In this article provided formulas of Surface Area and Volume of a Sphere and a Hemisphere with examples. Volume and surface area of a three dimensional (3D) solid geometrical shapes.
- 1 Sphere and Hemisphere Formulas with Examples
Sphere and Hemisphere Formulas with Examples
Sphere is a one of the three dimensional solid figure, Which made up of all points in the space, Which lie at a constant distance called the radius, from a fixed point called the center of a sphere. ( i.e The set of all points in the three space equidistant from a given points form sphere)
Surface Area of a Sphere
Surface area of the sphere will be covered completely fill the region of four circles, all of the same radius as of the sphere. So
Surface area of a Sphere with radius ( r ) = 4 x ( π r2 ) = 4 π r2
Volume of a Sphere
Volume of a sphere is equal to 4 π/3 times the cube of its radius.
Volume of a sphere =
The solid sphere is divided into two equal parts and its each half part is called a hemisphere.
Surface Area of a Hemisphere
Surface area of the hemisphere having two faces. There is curved face (Cap Area) and flat face ( base area).
Surface area of a Sphere with radius ( r ) = 4 π r2
Then cap area of hemisphere is half surface area of the sphere
i.e Cap Area or Curved surface area of the hemisphere = 1/2 ( 4 π r2 ) = 2 π r2
Flat surface area or base area of the hemisphere = Area of the circle with same radius = π r2
Total Surface Area of the Hemisphere = 3 π r2
Volume of a Hemisphere
Hemisphere is half of a sphere
Volume of a hemisphere = ( 1/2 ) ( 4 /3 π r2 ) =
Surface Area and Volume of Hemisphere shell
Take external radius is ‘R’ and inner radius is ‘r’ of hemisphere
Curved surface area of hemisphere shell = 2 π ( R2 + r2 ) ( Considered inside and outside area of hemisphere)
Volume of Hemisphere shell =
Volume of Hollow Sphere
Here Hollow sphere inner radius – r & outer radius – Rr
Volume of allow sphere =
Solid Sphere is the region in space bound by a sphere.
Examples on surface area and volume of sphere and hemisphere
Example -1 : Find the surface area and volume of sphere having the radius 7 mm
Solution: Here radius of sphere = r = 7 mm
Now Surface area of a Sphere = 4 π r2
= 4 x (22/7) x 7 x 7 = 616 mm2
Volume of a sphere = = (4312/3 ) mm3
Example-2 : A shot-put is a metallic sphere of radius 2.1 cm and density of the metal used for same is 7.8 gm/cm3. Find the weight of the shot-put
Solution: The shot-put is a solid sphere with the radius 2.1 cm
Volume of shot-put = = (4/3) x (22/7) x 2.1 x 2.1 x 2.1 = 38.8o8 cm3
Density of metal used for making shot-put is 7.8 gm/cm3
Weight of the shot-put = 38.8o8 x 7.8 ( gm x cm3 /cm3 ) = 302.7 gm
Example-3 : The volume of a solid hemisphere is 18π mm3. Find the total surface of the same.
Solution: Take the radius of sphere = r, then
Volume of a sphere =
⇒ = 18π
⇒ r = 3
Total Surface area of hemisphere = 3 π r2 = 27 π
Example-4: The three metallic spheres have radii 3cm, 4cm & 5cm respectively, are melted to form a single solid sphere. Find the radius of the resulting sphere.
Solution: Take the radius of final sphere is R
Volume of final sphere = volume of individual spheres
⇒ (4/3) x π x R3 = (4/3) x π x 33 + (4/3) x π x 43 + (4/3) x π x 53
⇒ R3 = ( 33 + 43 + 53 ) = 63
⇒ R = 6
Example-5 : A hemispherical bowl has a radius of 4.2 cm. What would be the volume of mercury it would contain?
Solution: Here radius of hemispherical bowl = r = 4.2 cm
Volume of a hemisphere = = (2/3) x (22/7) x 4.2 x 4.2 x 4.2 = 155.232 cm3
Note: Here asked only volume of mercury. In case of ask mass of the mercury then multiplying with density with volume)
Example-6: A metallic sphere have diameter of 6 cm. The metallic sphere is melted and manufacturing into a wire of uniform cross section. Find the radius of wire if the length of the wire is 36 m.
Solution: Here radius of sphere = r = 3 cm & Length of wire = 36 m = 3600 cm
Volume of Sphere = Volume of wire
⇒ (4/3) π r3 = π r2 h
⇒ ( 4/3) x ( 22/7) x 3 x 3 x 3 = (22/7) x r2 x 3600
⇒ r = 1/10 cm = 1 mm
Example-7: Volumes of two spheres are in the ratio 125:64. Then find the ratio of their surface areas.
Solution: Take radius of spheres are r1 and r2
Ratio of volume of Spheres = 125 : 64
⇒ = 125 : 64
⇒ r1 : r2 = 5 : 4
Ratio of surface area of spheres
⇒ = r1 2: r2 2 = 52 : 4 2 = 25 : 16
Example-8: A hemisphere tank is fabricated with an iron sheet of 1 cm thick. If the inner radius is 10 cm , then find the weight of the iron used to make the tank. (To be take Density of iron = 7 gm/cm3)
Solution: Hemisphere inner radius = 9 cm,then outer radius = 10 cm
Volume of shell of hemisphere =
= ( 2/3 )x (22/7) x ( 103 – 93) = ( 2/3 )x (22/7) x (1000 – 729) = (11924/21) cm3
Weight of the iron required = (11924/21) x 7 = (11924 /3) = 3974.66 gm
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