Tenth Grade Mathematics: Surface Area and Volume of Right Circular Cones and Spheres

While cylinders represent uniform vertical extrusion, many curved solids taper to a sharp vertex or exhibit perfect omnidirectional symmetry. A right circular cone tapers linearly from a circular base to an apex, forming a conical boundary governed by both vertical height and slant height. In contrast, a sphere represents the pinnacle of three-dimensional symmetry, with every surface point equidistant from a single center. Slicing a sphere in half creates hemispheres with distinct flat and curved surfaces. In this chapter, we derive the surface areas and volumetric formulas for cones, spheres, and hemispheres.

The Right Circular Cone

A right circular cone is a solid generated by rotating a right-angled triangle about one of its perpendicular legs.

Geometry of a Right Circular Cone:
                          Apex V
                            /                           /               Slant        / |  \ Slant Height (l)
             Height (l)  /  |h                          /   |                           +----+-----+ Circular Base (Radius r)
                            r

Parameters of a Cone:
- Radius of circular base: r
- Vertical (perpendicular) height: h
- Slant height: l
By the Pythagorean theorem in the cross-sectional right triangle:
                      l^2 = r^2 + h^2   ===>   l = sqrt( r^2 + h^2 )

Formulas for a Cone:
1. Curved (Lateral) Surface Area (CSA):
   Area of the unrolled conical sector:
   CSA = pi * r * l = pi * r * sqrt(r^2 + h^2)

2. Total Surface Area (TSA):
   Curved surface plus flat circular base:
   TSA = pi * r * l + pi * r^2 = pi * r * (l + r)

3. Volume (V):
   Exactly ONE-THIRD the volume of a cylinder with identical radius and height:
   V = (1/3) * pi * r^2 * h

Derivation of the Conical CSA Formula

When the curved lateral wall of a cone is sliced along its slant edge and flattened: The unrolled surface is a SECTOR of a large circle of radius R = l (the slant height). The curved arc length of this sector equals the circumference of the cone's base: C = 2 * pi * r. Recall from sector geometry: Area = (1/2) * (Arc Length) * (Radius): CSA = (1/2) * (2 * pi * r) * l = pi * r * l.

The Sphere

A sphere is the locus of all points in three-dimensional space that are at a constant distance r from a fixed center O.

Geometry of a Sphere:
                             +-----+
                            /   |                              |----+----| Radius r
                            \   |   /
                             +-----+

Archimedes discovered that the surface area of a sphere of radius r is exactly equal to the curved surface area of a circumscribing cylinder of radius r and height h = 2r: CSA(Cylinder) = 2 * pi * r * (2r) = 4 * pi * r^2.

Formulas for a Sphere of Radius r:
1. Surface Area (S):
   (A sphere has only one continuous surface, so CSA = TSA):
                     S = 4 * pi * r^2

2. Volume (V):
   V = (4/3) * pi * r^3

The Hemisphere

When a solid sphere is partitioned through its center by a flat plane, each half is termed a hemisphere:

Geometry of a Hemisphere:
                   +-------------------+ Flat Circular Base (Area = pi * r^2)
                    \                 /
                     \       r       /
                      \      .      /  Curved Boundary (Area = 2 * pi * r^2)
                       \           /
                        +---------+

Formulas for a Hemisphere of Radius r:
1. Curved Surface Area (CSA):
   Half of the sphere's surface area:
   CSA = (1/2) * (4 * pi * r^2) = 2 * pi * r^2

2. Total Surface Area (TSA):
   Curved surface PLUS the flat circular cross-section:
   TSA = 2 * pi * r^2 + pi * r^2 = 3 * pi * r^2

3. Volume (V):
   Half of the sphere's volume:
   V = (1/2) * [(4/3) * pi * r^3] = (2/3) * pi * r^3

CRITICAL DISTINCTION: For a hollow hemisphere (open at the top), the surface area is 2pir^2. For a solid hemisphere, the flat circular base adds pir^2, bringing the total surface area to 3pi*r^2.

Chapter Exercises

Exercise 1: Find the curved surface area and total surface area of a right circular cone of base radius 7 cm and slant height 25 cm. Also calculate its volume.

Exercise 2: The diameter of a metallic sphere is 4.2 cm. If the density of the metal is 8.9 g/cm^3, what is the mass of the sphere? (Use pi = 22/7).

Exercise 3: A hemispherical bowl made of brass has inner diameter 10.5 cm. Find the cost of tin-plating it on the inside at the rate of 16 dollars per 100 cm^2.

Solutions and Explanations

Solution 1: Given base radius r = 7 cm and slant height l = 25 cm. The curved surface area is: CSA = pi * r * l = (22/7) * 7 * 25 = 22 * 25 = 550 cm^2. The total surface area is: TSA = pi * r * (l + r) = (22/7) * 7 * (25 + 7) = 22 * 32 = 704 cm^2. To find the volume, first find vertical height h: h = sqrt(l^2 - r^2) = sqrt(25^2 - 7^2) = sqrt(625 - 49) = sqrt(576) = 24 cm. The volume is: V = (1/3) * pi * r^2 * h = (1/3) * (22/7) * 7^2 * 24 = (1/3) * (22/7) * 49 * 24 = (1/3) * 22 * 7 * 24 = 22 * 7 * 8 = 1232 cm^3.

Solution 2: The diameter is d = 4.2 cm, so radius is r = 2.1 cm. The volume of the sphere is: V = (4/3) * pi * r^3 = (4/3) * (22/7) * (2.1)^3 = (4/3) * (22/7) * 9.261 = (4/3) * 22 * 1.323 = 4 * 22 * 0.441 = 88 * 0.441 = 38.808 cm^3. The mass of the sphere is Mass = Volume * Density = 38.808 cm^3 * 8.9 g/cm^3 = 345.3912 grams. Thus, the mass of the sphere is approximately 345.39 grams.

Solution 3: Inner diameter is 10.5 cm, so inner radius is r = 10.5 / 2 = 5.25 cm = 21 / 4 cm. Since tin-plating is performed on the inside surface of the open bowl, we calculate the inner curved surface area: CSA = 2 * pi * r^2 = 2 * (22/7) * (21/4)^2 = 2 * (22/7) * (441 / 16) = (44/7) * (441 / 16) = (11 * 63) / 4 = 693 / 4 = 173.25 cm^2. The cost rate is 16 dollars per 100 cm^2, meaning 16 / 100 = 0.16 dollars per cm^2. The total cost is: Total Cost = 173.25 * (16 / 100) = 2772 / 100 = 27.72 dollars.