Surface Area of Cylinder
Surface Area of Cylinder
In geometry, the area of any shape is the region covered by it in a plane. A cylinder consists of two types of surface, one is a curved surface and the other is circular bases. The area of both circular bases is equal.
The Surface Area of the Cylinder can be Classified into Two Types:

Curved surface area (CSA)

Total surface area (TSA)
Curved Surface Area (CSA) of Cylinder
The curved surface area of a cylinder is defined as the area of its curved surface or the area obtained after excluding the area of its two circular bases. It is also called as Lateral surface area (LSA).
The CSA of cylinder having its base radius ‘r’ and height ‘h’ is given by:
Curved surface area (CSA) of cylinder = 2πrh sq. units.
Total Surface Area (TSA) of Cylinder
The total surface area of cylinder is defined as the sum of areas of its curved surface and two circular bases.
The TSA of cylinder having its base radius ‘r’ and height ‘h’ is given by:
Total surface area (TSA) of cylinder = Area of curved surface + Area of its two circular bases
Or, TSA of cylinder = CSA of cylinder + Area of its two circular bases
= 2πrh + 2πr2 sq. units
So, TSA of cylinder = 2πr (h + r) sq. units
Derivation of the Formula of Surface Area of Cylinder
Consider a situation in which a cylinder has to be covered with coloured papers. The covering has to be done with a minimum amount of paper. So, to cover the cylinder first take a rectangular sheet of paper, whose length (l) is just enough to go round the cylinder and the breadth is equal to the height (h) of the cylinder as shown in the figure below.
(images will be updated soon)
It is to be noted that the length (l) of the rectangular sheet is equal to the circumference of the circular base which is equal to 2πr.
The area of the rectangular sheet gives the curved surface area of the cylinder.
So, curved surface area of the cylinder = Area of the rectangular sheet of paper
= length × breadth
= Circumference of the circular base × h
= 2πr × h
Therefore, curved surface area of the cylinder = 2πrh sq. units
Where, r is the radius of the base and h is the height of the cylinder.
Now, if the top and the bottom of the cylinder are also to be covered with the coloured papers, then we need two circular regions to do that, each of radius r, and having an area of πr2 each.
So, total surface area of cylinder = Area of curved surface + Area of two circular regions
= CSA of cylinder + πr2 + πr2
= 2πrh + 2πr2
Therefore, Total surface area of cylinder = 2πr (h + r) sq. units
Where, r is the radius of the base and h is the height of the cylinder.
Solved Examples:
Q.1. A cylindrical pillar is 50 cm in diameter and 7 m in height. Find the cost of painting the curved surface of the pillar at the rate of ₹12 per sq. mts.
Solution: Given, height(h) of cylindrical pillar = 7 m
And, radius (r) of circular base = \[\frac{50 cm}{2}\] = 25 cm = 0.25 m
So, the curved surface area of the cylindrical pillar = 2πrh sq. units
= 2 × \[\frac{22}{7}\] × 0.25 × 7
= 2 × \[\frac{22}{7}\] ×\[\frac{25}{100}\] × 7
= 11 sq. mts
Given, the cost of painting 1 sq. mts of area is ₹12.
∴ the cost of painting 11 sq. mts of area = 11 × 12 = ₹132
Therefore, the cost of painting the curved surface of the pillar at the rate of ₹12 per sq. mts. is ₹132.
Q.2. The curved surface area of a right circular cylinder of base radius 7 cm is 110 cm2. Find the height of the cylinder.
Solution: Given, the base radius (r) of cylinder = 7 cm.
And, curved surface area of cylinder = 110 cm2
Let the height of the cylinder be h.
Then, 2πrh = 110 cm2
⇒ 2 × \[\frac{22}{7}\] × 7 × h = 110 cm2
⇒ h = 2.5 cm.
Therefore, the height of right circular cylinder is 2.5 cm
Q.3. How many square meters of metal sheet is required to make a closed cylindrical tank of height 1.8 m and base diameter 140 cm?
Solution: Given, height of closed cylindrical tank = 1.8 m
And, the radius (r) of circular base = \[\frac{140 cm}{2}\] = 70 cm = 0.7 m
So, area of metal sheet required to make closed cylindrical tank = total surface area of cylinder
= 2πr (h + r)
= 2 × \[\frac{22}{7}\] × 0.7 × (1.8 + 0.7)
= 2 × \[\frac{22}{7}\] × 0.7 × 2.5
= 2 × \[\frac{22}{7}\] × \[\frac{7}{10}\] × \[\frac{25}{10}\]
= 11 sq. mts
Therefore, the area of metal sheet required to make closed cylindrical tank is 11 sq. mts.