6.4 Surface Area & Volume
Key Takeaways
- Surface area is the sum of areas of all 2D outer faces bounding a 3D solid, measured in square units.
- Volume is the 3D space enclosed by a solid (V = B * h for prisms and cylinders), introduced conceptually via unit cube packing.
- Cylinder total surface area is SA = 2πr² + 2πrh and volume is V = πr²h.
- When linear dimensions of a solid scale by factor k, surface area increases by k² and volume increases by k³.
- Doubling linear dimensions (k = 2) multiplies surface area by 4 and volume by 8.
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.
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 ($ ext{in}^2, ext{cm}^2, ext{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$): The sum of the areas of the side faces only, excluding the area of the top and bottom bases.
Surface Area Formulas Summary Table
| 3D Solid | Surface Area Formula ($SA$) | Formula Breakdown |
|---|---|---|
| Cube | $SA = 6s^2$ | Sum of 6 identical square faces with side length $s$. |
| Rectangular Prism | $SA = 2lw + 2lh + 2wh$ | 2 front/back faces ($lh$) + 2 top/bottom faces ($lw$) + 2 side faces ($wh$). |
| Triangular Prism | $SA = 2(rac{1}{2}bh) + (s_1 + s_2 + s_3)H$ | 2 triangular bases + 3 rectangular lateral faces ($H$ is prism height). |
| Right Cylinder | $SA = 2\pi r^2 + 2\pi rh$ | 2 circular bases ($2 imes \pi r^2$) + 1 curved lateral rectangle ($2\pi r imes h$). |
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 ($ ext{in}^3, ext{cm}^3, ext{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 $1 ext{ unit}^3$.
- A rectangular box filled with 4 layers of unit cubes, where each layer contains 3 rows of 5 cubes, holds $5 imes 3 imes 4 = 60$ unit cubes, giving a volume of $60 ext{ units}^3$.
UNIT CUBE RECTANGULAR PRISM PACKING
┌─────┐ ┌─────┬─────┬─────┐
╱ ╱│ 1 unit ╱ ╱ ╱ ╱│
├─────┤ │ ├─────┼─────┼─────┤ │ h
1 unit │ │╱ 1 unit │ │ │ │╱
└─────┘ └─────┴─────┴─────┘
1 unit w
l
Volume = l × w × h
Volume Formulas Summary Table
| 3D Solid | General Base Volume Formula | Specific Volume Formula | Key Variable Definitions |
|---|---|---|---|
| Cube | $V = B imes h$ | $V = s^3$ | $s$ = edge length. |
| Rectangular Prism | $V = B imes h$ | $V = l imes w imes h$ | $l$ = length, $w$ = width, $h$ = height; $B = lw$ is base area. |
| Triangular Prism | $V = B imes h$ | $V = (rac{1}{2} b h_{ ext{tri}}) imes H_{ ext{prism}}$ | $b$ = triangle base, $h_{ ext{tri}}$ = triangle height, $H$ = prism length. |
| Right Cylinder | $V = B imes h$ | $V = \pi r^2 h$ | $r$ = circular radius, $h$ = cylinder height; $B = \pi r^2$ is base area. |
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 $5 ext{ cm}$ and a height of $8 ext{ cm}$. Find its exact total surface area and volume in terms of $\pi$.
Step-by-Step Solution:
- Total Surface Area ($SA$):
- Volume ($V$):
Example 2: Volume of a Triangular Prism
Problem: A triangular prism has a right-triangular base with legs of $6 ext{ ft}$ and $8 ext{ ft}$. The height (length) of the prism is $10 ext{ ft}$. Find the volume.
Step-by-Step Solution:
- Calculate the area of the triangular base ($B$): B = rac{1}{2} imes b imes h_{ ext{tri}} = rac{1}{2} imes 6 imes 8 = 24 ext{ sq ft}
- Multiply by the height of the prism ($H = 10 ext{ ft}$):
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 $1D
ightarrow 2D ightarrow 3D$ Scaling Principle If all linear dimensions of a 3D figure are multiplied by a constant scale factor $k$:
- Linear Dimensions (edges, perimeter, radius, height) scale by factor $k^1 = k$.
- Surface Area & Base Area scale by factor $k^2$.
- 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
| Scale Factor ($k$) | Linear Change | Surface Area Multiplier ($k^2$) | Volume Multiplier ($k^3$) |
|---|---|---|---|
| $k = 2$ (Doubled) | $2 imes$ | $2^2 = \mathbf{4 imes}$ | $2^3 = \mathbf{8 imes}$ |
| $k = 3$ (Tripled) | $3 imes$ | $3^2 = \mathbf{9 imes}$ | $3^3 = \mathbf{27 imes}$ |
| $k = 4$ (Quadrupled) | $4 imes$ | $4^2 = \mathbf{16 imes}$ | $4^3 = \mathbf{64 imes}$ |
| $k = rac{1}{2}$ (Halved) | $rac{1}{2} imes$ | $(rac{1}{2})^2 = \mathbf{rac{1}{4} imes}$ | $(rac{1}{2})^3 = \mathbf{rac{1}{8} imes}$ |
Scaling Proof Example
Consider a rectangular prism with dimensions $l = 2, w = 3, h = 4$:
- Original Volume: $V_1 = 2 imes 3 imes 4 = 24 ext{ units}^3$.
- Original Surface Area: $SA_1 = 2(2\cdot3 + 2\cdot4 + 3\cdot4) = 2(6 + 8 + 12) = 52 ext{ units}^2$.
Double all dimensions ($k = 2 \implies l'=4, w'=6, h'=8$):
- New Volume: $V_2 = 4 imes 6 imes 8 = 192 ext{ units}^3$. Note that $192 = 24 imes \mathbf{8}$ ($8 = 2^3$).
- New Surface Area: $SA_2 = 2(4\cdot6 + 4\cdot8 + 6\cdot8) = 2(24 + 32 + 48) = 208 ext{ units}^2$. Note that $208 = 52 imes \mathbf{4}$ ($4 = 2^2$).
[!IMPORTANT] Praxis Exam Shortcut: If a problem states "all dimensions of a solid are multiplied by 3," you do NOT need to calculate the actual new lengths! Simply multiply original area by $3^2 = 9$, or original volume by $3^3 = 27$.
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: Example: a prism that is $\frac{3}{2}$ ft by $\frac{2}{3}$ ft by $\frac{5}{4}$ ft has 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.
A rectangular storage box has a volume of 60 cubic inches. If a manufacturer doubles all three linear dimensions (length, width, height) of the box, what is the volume of the new larger box?
What is the total surface area of a closed right cylinder with a radius of 5 centimeters and a height of 8 centimeters?
A triangular prism has a right-triangular base with legs measuring 6 feet and 8 feet. The height (length) of the prism is 10 feet. What is the total volume of the triangular prism?