12.1 Volume and Surface Area of 3D Solids

12.1 Volume and Surface Area of 3D Solids

Three-dimensional geometry on the TABE 13&14 Mathematics assessment (spanning Levels M, D, and A) evaluates your ability to calculate spatial measurements in real-world trade, technical, and everyday applications. Test questions challenge adult learners to distinguish between volume (interior capacity in cubic units) and surface area (total exterior coverage in square units), interpret geometric nets, and apply geometric formulas to prisms, cylinders, pyramids, cones, and spheres.

---

3D Spatial Anatomy & Terminology

Before executing geometric calculations, you must understand the anatomical components that define three-dimensional solids:

Anatomical Structure of Polyhedra & Curved Solids:
├── Face: Any flat polygon surface that forms a boundary of a solid
├── Edge: A line segment where two distinct faces intersect
├── Vertex (plural: Vertices): A corner point where three or more edges intersect
├── Base(s): The foundational polygon(s) or circle(s) used to classify and orient the solid
├── Lateral Faces / Surface: The sides of a solid connecting the bases (excluding the bases)
├── Altitude / Height (h): The perpendicular distance between bases or from apex to base
└── Slant Height (l): The diagonal distance along the outer surface from apex to base edge

Polyhedra vs. Curved Solids

• Polyhedra (Prisms & Pyramids): 3D solids bounded strictly by flat polygon faces. For all convex polyhedra, Euler's Formula confirms the relationship: $F + V - E = 2$ (where $F = \text{faces}$, $V = \text{vertices}$, $E = \text{edges}$).
• Curved Solids (Cylinders, Cones, & Spheres): 3D solids featuring smooth, curved circular surfaces that cannot be classified as standard polyhedra.

---

Volume: Formulas & Principles of 3D Capacity

Volume ($V$) measures the amount of three-dimensional space enclosed inside a solid boundary, expressed in cubic units ($\text{in}^3, \text{ft}^3, \text{yd}^3, \text{cm}^3, \text{m}^3$).

1. Uniform Cross-Section Solids (Prisms & Cylinders)

For any solid whose cross-section remains constant from top to bottom, volume is simply the product of the Base Area ($B$) and the perpendicular Height ($h$):

$$\mathbf{V = B \cdot h}$$

2. Pointed Solids (Pyramids & Cones)

Solids that taper to a single point (apex) occupy exactly one-third of the capacity of a prism or cylinder with an identical base and height:

$$\mathbf{V = \frac{1}{3} B \cdot h}$$

• Right Square/Rectangular Pyramid: Base area $B = l \cdot w$, so $\mathbf{V = \frac{1}{3} lwh}$ (or $\frac{1}{3} s^2 h$ for a square base).
• Right Circular Cone: Circular base $B = \pi r^2$, so $\mathbf{V = \frac{1}{3} \pi r^2 h}$.

3. Spherical Solids

A sphere possesses neither edges nor bases. Its volume depends entirely on the distance from its center to its exterior shell, the radius ($r$):

$$\mathbf{V = \frac{4}{3} \pi r^3}$$

---

Surface Area: Formulas & Geometric Nets

Total Surface Area ($SA$) measures the total two-dimensional area covering the outside of a solid, expressed in square units ($\text{in}^2, \text{ft}^2, \text{cm}^2$).

Geometric Nets: Unfolding 3D into 2D

A geometric net is a two-dimensional flat pattern that folds into a three-dimensional solid without overlapping. Analyzing nets helps you visualize why surface area formulas work:

Unfolded Net of a Right Circular Cylinder:

       +-------------+        <-- Top Circular Base (Area = πr²)
       | (Top Base)  |
       +-------------+
              |
  +-----------------------+   <-- Lateral Rectangle
  |                       |       • Length = Base Circumference (2πr)
  |   Lateral Rectangle   | h     • Height = Cylinder Height (h)
  |                       |       • Area = 2πrh
  +-----------------------+
              |
       +-------------+        <-- Bottom Circular Base (Area = πr²)
       | (Bottom)    |
       +-------------+

Total Surface Area = Top Circle + Bottom Circle + Lateral Rectangle = 2πr² + 2πrh

Comprehensive Surface Area Reference Table

---

Step-by-Step Worked Applications

Worked Example 1: HVAC Air Duct & Material Estimation

Problem: An HVAC contractor fabricates a closed rectangular sheet-metal ventilation trunk measuring $6\text{ ft}$ long, $3\text{ ft}$ wide, and $2\text{ ft}$ tall.

1. What is the total volume of air the trunk holds?
2. How many square feet of sheet metal are needed to construct the complete closed box?

Solution:

• Step 1 (Volume): $$V = l \cdot w \cdot h = 6\text{ ft} \times 3\text{ ft} \times 2\text{ ft} = \mathbf{36\text{ cu ft}}$$
• Step 2 (Surface Area): $$SA = 2(lw + lh + wh) = 2((6 \times 3) + (6 \times 2) + (3 \times 2)) = 2(18 + 12 + 6) = 2(36) = \mathbf{72\text{ sq ft}}$$

Worked Example 2: Concrete Footing Foundation (Cubic Yard Conversion)

Problem: A builder pours 4 identical solid cylindrical concrete support pillars for an elevated deck. Each pillar has a diameter of $2\text{ ft}$ and a height of $9\text{ ft}$. Concrete is ordered by the cubic yard ($1\text{ yd}^3 = 27\text{ ft}^3$). Using $\pi \approx 3.14$, how many cubic yards of concrete must be delivered (round to the nearest tenth)?

Solution:

• Step 1 (Find radius): $r = \frac{d}{2} = \frac{2}{2} = 1\text{ ft}$.
• Step 2 (Volume of one pillar): $$V_{\text{one}} = \pi r^2 h = 3.14 \times (1)^2 \times 9 = 3.14 \times 9 = 28.26\text{ cu ft}$$
• Step 3 (Total volume for 4 pillars): $$V_{\text{total}} = 4 \times 28.26\text{ cu ft} = 113.04\text{ cu ft}$$
• Step 4 (Convert to cubic yards): $$\text{Cubic Yards} = \frac{113.04\text{ cu ft}}{27\text{ cu ft/yd}^3} \approx \mathbf{4.2\text{ cu yd}}$$

Worked Example 3: Composite Solid (Cylinder Silo with Conical Roof)

Problem: An agricultural grain silo consists of a cylindrical main body of radius $10\text{ ft}$ and height $30\text{ ft}$, capped with a conical roof of vertical height $12\text{ ft}$. What is the total storage volume of the silo in terms of $\pi$?

Solution:

• Step 1 (Cylinder body volume): $$V_{\text{cylinder}} = \pi r^2 h = \pi (10)^2 (30) = 100 \times 30 \times \pi = 3,000\pi\text{ cu ft}$$
• Step 2 (Cone roof volume): $$V_{\text{cone}} = \frac{1}{3}\pi r^2 h = \frac{1}{3}\pi (10)^2 (12) = \frac{1}{3} \times 100 \times 12 \times \pi = 400\pi\text{ cu ft}$$
• Step 3 (Sum components): $$V_{\text{total}} = 3,000\pi + 400\pi = \mathbf{3,400\pi\text{ cu ft}} \approx 10,681.4\text{ cu ft}$$