Classwork Exercise and Series (Mathematics-SS3): Surface area and volume of a sphere

Surface Area of a Sphere

A sphere is the locus of points in space equidistant from one point called the centre of the sphere.

Note: the difference between the sphere and the circle. The sphere is the locus of a point in space, while the circle is the locus of a point in a plane.

In general, surface area is the sum of all the shapes that cover the surface of an object. To calculate the surface area of a sphere we multiply 4 by pi by the radius of the sphere squared. Given this formula, we can find the surface area of a sphere when given the radius. Similarly, we can find the radius of a sphere is we are given the surface area.

The image below shows a shape of a sphere

The area of a sphere is mathematically given as 4πr²

The half of a sphere is called the hemisphere.

When we’re talking about the surface area of the sphere, you can think of it as how much paint would you need to cover a tennis ball or if you’d looked at a baseball and you took all the stitching apart, how much leather would you need to make that ball?

Well, to find the surface area of a sphere, you’re going to use the formula that surface area equals 4 times pi times the radius squared. Now, notice the dimensional here. We have r to the second power which agrees with what we know about surface area which is it’s a two dimensional property. So the only thing that you need to know in order to calculate the surface area of a sphere is this formula 4πr². Let’s look at a very basic example of this application.

If the radius of a sphere is 3 centimetres, what is the surface area? Well we’ll start off by writing our surface area formula. Surface area equals 4 pi r squared and then we’ll say our radius is 3 centimetres. So then we just need to substitute in and we’ll know our surface area.
4πr²

4 x 22/7 x 3²

4 x 3.142 x 9

113cm; thus gives the surface area of the sphere.

So when you have a surface area problem and they tell you the radius, all you need to do is to substitute into your formula and simplify.

Volume of a Sphere

In determining the volume of a sphere the following activities are carried out.

1. Cut a hollow ball vertically and horizontally into four equal parts with two open surfaces.

2. Take one part and join any of the two surfaces to form a solid that almost looks like a cone. Notice that the radius of then ball R is equal to the height h and then base r of the cone formed (i.e. R = h = r). But the volume of a cone = 1/3 πr²h, where h is the height and r is the base radius of the cone.

So, the volume of the cone formed = 1/3 πr³ (since R = h =r)

Since there four equal parts of the ball there are four such cones.

 Volume of the entire ball = 4 x1/3 πR³ = 4/3 πR³ cubic units

 Since a ball is spherical in nature it is likely that the volume of a sphere is derived thus.

 Volume of sphere = 4/3 πR³ cubic units where R is the radius of the sphere.

Examples

Given a sphere with the radius 7cm. Find the volume of this sphere. Take π as 3.14.

Since the values for r, and π are given, we can substitute r with 5 and π with 3.14. After substituting these values, we can calculate V as shown below:

V = 4/3 πr³

V =4/3(3.14)(7)³

V = 4/3(3.14)(343)

= 4/3(1077.02)

= 4(1077.03)/3

= 4308.08/3

= 1436.03cm³

Surface area of Hemisphere is S = ½(4πr²)

Surface area and Volume of composite shape

Geometric Shape Surface AreaB = area of the baseP = perimeter of the base VolumeB = area of the baseP = perimeter of the base prism (general)   SA = 2B + Ph V = Bh Triangular Prism   SA = 2B + PhSA = 2(1/2ab) + (b + c + d)hSA = ab + (b + c + d)h V = BhV = 1/2 abh Rectangular prism SA = 2B + PhSA = 2(lw) + (2l + 2w)h V = lwh Regular square prism SA = 2B + PhSA = 2(s2) + (4s)h V = BhV = s2h Regular pentagonal prism SA = 2B + PhSA = 2(1/2ans) + nshSA = 2(1/2a)(5)s + 5shSA = 5as + 5sh V = BhV = 1/2anshV = 1/2a(5)shV = 5/2ash Regular hexagonal prism SA = 2B + PhSA = 2(1/2ans) + nshSA = 2(1/2a)(6)s + 6shSA =6as + 6sh V = BhV = 1/2anshV = 1/2a(6)shV = 3ash Cube SA = 2B + PhSA = 2(s2) + (4s)s = 6s2 V = BhV = s3 Regular pyramid (general) SA = B + n(1/2sl)           l = slant height V = 1/3Bh Regular triangular pyramid SA = B + n(1/2sl)SA = 1/2as + (3)(1/2sl)SA = 1/2as + 3/2sl               l = slant height   V = 1/3BhV = 1/3(1/2 ab)hV = 1/6 abh Regular square pyramid SA = B + n(1/2sl)SA = s2 + (4)(1/2sl)SA = s2 + 2sl               l = slant height V = 1/3BhV = 1/3(b2)h= 1/3b2h Regular pentagonal pyramid SA = B + n(1/2sl)SA = 1/2a(5)s + (5)(1/2sl)SA = 5/2as + 5/2sl               l = slant height   V = 1/3BhV = 1/3(1/2anb)hV = 1/6 anbh Regular hexagonal pyramid SA = B + n(1/2sl)SA = 1/2a(6)s + (6)(1/2sl)SA = 3as + 3sl              l = slant height   V = 1/3BhV = 1/3(1/2anb)hV = 1/6 anbh Cylinder SA = 2B + phSA = 2(π r2) + (2πr)h V = BhV = π r2h

 

Questions

1. Calculate the volume of the sphere with radius 3.2. Take π =3.14.

A. 207.04 cm³ B. 137.2 cm³ C. 189. 5 cm³ D. 200 cm³

Find the volume of the following spheres with

2. Diameter 8.6 cm

A. 432 cm³ B. 333 cm³ C. 453 cm³ D. 234 cm³

Calculate the surface area of the spheres with

3. Radius 3.2 cm

A. 125 cm B. 128 cm C. 121 cm D. 230 cm

4. Calculate the surface area of the sphere, given diameter 5.7 cm

A. 122 cm B. 102.01 cm C. 201.10 cm D. 98.4 cm

5. Which of these is correct for the Volume of a Regular Square Pyramid

A. v = ½ b²h B. v = 1/3 b²h C. b v = ½ b³h D. v = 1/3 b³h

Answer

1. B  2. B  3. C  4. B  5. B

For more notes;see: http://passnownow.com/classwork-support/