# Total Surface Area of Right Circular Cylinder

## Definition of Total Surface Area of Right Circular Cylinder

The Total Surface Area of Right Circular Cylinder may be defined as the surface area of the total cylinder including the two bases (top and the bottom). Therefore, the Total Surface Area is the sum of the Curved Surface area and the area of the two circular faces which are at the top and the bottom.

## Total Surface Area of Right Circular Cylinder Formula

The figure below represents the Total Surface Area of Right Circular Cylinder.

Therefore, the Total Surface Area of Right Circular Cylinder will be,

Total Surface Area = Curved Surface Area + Area of Top Face + Area of Bottom Face

Let the radius of the cylinder is $r$

Therefore,

Area of the top face $=\pi r^2$

Area of the bottom face $= \pi r^2$

Therefore, the Total Surface Area will be,

Total Surface Area $= 2\pi r h+\pi r^2+\pi r^2$

Total Surface Area $=2\pi r h+2\pi r^2$

Total Surface Area $=2\pi r(r+h)$

## Solved Examples

### 1. The Curved Surface Area of a Right Circular Cylinder of height 21 cm is 100 sq.cm. Find the diameter of the base of the cylinder?

Solution:
Given,
Height of the Cylinder, $h=21\,cm$
Curved Surface Area of the Cylinder, $CSA=100 cm^2$
We know that,
Curved Surface Area$\,=2\pi rh$
Therefore,
$100 cm^2=2\pi rh$
$100 cm^2=2 \times \frac{22}{7} \times r\times 21$
$r=\frac{25}{33}$
$r=0.75 cm$
The diameter of the cylinder is, $D = 2\times r= 2\times 0.75 =1.5\,cm$

### 2. It is required to make a closed cylindrical tank of height 2 meters and base diameter of 280 cm from a metal sheet. How many square meters of sheet is required for the same?

Solution:
Given,
Height of the cylinder, $h = 2 m$
Diameter of the cylinder, $D = 280 cm = 0.28 m$
Therefore, the Radius of the Cylinder, $r= 0.14 m$
Since the cylindrical tank is closed, the amount of sheet is required is equal to the total surface area of the cylinder. Therefore,
Total Surface Area,
$TSA=2\pi r (r+h)$
$TSA=2\times \frac{22}{7} \times 0.14 (0.14 + 2)$
$TSA=1.88 m^2$
Therefore, the amount of sheet required for making the closed cylindrical tank is $1.88 m^2$

### i. Inner Curved Surface Areaii. Outer Curved Surface Areaiii. Total Surface Area

Solution:
Given,
Height of the metal, $h=49 cm$
Inner Diameter, $d_1=5 cm$
Outer Diameter, $d_2=5.5 cm$
Therefore,
Inner Radius, $r_1=2.5 cm$
Outer Radius, $r_2=2.75 cm$

i. Inner Curved Surface Area
$CSA_1=2\pi r_1 h$
$CSA_1=2\times \frac{22}{7}\times 2.5\times 49$
$CSA_1=770 cm^2$

ii. Outer Curved Surface Area
$CSA_2= 2\pi r_2h$
$CSA_2=2\times \frac{22}{7}\times 2.75\times 49$
$CSA_2=847 cm^2$

iii. Total Surface Area
$TSA=2\pi r_2(r_2+h)$
$TSA=2\times \frac{22}{7}\times 2.75\times (2.75+49)$
$TSA=894.54 cm^2$

### 4. The diameter of a roller is 70 cm and its length is 140 cm. It takes 600 complete revolutions to move once over to level a playground. Find the area of the playground in m2?

Solution:
Given,
Diameter of the Roller, $D=70 cm$
The Radius of the Roller will be, $r= 35 cm$
Height of the Roller, $h=140 cm$
One revolution of the roller is equal to the Curved Surface Area of the roller. Therefore,
$CSA=2\pi rh$
$CSA=2\times \frac{22}{7}\times 35\times 140$
$CSA=30800 cm^2$
$CSA=3.08 m^2$
Since the roller takes 600 revolutions to move once over to level the playground, the area of the playground will be,
$A=600\times CSA$ of the Roller
$A=600×3.08 m^2$
$A=1848 m^2$

### 5. A Cylindrical pillar is 42 cm in diameter and 2 m in height. Find the cost of painting the curved surface of the pillar at the rate of Rs. 10 per m2?

Solution:
Given,
Diameter of the pillar, $D = 42\,cm$
The Radius of the pillar will be, $r = 21 cm = 0.21m$
Height of the Pillar, $h = 2 m$
The curved Surface Area of the Pillar is,
$CSA=2\pi rh$
$CSA=2\pi \frac{22}{7} \times 0.21 \times 2$
$CSA=2.64 m^2$
The cost of Painting per $m^2$ is $= Rs.\,10$
The cost of Painting $2.64 m^2$ is
$= Rs. 10 × 2.64 m^2$
$= Rs. 26.4$

### 6. Curved Surface Area of a Right Circular Cylinder is 5 m2. If the radius of the base of the cylinder is 1.4 m. Find its height.

Solution:
Given,
CSA of the Cylinder$= 5 m^2$
The radius of the Cylinder, $r=1.4 m$
We know that,
$CSA=2\pi rh$
$h=\frac{CSA}{2\pi r}$
$h=\frac{5}{8.8}$
$h=0.56m$

### 7. The inner diameter of a circular well is 4.2 m. It is 10 m deep. Findi. Its inner curved surface areaii. The cost of plastering this curved surface area at the rate of Rs. 40 per m2.

Solution:
Given,
Diameter of the Circular well, $D=4.2 m$
The Radius of the well will be, $r= 2.1 m$
Height of the Circular Well, $h= 10 m$
i. The Inner Curved Surface Area
We know that,
$CSA=2\pi rh$
$CSA=2\times \frac{22}{7}\times 2.1\times 10$
$CSA=132 m^2$
ii. Cost of Painting in $1 m^2$ is $= Rs.\, 40$
Cost of Painting $132 m^2$ is $= Rs.\, 40\times 132= Rs.\, 5280$

### 8. In a hot water heating system there is a cylindrical pipe of length 42 m and diameter 10 cm. Find the total radiating surface of the system.

Solution:
Given,
Diameter of the Cylindrical pipe, $D= 10\,cm$
The radius of the Cylindrical Pipe, $r=5cm=0.05m$
Height of the Cylindrical Pipe, $h=42m$
The total radiating surface of the system is the curved surface area of the pipe.
Therefore,
$CSA=2\pi rh$
$CSA=2\times \frac{22}{7}\times 0.05\times 42$
$CSA=13.2 m^2$