Cylinder Volume Formula
Students who are searching for the Cylinder Volume Formula and solved examples based on volume formulas for cylinder then you are at the right place. It is compulsory for the candidates who are going to appear in the exams that they should have the knowledge of Cylinder Volume Formula and also know the procedure how to calculate the big numbers in less time. In many competitive exams like SSC, UPSC, Bank PO etc. there is Quantitative Aptitude section. This section is fully based on Maths formulas, shortcut tricks and tips. This section requires 20 minutes but those aspirants who don’t have the knowledge about Cylinder Volume Formula could not be able to finish the section in a given time period.
For them we the team of resultinbox.com are providing some cylinder volume formulas and some solved illustrations for the convenience of all candidates. Keep on practicing the formulas and important topics daily so that you will achieve success. Make a time table so that you will complete your preparation before 10-15 days from exams dates.
Surface Area and Volume of a Cylinder
How to find the Volume of a Cylinder
Although a cylinder is technically not a prism, it shares many of the properties of a prism. Like prisms, the volume is found by multiplying the area of one end of the cylinder (base) by its height.
Since the end (base) of a cylinder is a circle, the area of that circle is given by the formula:
Multiplying by the height h we get
π is Pi, approximately 3.142
r is the radius of the circular end of the cylinder
h height of the cylinder
A cylinder is nothing but a set of circular discs stacked one upon the other. So if we compute the space occupied by each of these discs and add them up, what we get is the volume of the cylinder.
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Volume Formulas for Basic Shapes
|Rectangular Solid or Cuboid||l.w.h||l = Length,
w = Width,
h = Height
|Cube||a3||a = length of edge or side|
|Cylinder||πr2h||r = radius of the circular edge,
h = height
|Prism||B. h||B = area of base, (B = side2 or length.breadth)
h = height
|Sphere||4/3 πr3||r = radius of sphere|
|Pyramid||1/3 B.h||B = area of base,
h = height of pyramid
|Right Circular Cone||1/3πr2h||r = radius of the circular base,
h = height (base to tip)
|Square or Rectangular Pyramid||1/3lwh||l = length of base,
w = width of base,
h = height (base to tip)
|Ellipsoid||4/3 πabc||a, b, c = semi – axes of ellipsoid|
|Tetrahedron||√2/12 a3||a = length of the edge|
Cylinder Volume Formula
Solved Examples based on Cylinder Volume Formula
1) What is the volume of the cylinder with a radius of 2 and a height of 6?
Solution: Volume =π*(r)2 (h)
Volume = π*(2)2 (6) = 24
2) What is the volume of the cylinder with a radius of 3 and a height of 5?
Solution: Volume = π *(r)2 (h)
Volume = π *(3)2 (5) = 45 π
3) What is the area of the cylinder with a radius of 6 and a height of 7?
Solution: Volume= π *(r)2 (h)
Volume = π *(6)2 (7)= 252 π
4) The volume and height of a cylindrical container are 440 m3 and 35m respectively. Calculate its radius of the base.
Solution: Volume of the cylindrical container = 440 m³
Height of the cylindrical container = 35 m
Or 440 m³= πr2h
=> r² = 4m
=> r = 2 m
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The radius of the base of the cylindrical container is 2 m.
5) How many litres of water can a cylindrical water tank with base radius 20 cm and height 28 cm hold?
Solution: Base radius of the cylindrical water tank, r = 20 cm
Height of the cylindrical water tank, h = 28 cm
Volume of the cylindrical water tank = πr2h
1 cubic centimeter = 0.001 litre = litre
∴ 35200 cubic centimeter = 35.2 litres
The cylindrical water tank can hold 35.2 litres of water.
6) The figure shows a section of a metal pipe. Given the internal radius of the pipe is 2 cm, the external radius is 2.4 cm and the length of the pipe is 10 cm. Find the volume of the metal used.
Solution: The cross section of the pipe is a ring:
Area of ring = [π (2.4)2– π (2)2]= 1.76 π cm2
Volume of pipe = 1.76 π × 10 = 55.3 cm3
Volume of metal used = 55.3 cm3
7) Find the volume of a cylinder if its radius is of 4 cm and the height is of 5 cm.
