Surface Area and Volume of Sphere, Hemisphere, Hollow Sphere Formulas, Examples

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.

Sphere and Hemisphere Formulas with Examples

Sphere

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 =

Hemisphere

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

then

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

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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