We can find the volume V of both a prism and a cylinder by multiplying the height by the area of the base.
Let the area of the base of a cylinder be B and the height of the cylinder be h. Write a formula for the cylinder’s volume V.
V = Bh
The base of a cylinder is a circle, so for a cylinder,
B = πr2
Then, we have
V = πr2h cubic units
Example 1 :
Find the volume of a solid cylinder whose radius is 14 cm and height is 30 cm.
Solution :
We need to find the volume of the solid cylinder.
Radius of the cylinder = 14 cm
Height of the cylinder = 30 cm
Volume of the right circular cylinder = πr2h
= (22/7) · (14)2 · 30
= (22/7) · 14 · 14 · 30
= 18480 cubic.cm
Volume of the cylinder = 18480 cubic.cm
Example 2 :
A patient in a hospital is given soup daily in a cylindrical bowl of diameter 7 cm. If the bowl is filled with soup to a height of 4 cm, then find the quantity of soup to be prepared daily in the hospital to serve 250 patients?
Solution :
Radius of the cylinder = 7/2 cm
Height of the cylinder = 4 cm
To find quantity of soup in one bowl, we have to find the quantity of each bowl.
Volume of the right circular cylinder = πr2h
= (22/7) · (7/2)2 · 4
= (22/7) · (7/2) · (7/2) · 4
= 154 cm3
Volume of soup in one cylindrical bowl = 154 cm3
Volume of soup in 250 cylindrical bowl = 250 · 154
= 38500 cm3
1000 cm3 = 1 L
Therefore required quantity of soup = 38500/1000
= 38.5 L
Required quantity of soup for 25 patients = 38.5 L
Example 3 :
The sum of the base radius and the height of a solid cylinder is 37 cm. If the total surface area of the cylinder is 1628 sq.cm, then find the volume of the cylinder.
Solution :
Let r and h are the radius and height of the cylinder respectively
Sum of radius and height = 37
r + h = 37 cm
Total surface area of cylinder = 1628 sq.cm
2πr(h + r) = 1628
2πr(37) = 1628
2πr = 1628/37
2 · (22/7) · r = 44
r = 44 · (1/2) · (7/22)
r = 7
7 + h = 37
h = 30 cm
Volume of the right circular cylinder = πr2h
= (22/7) ·72·30
= (22) x (7) x (30)
= 4620 cm3
Volume of cylinder = 4620 cm3
Kindly mail your feedback to v4formath@gmail.com
We always appreciate your feedback.
©All rights reserved. onlinemath4all.com
Oct 01, 24 12:04 PM
Oct 01, 24 11:49 AM
Sep 30, 24 11:38 AM