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## Surface Area and Volume of a Sphere

Sphere: a round ball-shaped object; a set of points in space that are a given distance (radius) from a fixed point (centre).

Surface Area =4r2 If the radius r is given

d=2r

d2=r

Surface Are=4d24

=d2 if the diameter d is given

#### Example 1: Calculate the Surface Area.

Solution: r=3.5m

∴ The surface area is approx. 153.94m2

#### Example 2: A baseball has a diameter of 7.3 cm

a) How much leather is needed to cover this baseball?

b) If the leather is being cut from a square sheet of leather with a side length of 7.

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3 cm,

how much leather is leftover?

Leather left over=Asquare-S.A.

=7.37.3-167.42

=53.29-167.42

=-114.13

∴ We do not have enough leather.

c) If a sheet of leather has an area of 1 m2, how many balls can be covered using this sheet of leather?

1m=100cm

1m2=1002cm2

1m2=10000cm2

Number of baseball=10000167.42

∴ 59 balls can be covered with a sheet of leather that has an area of 1m2

Volume of sphere

VSphere=32VCylinder

=23r2h

=32r22r

=43r3

VSphere=43r3

#### Example 3: Calculate the volume

Solution:

#### Example 4: A sphere has a surface area of 100 cm .

Calculate the radius of the sphere, to the nearest tenth of a centimetre.

Solution:

S.A.=100cm2

S.A.=4r2sub S.A.=100

100=3r2Solve for r

1004=4r24

r2=1004

r2=1004

r=2.82

So, r=2.82cm

∴ The radius of the sphere is approx. 2.82cm

#### Example 5: A baseball has a diameter of 7.3 cm. They are packed in a cardboard box that is just large enough to hold 12 baseballs.

a) What shape is the box?

The box is a rectangular prism.

b) What are the possible dimensions?

h=47.3

=29.2cm

w=7.3cm

l=37.3

=21.9cm

∴ The possible dimensions are 21.9cm by 7.3cm by 29.2cm

c) What is the volume of the 12 baseballs?

∴V12basketballs=12203.69

=2444.28cm3

d) How much airspace is there inside the box?

V=Vbox-V12baseballs

=lwh-2444.28

=21.97.329.2-2444.28

=4668.20-2444.28

=2223.92

∴ There is approx. 2223.92cm3 of airspace inside the box

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