9.2 Volume of 3D Geometric Solids

Volume of 3D Geometric Solids

Quick Summary: Volume ($V$) quantifies the amount of three-dimensional space enclosed inside a solid boundary or the liquid capacity a container can hold. While surface area measures the 2D outer shell in square units ($in^2$), volume measures 3D interior capacity in cubic units ($in^3, cm^3, ft^3, m^3$). On the HiSET exam, you will compute volumes of prisms, cylinders, pyramids, cones, and spheres using official formulas, apply Cavalieri's principle to oblique solids, calculate composite/hollow shapes, and solve dimensional scaling problems.

Volume problems frequently appear in real-world contexts such as filling swimming pools, pouring concrete slabs, packaging shipping containers, and calculating fuel tank storage.

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The Base-Extension Principle: Prisms and Cylinders

For any three-dimensional solid with a uniform cross-section that extends straight upward along a perpendicular height $h$, the volume is computed by taking the two-dimensional area of the base ($B$) and multiplying it by the height ($h$): $\mathbf{V = B \cdot h}$

   1. Rectangular Prism                    2. Right Circular Cylinder
      ┌─────────────────────────┐               ╭───────────────╮
     /                         /│              │  Top Circle   │
    ┌─────────────────────────┐ │              │   (Area = B)  │
    │                         │ │              ╰───────┬───────╯
    │       Base Area (B)     │ │                      │
    │        (B = l × w)      │ │ Height (h)           │ Height (h)
    │                         │ │                      │
    │                         │/               ╭───────┴───────╮
    └─────────────────────────┘                │  Base Circle  │ (B = πr²)
     Length (l) × Width (w)                    ╰───────────────╯
    Volume = (l × w) × h                       Volume = (πr²) × h

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Official HiSET Volume Formulas Reference

Volume is the one area where the formula sheet genuinely helps, but it prints only three general entries — Prism/Cylinder $= (\text{area of the base})(\text{height})$, Pyramid/Cone $= \frac{1}{3}(\text{area of the base})(\text{height})$, and Sphere $= \frac{4}{3}\pi r^3$. You must supply the base area yourself. The table below expands those three entries into the specific forms you will use:

Cavalieri's Principle & Tapered Solids

1. Cavalieri's Principle: Right vs. Oblique Solids

Cavalieri's Principle states that if two three-dimensional solids have equal heights and identical cross-sectional areas at every level parallel to their bases, then the two solids have identical volumes.

Intuitive Analogy: Imagine a stack of 50 playing cards neatly aligned in a vertical stack (a right prism). If you push the side of the deck so the stack leans over diagonally (an oblique prism), the shape changes, but the total amount of card material and volume remains exactly unchanged.

$\text{Volume of Oblique Prism/Cylinder} = B \cdot h_{\text{perpendicular}}$

Exam Rule: When calculating the volume of a slanted (oblique) cylinder or prism, always use the perpendicular altitude $h$ (the vertical distance between base planes), never the slanted edge length.

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2. Tapered Solids: The One-Third Principle (Cones & Pyramids)

When a solid tapers smoothly from a 2D base up to a point (apex), it encloses exactly one-third of the volume of a prism or cylinder with the same base and vertical height: $V_{\text{pyramid}} = \frac{1}{3} B h$ $V_{\text{cone}} = \frac{1}{3} \pi r^2 h$

   Prism vs Pyramid Relationship              Cylinder vs Cone Relationship
   ┌───────────────────────┐                  ╭───────────────────────╮
   │                       │                  │                       │
   │      Full Prism       │                  │     Full Cylinder     │
   │       (V = Bh)        │                  │       (V = πr²h)      │
   │      ▲        ▲       │                  │      ▲        ▲       │
   │     / \      / \      │                  │     / \      / \      │
   │    /   \    /   \     │                  │    /   \    /   \     │
   └───────────────────────┘                  ╰───────────────────────╯
    Pyramid Volume = 1/3 (Bh)                  Cone Volume = 1/3 (πr²h)

Slant Height vs. Vertical Height in Volume Calculations

In surface area, slant height $l$ is required. In volume, vertical height $h$ is strictly required: $\text{Given radius } r \text{ and slant height } l \implies h = \sqrt{l^2 - r^2}$

Worked Example 1: Volume of a Conical Funnel

A conical funnel has a base diameter of $12\text{ inches}$ and a vertical height of $10\text{ inches}$. What is the volume of the funnel in cubic inches?

