6.4 Surface Area & Volume

Surface Area & Volume

Surface area and volume measure the spatial boundaries and 3D space occupancy of solid figures. Candidates taking Praxis 5003 must master multi-step 3D measurement calculations and understand the geometric principles governing unit cube packing and scale factor effects.

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1. Surface Area: Concepts & Net Summation

Surface Area ($SA$) is the total area of all outer 2D faces bounding a 3D solid figure. Surface area is measured in square units ($\text{in}^2, \text{cm}^2, \text{m}^2$).

Total Surface Area vs. Lateral Surface Area

• Total Surface Area: The sum of the areas of ALL faces (including top and bottom bases).
• Lateral Surface Area ($LSA$): The sum of the areas of the side faces only, excluding the area of the top and bottom bases.

Surface Area Formulas Summary Table

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2. Volume Concepts & Unit Cubes

Volume ($V$) is the measure of the amount of three-dimensional space enclosed by a solid figure. Volume is measured in cubic units ($\text{in}^3, \text{cm}^3, \text{m}^3$).

Unit Cube Packing

In elementary education, volume is introduced concretely by counting unit cubes—cubes measuring 1 unit on each edge with a volume of $1\text{ unit}^3$.

• A rectangular box filled with 4 layers of unit cubes, where each layer contains 3 rows of 5 cubes, holds $5 \times 3 \times 4 = 60$ unit cubes, giving a volume of $60\text{ units}^3$.

           UNIT CUBE                        RECTANGULAR PRISM PACKING
            ┌─────┐                              ┌─────┬─────┬─────┐
           ╱     ╱│ 1 unit                      ╱     ╱     ╱     ╱│
          ├─────┤ │                            ├─────┼─────┼─────┤ │ h
   1 unit │     │╱ 1 unit                      │     │     │     │╱
          └─────┘                              └─────┴─────┴─────┘
          1 unit                                       w
                                                  l
                                            Volume = l × w × h

Volume Formulas Summary Table

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3. Step-by-Step Worked Volume & Surface Area Problems

Example 1: Cylinder Surface Area & Volume

Problem: A closed cylindrical container has a base radius of $5\text{ cm}$ and a height of $8\text{ cm}$. Find its exact total surface area and volume in terms of $\pi$.

Step-by-Step Solution:

1. Total Surface Area ($SA$): $$SA = 2\pi r^2 + 2\pi rh = 2\pi (5)^2 + 2\pi (5)(8)$$ $$SA = 2\pi (25) + 2\pi (40) = 50\pi + 80\pi = 130\pi\text{ cm}^2$$
2. Volume ($V$): $$V = \pi r^2 h = \pi (5)^2 (8) = \pi (25)(8) = 200\pi\text{ cm}^3$$

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Example 2: Volume of a Triangular Prism

Problem: A triangular prism has a right-triangular base with legs of $6\text{ ft}$ and $8\text{ ft}$. The height (length) of the prism is $10\text{ ft}$. Find the volume.

Step-by-Step Solution:

1. Calculate the area of the triangular base ($B$): $$B = \frac{1}{2} \times b \times h_{\text{tri}} = \frac{1}{2} \times 6 \times 8 = 24\text{ sq ft}$$
2. Multiply by the height of the prism ($H = 10\text{ ft}$): $$V = B \times H = 24 \times 10 = 240\text{ cu ft}$$

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4. Dimensional Scaling Effects (Scale Factor $k$)

One of the most frequently tested concepts on Praxis 5003 is how changing the linear dimensions of a 3D figure affects its surface area and volume.

The $1D \rightarrow 2D \rightarrow 3D$ Scaling Principle

If all linear dimensions of a 3D figure are multiplied by a constant scale factor $k$:

1. Linear Dimensions (edges, perimeter, radius, height) scale by factor $k^1 = k$.
2. Surface Area & Base Area scale by factor $k^2$.
3. Volume scales by factor $k^3$.

SCALE FACTOR EFFECTS WHEN DOUBLING ALL DIMENSIONS (k = 2)

Dimension Type       Scale Factor Power     Multiplier for k = 2      Example Result
────────────────────────────────────────────────────────────────────────────────────────
Linear (Length)      k¹ = 2¹               × 2                      10 cm → 20 cm
Area (Surface Area)  k² = 2²               × 4                      50 cm² → 200 cm²
Volume               k³ = 2³               × 8                      60 cm³ → 480 cm³

Scaling Effects Table Summary

Scaling Proof Example

Consider a rectangular prism with dimensions $l = 2, w = 3, h = 4$:

• Original Volume: $V_1 = 2 \times 3 \times 4 = 24\text{ units}^3$.
• Original Surface Area: $SA_1 = 2(2\cdot3 + 2\cdot4 + 3\cdot4) = 2(6 + 8 + 12) = 52\text{ units}^2$.

Double all dimensions ($k = 2 \implies l'=4, w'=6, h'=8$):

• New Volume: $V_2 = 4 \times 6 \times 8 = 192\text{ units}^3$. Note that $192 = 24 \times \mathbf{8}$ ($8 = 2^3$).
• New Surface Area: $SA_2 = 2(4\cdot6 + 4\cdot8 + 6\cdot8) = 2(24 + 32 + 48) = 208\text{ units}^2$. Note that $208 = 52 \times \mathbf{4}$ ($4 = 2^2$).

Fractional Edge Lengths and Surface Area from Nets

Right rectangular prisms with fractional edges

ETS explicitly includes volume and surface area of right rectangular prisms whose edge lengths are fractions. Use the same formulas with fractional factors:

$$V = \ell \times w \times h.$$

Example: a prism that is $\frac{3}{2}$ ft by $\frac{2}{3}$ ft by $\frac{5}{4}$ ft has

$$V = \frac{3}{2} \times \frac{2}{3} \times \frac{5}{4} = \frac{5}{4}\ \text{cubic feet}.$$

Surface area still sums the areas of all six faces (or three unique face areas doubled): $SA = 2(\ell w + \ell h + wh)$ with the same fractional lengths.

Computing surface area from a net

A net unfolds the prism into rectangles (and possibly squares). To find surface area from a net, compute the area of each rectangular region in the net and add them—no missing faces, no double-counting overlapping flaps that are not part of the solid’s surface. This matches packing the faces that enclose the solid.