Surface Area of Hollow Cylinder

Concept Explanation

Surface Area of Hollow Cylinder

Hollow Cylinder: A solid bounded by two co-axial cylinders of the same height and different radii is called a hollow cylinder. Solids like iron pipes, rubber tubes etc. are hollow cylinder.

Let R and r be the external and internal radii of a hollow cylinder and h be its height as shown in the following figure.

Then we have the following results:

(i) Each base Surface Area = $pi (R^{2}-r^{2});sq;units$

(ii) Curved(Lateral) surface Area = (External surface area) + (Internal surface Area)

$=2pi Rh+2pi rh$

$=2pi h(R+r);sq; units$

(iii) Total Surface Area $=2pi Rh+2pi rh+2pi (R^{2}-r^{2})$

$=2pi h(R+r)+2pi (R+r)(R-r);sq; units$

$=2pi (R+r)(h+R-r);sq; units$

Illustration: Find the total surface area of the hollowed out cylinder shown in the adjacent figure.

Solution: Total surface area of hollow cylinder = Outer CSA + Inner CSA + top and bottom ringed area

$=2pi r_{1}h+2pi r_{2}h+2pi (r_{1}^{2}-r_{2}^{2})$

$=2pi [5times 18+4times 18+(5^{2}-4^{2})];cm^{2}$

$=2pi(90+72+9);cm^{2}$

$=342times 3.14;cm^{2}$

$=1073.88;cm^{2}$

Illustration: An iron pipe 20 cm long has exterior diameter equal to 25 cm. If the thickness of the pipe is 1 cm, Find the whole surface area of the pipe.

Solution: We have,

R = External radius = 12.5 cm

r = Internal radius

= (external radius - Thickness)

= 12.5 - 1

= 11.5 cm

h = length of the pipe = 20 cm

So, Total Surface Area of the Pipe = (External curved surface) + (Internal curved surface) + 2(Area of the base of the ring)

$=2pi Rh +2pi rh+2(pi R^{2}-pi r^{2})$

$=2pi(R+r)h +2pi (R^{2}-r^{2})$

$=2pi(R+r)h +2pi (R-r)(R+r)$

$=2pi(R+r) (h+R-r)$

$=2times frac{22}{7}times (12.5+11.5)times (20+12.5-11.5);cm^{2}$

$=2times frac{22}{7}times (24)times (21);cm^{2}$

$=3168;cm^{2}$

Illustration: A metal pipe is 77 cm long. The inner diameter of a cross section is 4 cm, the outer diameter being 4.4 cm. Find its

(i) inner curved surface area

(ii) outer curved surface area

(iii) total surface area

Solution: Length of the metal pipe = 77 cm(h)

Inner radius $(r_{2})=frac{4}{2}=2;cm$

Outer radius $(r_{1})=frac{4.4}{2}=2.2;cm$

(i) Inner Curved Surface Area $=2pi r_{2}h$

$=2times frac{22}{7}times 2times 77$

$=968;cm^{2}$

(ii) Outer Curved Surface Area $=2pi r_{1}h$

$=2times frac{22}{7}times 2.2times 77$

$=1064.8;cm^{2}$

(iii) Total surface Area = Inner Curved surface Area + outer curved surface Area + top and bottom ringed area

$=968;cm^{2}+1064.8;cm^{2}+2pi (r_{1}^{2}-r_{2}^{2})$

$=2032.8;cm^{2}+2times frac{22}{7}times left ( 4.84-4 right );cm^{2}$

$=(2032.8+5.28);cm^{2}$

$=2038.08;cm^{2}$

Sample Questions

(More Questions for each concept available in Login)

Question : 1

Find the total surface area of the hollowed out cylinder having inner and outer radius 3 cm and 5 cm respectively and height 7 cm.
A. $422.26;cm^{2}$	B. $452.16;cm^{2}$	C. $452.26;cm^{2}$	D. $451.26;cm^{2}$
Right Option : B
View Explanation

Question : 2

Find the total surface area of the hollowed out cylinder having inner and outer radius 2 cm and 3 cm respectively and height 5 cm.
A. $188.4;cm^{2}$	B. $108.4;cm^{2}$	C. $188.6;cm^{2}$	D. $178.6;cm^{2}$
Right Option : A
View Explanation

Question : 3

The total surface area of a hollow cylinder which is open from both sides is 4620 sq cm, area of the base ring is 115.5 sq cm and height 7 cm. Find the thickness of the cylinder.
A. $frac{7}{19};cm$	B. $frac{9}{19};cm$	C. $frac{3}{19};cm$	D. none of these
Right Option : A
View Explanation

Chapters

Real Numbers I

Real Numbers II

Pair of Linear Equations I

Pair of Linear Equations II

Polynomial I

Polynomials II

Quadratic Equations I

Quadratic Equations II

Introduction to Trigonometry I

Introduction to Trigonometry II

Arithmetic Progression I

Arithmetic Progression II

Triangles I

Triangles II

Some Applications of Trigonometry

Coordinate Geometry I

Coordinate Geometry II

Introduction to Trigonometry

Direct and Inverse Variation

Surface Area and Volume I

Surface Area and Volume II

Areas Related to Circles

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