1. Convert diameter to radius: $r = \frac{d}{2} = \frac{12}{2} = 6\text{ in}$
2. Apply cone volume formula: $V = \frac{1}{3}\pi r^2 h$
3. Substitute values: $V = \frac{1}{3}\pi (6)^2 (10) = \frac{1}{3}\pi (36)(10) = \frac{1}{3}\pi (360) = 120\pi\text{ cu in}$
4. Decimal approximation: $V \approx 120 \times 3.14159 \approx 376.99\text{ cu in}$

Spherical Volume & The Dimensional Scaling Law

1. Volume of Spheres and Hemispheres

The volume of a sphere depends exclusively on its radius $r$, raised to the third power: $V_{\text{sphere}} = \frac{4}{3}\pi r^3$

For a hemisphere (half of a sphere): $V_{\text{hemisphere}} = \frac{1}{2} \left(\frac{4}{3}\pi r^3\right) = \frac{2}{3}\pi r^3$

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The Dimensional Scaling Law ($k, k^2, k^3$)

A major HiSET conceptual topic is how expanding or shrinking linear dimensions affects perimeter, area, and volume. When all linear dimensions (length, width, height, radius) of a solid are multiplied by a scale factor $k$:

$\text{Linear Dimensions (Length, Perimeter, Radius)} \propto k^1$ $\text{Surface Area / Base Area} \propto k^2$ $\mathbf{\text{Volume}} \mathbf{\propto k^3}$

Scaling Behavior Comparison Table

Worked Example 2: Volume Scaling in Packaging

A rectangular cardboard shipping box has length $4\text{ ft}$, width $3\text{ ft}$, and height $2\text{ ft}$ (original volume $= 4 \times 3 \times 2 = 24\text{ cu ft}$). If a larger version of the box is manufactured with every linear dimension doubled ($k = 2$), what is the volume of the new box?

1. Method A (Direct Formula Calculation): $l_{\text{new}} = 8\text{ ft}, \quad w_{\text{new}} = 6\text{ ft}, \quad h_{\text{new}} = 4\text{ ft}$ $V_{\text{new}} = 8 \times 6 \times 4 = 192\text{ cu ft}$
2. Method B (Using the Scaling Rule $k^3$): $V_{\text{new}} = k^3 \times V_{\text{original}} = 2^3 \times 24 = 8 \times 24 = 192\text{ cu ft}$

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Composite and Subtractive Volume

Real-world solids are frequently formed by combining or coring out simple geometric shapes.

1. Additive Volume (Combined Solids)

Total volume is simply the direct sum of the individual solid components: $V_{\text{total}} = V_1 + V_2$

Example: A grain storage silo consisting of a cylinder of height $h$ capped by a hemispherical dome of radius $r$: $V_{\text{silo}} = V_{\text{cylinder}} + V_{\text{hemisphere}} = \pi r^2 h + \frac{2}{3}\pi r^3$

2. Subtractive Volume (Hollow / Drilled Solids)

When a hole or hollow core is cut out of a solid, subtract the interior removed volume from the outer total volume: $V_{\text{remaining}} = V_{\text{outer}} - V_{\text{inner}}$

Worked Example 3: Volume of a Hollow Cylindrical Steel Tube

A cylindrical steel pipe is $20\text{ cm}$ long. The outer radius is $5\text{ cm}$ and the inner hollow radius is $3\text{ cm}$. What is the volume of steel used to manufacture the pipe?

1. Calculate total outer cylinder volume: $V_{\text{outer}} = \pi R^2 h = \pi (5)^2 (20) = \pi (25)(20) = 500\pi\text{ cm}^3$
2. Calculate inner hollow cylinder volume: $V_{\text{inner}} = \pi r^2 h = \pi (3)^2 (20) = \pi (9)(20) = 180\pi\text{ cm}^3$
3. Subtract inner volume from outer volume: $V_{\text{steel}} = 500\pi - 180\pi = 320\pi\text{ cm}^3 \approx 1,005.31\text{ cm}^3$