:
- TSA = 2πr(r + h) = 2 × 3.14 × 5 × (5 + 10)
- = 2 × 3.14 × 5 × 15
- = 471 cm²
Difference:
- TSA − CSA = 471 − 314 = 157 cm²
- This equals 2πr² = 2 × 3.14 × 25 = 157 cm² ✓
Answer: CSA = 314 cm², TSA = 471 cm².
Example 8: Example 8: Roller problem
Problem: A road roller has diameter 84 cm and length 1.2 m. Find the area covered in one complete revolution.
Solution:
Given:
- d = 84 cm, so r = 42 cm
- Length (= height of cylinder) = 1.2 m = 120 cm
Area covered in one revolution = CSA:
- CSA = 2πrh = 2 × (22/7) × 42 × 120
- = 2 × 22 × 6 × 120
- = 31,680 cm²
- = 3.168 m²
Answer: Area covered = 31,680 cm² (or 3.168 m²).
Example 9: Example 9: Material required
Problem: How many square metres of metal sheet is needed to make a closed cylindrical tank of radius 1.4 m and height 2 m?
Solution:
Given:
- r = 1.4 m, h = 2 m
TSA = 2πr(r + h):
- = 2 × (22/7) × 1.4 × (1.4 + 2)
- = 2 × (22/7) × 1.4 × 3.4
- = 2 × 22 × 0.2 × 3.4
- = 2 × 22 × 0.68
- = 29.92 m²
Answer: Metal sheet required = 29.92 m².
Example 10: Example 10: Ratio of CSA to TSA
Problem: The radius and height of a cylinder are equal (r = h). Find the ratio of CSA to TSA.
Solution:
Let r = h = a:
- CSA = 2πah = 2πa²
- TSA = 2πa(a + a) = 2πa × 2a = 4πa²
Ratio:
- CSA : TSA = 2πa² : 4πa²
- = 2 : 4
- = 1 : 2
Answer: CSA : TSA = 1 : 2.
Real-World Applications
Real-world applications of cylinder surface area:
- Labelling cans: The label on a tin can covers the CSA. Knowing CSA helps calculate label size.
- Painting pillars: The paint required for a cylindrical pillar depends on its CSA (if top and bottom are not painted).
- Metal fabrication: Making cylindrical tanks, drums, and containers requires knowing the TSA to calculate the amount of metal sheet.
- Road rollers: The area flattened per revolution equals the CSA of the roller.
- Pipes and tubes: Insulation material for pipes is calculated using CSA.
- Packaging: Wrapping paper for cylindrical gifts or products uses CSA calculations.
- Construction: Estimating plaster, paint, or tiling for cylindrical columns and water tanks.
Key Points to Remember
- CSA (Curved Surface Area) = 2πrh — area of only the curved part.
- TSA (Total Surface Area) = 2πr(r + h) — curved surface + two circular bases.
- TSA = CSA + 2πr² (adding two base areas to CSA).
- The curved surface unrolls into a rectangle with dimensions 2πr × h.
- Use CSA for open cylinders, rollers, and labels.
- Use TSA for closed cylinders, cans, and tanks.
- Always express surface area in square units (cm², m²).
- If diameter is given, convert to radius first: r = d/2.
- Ensure radius and height are in the same unit before substituting.
- Open-at-one-end surface area = 2πrh + πr² (CSA + one base).
Practice Problems
- Find the CSA of a cylinder with radius 14 cm and height 20 cm.
- Find the TSA of a cylinder with radius 3.5 cm and height 10 cm.
- A cylinder has diameter 42 cm and height 25 cm. Find its CSA.
- The CSA of a cylinder is 880 cm² and its radius is 7 cm. Find the height.
- How much sheet metal is needed for a closed tank with radius 2.1 m and height 3 m?
- A roller of diameter 70 cm and length 100 cm makes 20 revolutions. Find the area covered.
- A cylinder open at the top has r = 10.5 cm and h = 15 cm. Find its surface area.
- The radius and height of a cylinder are in the ratio 5 : 7. If its CSA is 440 cm², find the radius and height.
Frequently Asked Questions
Q1. What is the curved surface area of a cylinder?
CSA = 2πrh. It is the area of the curved surface only, excluding the two circular bases. Think of it as the area of the label on a can.
Q2. What is the total surface area of a cylinder?
TSA = 2πr(r + h) = 2πrh + 2πr². It is the area of the entire surface — curved part plus both circular bases.
Q3. How is CSA derived?
Cut the cylinder along its height and unroll it. The curved surface becomes a rectangle with length = 2πr (circumference) and width = h (height). Area = 2πr × h = 2πrh.
Q4. When do you use CSA vs TSA?
Use CSA when the cylinder is open (like a pipe) or when you need only the lateral area (like a label). Use TSA when the cylinder is closed (like a sealed tin can) and you need all surfaces.
Q5. What is the unit of surface area?
Surface area is measured in square units — cm², m², mm², etc. If the radius is in cm and height in cm, the area is in cm².
Q6. What if the cylinder is open at one end?
Surface area = CSA + one circular base = 2πrh + πr². You add only one base instead of two.
Q7. How does doubling the radius affect the CSA?
If radius doubles (and height stays the same), CSA = 2π(2r)h = 4πrh, which is double the original CSA. CSA is directly proportional to r.
Q8. How does doubling the height affect the CSA?
If height doubles (and radius stays the same), CSA = 2πr(2h) = 4πrh, which is double the original CSA. CSA is directly proportional to h.
Q9. What is a hollow cylinder?
A hollow cylinder has two radii — an outer radius R and inner radius r (like a pipe). Its CSA includes both inner and outer curved surfaces: 2π(R+r)h.
Q10. What shape does the net of a cylinder form?
The net of a cylinder consists of one rectangle (the curved surface) and two circles (the bases). The rectangle's length equals the circumference (2πr) and its width equals the height (h).