The volume of the cylinder is
V= π r2 h= 3.14159 * 42 * 5 =3.14159*16*5 = 3.14159*80 = 251.33 cm2
The volume of the cylinder is 251.33 cm2 (approximately).
8) Two cylinders are joined in a way that the base of one cylinder is overposed on the base of the other as shown in the Figure 2a.
The radius of one cylinder is 10 cm and the height is 4 cm. The radius of the other cylinder is 4 cm and the height is 10 cm.
Find the volume of the composite body.
The Figure 2b represents the side view of the two cylinders.
The common axis is shown in blue in the Figures 2a and 2b.
The volume under consideration is composed of the volume of the first
cylinder and the volume of the second cylinder:
V = π.r12 h1 + π.r22.h2 = π.102.4+ π.42.10= π (10.4+16.10)
The volume of the composite body is 1759.29 cm3 (approximately).
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Note The assumption that the cylinders are co-axial is not necessary. The result is valid for non-axial cylinders too.
9) Find the volume of the solid body concluded between two co-axial cylindrical surfaces (Figure 3) of the radii of 8 cm and 4 cm respectively if the common height of the two cylindrical shells is of 10 cm.
The Figure 3 represents the solid body concluded between two co-axial cylindrical
Their common axis is shown in blue in this Figure.
The volume under consideration is the volume of the larger cylinder minus the volume
of the smaller one, i.e
V = π.82.10 – π.42.10 = π. (64-16).10=3.14159*48*10=1507.96cm3 (approximately).
The volume of the solid body under consideration is 1507.96 cm%5E3 (approximately).
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10) Four through cylindrical holes are made in the solid cylinder parallel to its axis of symmetry (Figure 4).
Find the volume of the obtained solid body if the diameter of the original cylinder is 20 cm, its height is 16 cm and the diameter of each hole is 4 cm.
The volume of the original solid cylinder is V= π.r2 h = π.*102 * 16 = 1600. π.
Hence the volume of the solid body under consideration is
V – 4.v Hole = 1600. π – 4.64. π = 1344. π = 4220.16 cm3 (approximately).
The volume of the solid body under consideration is 4220.16 cm3 (approximately).
11) A through cylindrical hole is made in a rectangular prism (rectangular box) of dimensions 3x4x5 cm along its axis of symmetry parallel to the shortest edge (Figure 5).
Find the volume of the obtained solid body if the diameter of the hole is 2 cm.
The volume of the original rectangular prism is
V prism= 3. 4. 5=60 cm3
The volume of the cylindrical hole is
V Hole = π r 2 h = π * 12 * 3 = 3. π
Hence, the volume of the solid body under consideration is
V = V prism – V Hole = 60 – 3. π = 60-3.3.14=50.58 cm3
The volume of the solid body under consideration is 50.58 cm3 (approximately).
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12) A swimming pool has a cylindrical shape. Find the volume of the pool if its diameter is of 40 ft and the depth is of 8 ft.
Solution: The volume of the swimming pool is equal to the volume of a cylinder with the radius of 20 ft and the height of 8 ft, i.e.
V = π r 2 h = π* 20 2 * 8 = 3200 π
Answer. The volume of diving pool is 10048 cubic ft (approximately).
13) Calculate Volume of cylinder if r = 4 inches and h = 8 inches
= pi × r2× h
= 3.14 × 42× 8
= 3.14 × 16 × 8
= 3.14 × 128
Volume of cylinder = 401.92 inches3
14) Calculate the Volume of cylinder if r = 2 cm and h = 5 cm
= pi × r2× h
= 3.14 × 22× 5
= 3.14 × 4 × 5
= 3.14 × 20
= 62.8 cm3
15) An oil storage tank has a cylindrical shape. Find the volume of the storage tank if its diameter is of 60 m and the height is of 20 m.
V = π r 2 h = π*30 2 * 20 = 1800 π = 56520 m 3 (approximately).
Apply the formula for the volume of a cylinder.
The volume of the storage is equal to
The volume of diving pool is 56520 m3 (approximately).
We like to say all the very best to the students who are going to be appear in the competitive exam. Aspirants don’t get confused among the formulas, read and understand it. Write the formulas in your notebook for practice and be confident at the time of examination.