Processing math: 100%

Right Circular Cylinder

A cylinder, whose uniform cross section perpendicular to its height (or length) is a circle, is called a right circular cylinder.

A Right Circular Cylinder



A Cylinder but Not a Right Circular One

A right circular cylinder has two plane faces which are circular and curved surface.

A right circular cylinder is a solid generated by the revolution of a rectangle about one of its sides.

Right Circular Cylinder Revolution of a Rectangle

Here, the rectangle PQRS has been revolved about the side SR. The length of the side SR will be the height and the length of the side PS will be the radius of the cross section.



Volume of a Right Circular Cylinder

= (Area of the Cross Section) × Height

= (Area of the Base) × Height

= (Area of the Circle whose Radius is r) × h

= πr2h




Lateral Surface Area or Curved Surface Area of a Right Circular Cylinder

Curved Surface Area of a Right Circular Cylinder

= (Perimeter of the Cross Section) × Height

= 2πrh










Total Surface Area of a Right Circular Cylinder

Total Surface Area of a Right Circular Cylinder

= Lateral Surface Area + 2 × (Area of the Cross Section)

= 2πrh + 2πr2

= 2πr(h + r)









Solved Examples on Volume and Surface Area of Right Circular Cylinders:

1. The radius of the base of a solid right circular cylinder is 7 cm and its height is 20 cm. Find its (i) volume, (ii) curved surface area, and (iii) total surface area.

Solution:

Here, the radius of the base of a solid right circular cylinder r = 7 cm and height h = 20 cm.

Solid Right Circular Cylinder

(i) Volume of the right circular cylinder = πr2h

                                                         = 227 ∙ 72 ∙ 20 cm3

                                                         = 3,080 cm3.


(ii) Curved surface area of the right circular cylinder

                                                         = 2πr × h

                                                         = 2 ∙ 227 ∙ 7 ∙ 20 cm2

                                                         = 880 cm2.


(iii) Total surface area of the right circular cylinder

                                                          = 2πrh + 2πr2

                                                          = 2πr(h + r)

                                                          = 2 ∙ 227 ∙ 7(20 + 7) cm2 

                                                          = 1,188 cm2.


2. The volume of the right circular cylinder is 308 cm3 and its height is 8 cm. Find (i) the radius of its cross section, and (ii) the area of its curved surface.

Solution:

Let r be the radius of the cross section (or base).

Here h = 8 cm.

Then, the volume of the right circular cylinder = πr2h = πr2 ∙ 8 cm.

Therefore, 308 cm3 = πr2 ∙ 8 cm

⟹ r= 308cm2π8

⟹ r= 308cm22278

⟹ r= 308×722×8 cm2.

⟹ r= 494 cm2.

Therefore, r = 72 cm.

Therefore, (i) radius of the cross section of the right circular cylinder

                                                                = 72 cm

                                                                = 3.5 cm,

(ii) area of the curved surface of the right circular cylinder

                                                                = 2πrh

                                                                = 2 × 227 × 72 × 8 cm2.

                                                                = 176 cm2.


3. The radius of the base of a right circular cylinder is increased by 75% and the height is decreased by 50%. Find the per cent increase or decrease in the (i) volume, and (ii) curved surface area.

Solution:

Let the radius of the right circular cylinder = r,

Height of the right circular cylinder = h,

Volume of the right circular cylinder = V, and

Curved surface area of the right circular cylinder = S.

Therefore, V = πr2h and S = 2πrh.

After the changes, radius = R = r + (75% of r)

                                           = r + ¾ r

                                           = 74r;

Height = H = h – (50% of h)

                 = h – ½ h

                 = ½ h.

(i) The changed volume V’ = πR2H

                                       = π ∙ (74r)2  ∙ 12h

                                       = 4932 πr2h

                                       = 4932V.

Therefore, the volume increases by 4932V – V, i.e., 1732V.

Therefore, the percent increase in volume = 1732VV × 100%

                                                             = 17×10032%

                                                             = 5318%.


(ii) The changed curved surface area S’ = 2πRH

                                                         = 2π ∙ 74r ∙ 12 h

                                                         = 78 ∙ 2πrh

                                                         = 78 S.

Therefore, the curved surface area decreases by S - 78 S , i.e., 18 S.

Therefore, the per cent decreases in curved surface area = 18SS × 100%

                                                                                  = 1212%.




9th Grade Math

From Right Circular Cylinder to HOME PAGE




Didn't find what you were looking for? Or want to know more information about Math Only Math. Use this Google Search to find what you need.



New! Comments

Have your say about what you just read! Leave me a comment in the box below. Ask a Question or Answer a Question.




Share this page: What’s this?

Recent Articles

  1. Formation of Numbers | Smallest and Greatest Number| Number Formation

    Jul 15, 25 11:46 AM

    In formation of numbers we will learn the numbers having different numbers of digits. We know that: (i) Greatest number of one digit = 9,

    Read More

  2. Formation of Square and Rectangle | Construction of Square & Rectangle

    Jul 15, 25 02:46 AM

    Construction of a Square
    In formation of square and rectangle we will learn how to construct square and rectangle. Construction of a Square: We follow the method given below. Step I: We draw a line segment AB of the required…

    Read More

  3. 5th Grade Quadrilaterals | Square | Rectangle | Parallelogram |Rhombus

    Jul 15, 25 02:01 AM

    Square
    Quadrilaterals are known as four sided polygon.What is a quadrilateral? A closed figure made of our line segments is called a quadrilateral. For example:

    Read More

  4. 5th Grade Geometry Practice Test | Angle | Triangle | Circle |Free Ans

    Jul 14, 25 01:53 AM

    Name the Angles
    In 5th grade geometry practice test you will get different types of practice questions on lines, types of angle, triangles, properties of triangles, classification of triangles, construction of triang…

    Read More

  5. 5th Grade Circle Worksheet | Free Worksheet with Answer |Practice Math

    Jul 11, 25 02:14 PM

    Radii of the circRadii, Chords, Diameters, Semi-circles
    In 5th Grade Circle Worksheet you will get different types of questions on parts of a circle, relation between radius and diameter, interior of a circle, exterior of a circle and construction of circl…

    Read More