Solid Geometry Formula Bank

1. Cylinder

Formulas

1. Volume:
\[
V = \pi r^2 h
\]
2. Surface Area:
\[
A = 2\pi r h + 2\pi r^2
\]

Worked Example

A cylinder has a radius of \(r = 3 \, \text{units}\) and height \(h = 7 \, \text{units}\). Find its volume and surface area.

Solution:
1. Volume:
\[
V = \pi r^2 h = \pi (3)^2 (7) = 63\pi \approx 197.92 \, \text{units}^3
\]

2. Surface Area:

\[
A = 2\pi r h + 2\pi r^2 = 2\pi (3)(7) + 2\pi (3)^2 = 42\pi + 18\pi = 60\pi \approx 188.40 \, \text{units}^2
\]

2. Cone

Formulas

1. Volume

\[
V = \frac{1}{3} \pi r^2 h
\]
2. Surface Area:

\[
A = \pi r \sqrt{r^2 + h^2} + \pi r^2
\]

Worked Example:

A cone has a radius \(r = 4 \, \text{units}\) and height \(h = 9 \, \text{units}\). Find its volume and surface area.

Solution:
1. Volume:
\[
V = \frac{1}{3} \pi r^2 h = \frac{1}{3} \pi (4)^2 (9) = \frac{144\pi}{3} = 48\pi \approx 150.80 \, \text{units}^3
\]

2. Surface Area:
\[
A = \pi r \sqrt{r^2 + h^2} + \pi r^2 = \pi (4) \sqrt{4^2 + 9^2} + \pi (4)^2
\]
\[
= \pi (4) \sqrt{16 + 81} + 16\pi = \pi (4)(9.49) + 16\pi = 37.96\pi + 16\pi = 53.96\pi \approx 169.57 \, \text{units}^2
\]

3. Pyramid

Formulas:
1. Volume:
\[
V = \frac{1}{3} \text{Base Area} \times h
\]
2. Surface Area:
\[
A = \text{Sum of lateral areas} + \text{Base Area}
\]

Worked Example:
A pyramid has a square base with side length \(a = 6 \, \text{units}\) and height \(h = 10 \, \text{units}\). Find its volume.

Solution:
1. Base Area:
\[
\text{Base Area} = a^2 = 6^2 = 36 \, \text{units}^2
\]

2. Volume:
\[
V = \frac{1}{3} \text{Base Area} \times h = \frac{1}{3} (36) (10) = 120 \, \text{units}^3
\]

3. Surface Area (with slant height \(l = 12\)):
\[
A = \text{Base Area} + 4 \times \frac{1}{2} a l = 36 + 4 \times \frac{1}{2} (6)(12)
\]
\[
A = 36 + 144 = 180 \, \text{units}^2
\]

4. Prism

Formulas:
1. Volume:
\[
V = \text{Base Area} \times h
\]
2. Surface Area:
\[
A = 2 \times \text{Base Area} + \text{Lateral Area}
\]

Worked Example:
A triangular prism has a base with sides \(a = 3\), \(b = 4\), and \(c = 5\) and height \(h = 8\).

Solution:
1. Base Area (using Heron’s formula):
\[
s = \frac{a + b + c}{2} = \frac{3 + 4 + 5}{2} = 6
\]
\[
\text{Base Area} = \sqrt{s(s – a)(s – b)(s – c)} = \sqrt{6(6 – 3)(6 – 4)(6 – 5)} = \sqrt{6 \times 3 \times 2 \times 1} = \sqrt{36} = 6
\]

2. Volume:
\[
V = \text{Base Area} \times h = 6 \times 8 = 48 \, \text{units}^3
\]

3. Surface Area (adding rectangular lateral faces):
\[
A = 2 \times 6 + 8(3 + 4 + 5) = 12 + 96 = 108 \, \text{units}^2
\]

5. Frustum of a Cone

Formulas:
1. Volume:
\[
V = \frac{1}{3} \pi h \left(r_1^2 + r_1r_2 + r_2^2\right)
\]
2. Lateral Surface Area:
\[
A = \pi (r_1 + r_2) \sqrt{(r_1 – r_2)^2 + h^2}
\]

Worked Example:
A frustum has top radius \(r_1 = 3\), bottom radius \(r_2 = 5\), and height \(h = 7\).

Solution:
1. Volume:
\[
V = \frac{1}{3} \pi (7) \left(3^2 + 3 \times 5 + 5^2\right)
\]
\[
V = \frac{1}{3} \pi (7) \left(9 + 15 + 25\right) = \frac{1}{3} \pi (7)(49) = \frac{343\pi}{3} \approx 359.19 \, \text{units}^3
\]

2. Lateral Surface Area:
\[
A = \pi (3 + 5) \sqrt{(5 – 3)^2 + 7^2} = \pi (8) \sqrt{4 + 49} = \pi (8) \sqrt{53}
\]
\[
A = 8 \pi \times 7.28 \approx 182.35 \, \text{units}^2
\]

6. Hyperboloid

One-Sheet Formula:
\[
\frac{x^2}{a^2} + \frac{y^2}{b^2} – \frac{z^2}{c^2} = 1
\]

Two-Sheet Formula:
\[
\frac{x^2}{a^2} – \frac{y^2}{b^2} – \frac{z^2}{c^2} = 1
\]

7. Surface Area of a Torus

For a torus with inner radius \(a\) and outer radius \(b\):

\[
\text{Surface Area} = 2\pi^2 (b^2 – a^2)
\]

Worked Example

Calculate the surface area of a torus with inner radius \(a = 3\) units and outer radius \(b = 6\) units.

\[
\text{Surface Area} = 2\pi^2 ((6)^2 – (3)^2)
\]
\[
\text{Surface Area} = 2\pi^2 (36 – 9) = 2\pi^2 (27)
\]
\[
\text{Surface Area} \approx 2 \times 9.87 \times 27 \approx 532.92 \, \text{units}^2
\]

8. Volume of an Ellipsoid

For an ellipsoid with semi-axes \(a\), \(b\), and \(c\):

\[
\text{Volume} = \frac{4}{3} \pi a b c
\]

Worked Example for 7.47
Find the volume of an ellipsoid with semi-axes \(a = 4\), \(b = 3\), and \(c = 5\).

\[
\text{Volume} = \frac{4}{3} \pi (4)(3)(5)
\]
\[
\text{Volume} = \frac{4}{3} \pi (60) = \frac{240\pi}{3}
\]
\[
\text{Volume} \approx \frac{240 \times 3.14}{3} \approx 251.20 \, \text{units}^3