Ninth Grade Mathematics: Surface Area & Volume of Spheres and Hemispheres

The sphere is the supreme three-dimensional solid of spatial geometry, representing the set of all points in space equidistant from a fixed center. Possessing continuous spherical symmetry, the sphere has no vertices, no edges, and no flat faces. In this chapter, you will master the surface area and volume of spheres and hemispheres, exploring Archimedes' famous formulas.

The Geometry of the Sphere

A sphere is formed by revolving a semicircle around its diameter through 360 degrees.

Anatomy of a Sphere:
                                   . - ~ ~ - .
                               . '      |      ' .
                             /          |          \
                            /           | r         \
                           ;      r     v            ;
                           | <----------O----------> | Diameter = 2*r
                           ;            * Center     ;
                            \                       /
                             \                     /
                               . _             _ '
                                   ' - ~ ~ - .

The fixed center is O, and the distance from the center to any point on the spherical surface is the radius r. Any plane passing through the center cuts the sphere into two equal halves along a great circle of area pi * r^2.

Surface Area of a Sphere

Archimedes of Syracuse proved in 'On the Sphere and Cylinder' that the surface area of a sphere is exactly equal to the area of four of its great circles:

Surface Area of a Sphere:

Surface Area = 4 * pi * r^2

Archimedes also proved that the surface area of a sphere is identical to the curved surface area of a cylinder that circumscribes it tightly (with radius r and height 2r: CSA = 2pir(2r) = 4pi*r^2).

Hemispheres: Curved and Total Surface Area

When a sphere is sliced through its center along a great circle, it divides into two identical hemispheres.

                               Hemisphere
                               . - ~ ~ - .
                           . '             ' .
                         /                     \   Curved Surface = 2 * pi * r^2
                        *-----------------------*
                          . '      r      ' .       Circular Base = pi * r^2
                              ' - ~ ~ - .

A hemisphere possesses two distinct surface areas:

Curved Surface Area (CSA):
Half of the sphere's total surface:
CSA = (1/2) * (4 * pi * r^2) = 2 * pi * r^2

Total Surface Area (TSA):
Sum of the curved surface and the flat circular base:
TSA = 2 * pi * r^2 + pi * r^2

TSA = 3 * pi * r^2

A hollow hemisphere (like an open bowl) has only the curved surface area 2pir^2, whereas a solid hemisphere includes the flat base, requiring 3pir^2.

Volume of a Sphere and Hemisphere

Archimedes' crowning achievement was proving that the volume of a sphere is two-thirds the volume of its circumscribing cylinder:

Volume of Sphere = (2/3) * (pi * r^2 * (2*r))

Volume of Sphere = (4/3) * pi * r^3

Because a hemisphere is exactly half of a sphere:

Volume of Hemisphere = (1/2) * ( (4/3) * pi * r^3 )

Volume of Hemisphere = (2/3) * pi * r^3

Notice the dimensional consistency: surface area involves r^2 (two dimensions), while volume involves r^3 (three dimensions).

Chapter Exercises

Exercise 1: Find the surface area and volume of a spherical cannonball of radius 10.5 cm (use 22/7 for pi).

Exercise 2: A solid hemisphere has a radius of 7 cm. Calculate its Curved Surface Area, its Total Surface Area, and its Volume (use 22/7 for pi).

Exercise 3: Twenty-seven identical solid metallic spheres, each of radius r, are melted down and recast to form a single large solid sphere of radius R. Find the radius R of the new sphere in terms of r, and determine the ratio of the surface area of the original small sphere to that of the large sphere.

Solutions and Explanations

Solution 1: Given radius r = 10.5 cm = 21/2 cm. Surface Area is SA = 4 * pi * r^2 = 4 * (22/7) * (21/2) * (21/2) = 4 * (22/7) * (441 / 4) = 22 * 63 = 1,386 square cm. Volume is V = (4/3) * pi * r^3 = (4/3) * (22/7) * (21/2) * (21/2) * (21/2) = (4/3) * (22/7) * (9261 / 8) = (88 / 21) * (9261 / 8) = 11 * 441 = 4,851 cubic cm.

Solution 2: For a solid hemisphere with r = 7 cm: Curved Surface Area is CSA = 2 * pi * r^2 = 2 * (22/7) * 7 * 7 = 2 * 22 * 7 = 308 square cm. Total Surface Area is TSA = 3 * pi * r^2 = 3 * (22/7) * 7 * 7 = 3 * 22 * 7 = 462 square cm. Volume is V = (2/3) * pi * r^3 = (2/3) * (22/7) * 7 * 7 * 7 = (2/3) * 22 * 49 = 2,156 / 3 = 718.67 cubic cm.

Solution 3: When metal is melted and recast, the total volume of metal is conserved. Volume of 27 small spheres = Volume of 1 large sphere: 27 * ((4/3) * pi * r^3) = (4/3) * pi * R^3. Dividing both sides by (4/3)pi gives 27 * r^3 = R^3. Taking the cube root of both sides yields R = 3 * r. The radius of the new large sphere is exactly 3 times the radius of a small sphere. To find the ratio of surface areas: Surface Area of small sphere s = 4 * pi * r^2; Surface Area of large sphere S = 4 * pi * R^2 = 4 * pi * (3r)^2 = 4 * pi * 9*r^2 = 36 * pi * r^2. The ratio of small to large surface area is s / S = (4 * pi * r^2) / (36 * pi * r^2) = 4 / 36 = 1 / 9 (or 1 : 9).