8.3 Three-Dimensional Figures, Surface Area & Volume

8.3 Three-Dimensional Figures, Surface Area & Volume

Quick Answer: Three-dimensional figures are classified as polyhedra (flat polygonal faces meeting at straight edges and vertices) or non-polyhedra (curved surfaces such as cylinders, cones, and spheres). Convex polyhedra satisfy Euler's Formula: $V - E + F = 2$. Surface area represents the total two-dimensional area of a solid's unfolded net. Volume quantifies internal capacity: right prisms and cylinders satisfy $V = Bh$ (base area times perpendicular height); tapered cones and pyramids hold one-third the volume of their corresponding prism/cylinder ($V = \frac{1}{3}Bh$); and spheres satisfy $V = \frac{4}{3}\pi r^3$ with surface area $SA = 4\pi r^2$. Scaling linear dimensions by $k$ scales surface area by $k^2$ and volume by $k^3$.

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Structural Classification: Polyhedra vs. Non-Polyhedra

Three-dimensional space ($\mathbb{R}^3$) contains solids that are categorized based on whether their bounding surfaces are planar polygons or curved manifolds.

Polyhedra and Their Anatomical Features

A polyhedron (plural: polyhedra) is a closed three-dimensional solid whose boundary consists exclusively of flat polygonal regions called faces. Polyhedra possess three defining geometric elements:

1. Face ($F$): A two-dimensional polygonal region bounding the solid.
2. Edge ($E$): A one-dimensional straight line segment where two polygonal faces intersect.
3. Vertex ($V$, plural vertices): A zero-dimensional point where three or more edges intersect.

Two Main Families of Polyhedra

• Prisms: Polyhedra possessing two congruent, parallel polygonal bases. The remaining faces connecting the two bases are called lateral faces, which are always parallelograms (or rectangles in a right prism). Prisms are classified by the shape of their base polygon:
  • Rectangular Prism: Two parallel rectangular bases with four rectangular lateral faces.
  • Triangular Prism: Two parallel triangular bases with three rectangular lateral faces.
• Pyramids: Polyhedra possessing one polygonal base and triangular lateral faces that converge upward at a single shared point called the apex. Pyramids are also named by their base polygon (e.g., square pyramid, triangular pyramid/tetrahedron).

Non-Polyhedra (Curved Solids)

Solids that possess curved boundary surfaces are non-polyhedra:

• Cylinder: Features two parallel, congruent circular bases connected by a smooth curved lateral surface.
• Cone: Features one circular base tapering smoothly to a single apex vertex.
• Sphere: The locus of all points in three-dimensional space equidistant from a fixed center point.

Euler's Formula for Convex Polyhedra

For any convex polyhedron, the relationship between vertices ($V$), edges ($E$), and faces ($F$) is invariant, governed by Euler's Formula:

$$V - E + F = 2$$

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Nets & Total Surface Area of Prisms and Cylinders

A net is a two-dimensional planar unfolding of a three-dimensional solid that can be folded along line segments to reconstruct the original solid without gaps or overlapping faces. Total Surface Area ($SA$) is the sum of the areas of all faces comprising the solid's net.

Right Rectangular Prisms

The net of a right rectangular prism consists of six rectangular faces organized into three congruent opposite pairs:

• Top and Bottom: $2 \times (lw)$
• Front and Back: $2 \times (lh)$
• Left and Right Sides: $2 \times (wh)$

$$SA = 2lw + 2lh + 2wh = 2(lw + lh + wh)$$

Lateral Surface Area ($LSA$): The area of the four vertical sides excluding the top and bottom bases:

$$LSA = 2lh + 2wh = (2l + 2w)h = P_{\text{base}} \cdot h$$

Right Triangular Prisms

The net consists of two congruent triangular bases and three rectangular lateral faces:

$$SA = 2 \cdot A_{\text{base}} + \text{Lateral Area} = 2\left(\frac{1}{2} b_{\text{tri}} h_{\text{tri}}\right) + (s_1 + s_2 + s_3)H_{\text{prism}}$$

Notice that the two triangular base areas simplify to $b_{\text{tri}} h_{\text{tri}}$. The lateral area is the perimeter of the triangular base multiplied by the overall prism height ($H_{\text{prism}}$).

Right Circular Cylinders

Unfolding a cylinder's net reveals two congruent circles (the bases) and one large rectangular lateral sheet. The height of this rectangle is the cylinder height ($h$), while the width of the rectangle unrolls to match the circumference of the circular base ($2\pi r$):

$$\text{Lateral Area} = 2\pi r h$$ $$SA = 2 \cdot A_{\text{base}} + \text{Lateral Area} = 2\pi r^2 + 2\pi r h$$

Worked Example: Calculate the total surface area of a closed cylindrical container with radius $r = 7\text{ cm}$ and height $h = 10\text{ cm}$ using $\pi \approx \frac{22}{7}$:

• Base Area ($2$ circles): $2 \times \left(\frac{22}{7} \times 7^2\right) = 2 \times (22 \times 7) = 308\text{ cm}^2$.
• Lateral Area: $2 \times \frac{22}{7} \times 7 \times 10 = 440\text{ cm}^2$.
• Total Surface Area: $SA = 308 + 440 = 748\text{ cm}^2$.

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Volume of Right Prisms & Cylinders: The Uniform Base Principle ($V = Bh$)

Volume ($V$) quantifies the three-dimensional capacity enclosed within a solid, measured in cubic units (e.g., $\text{cm}^3, \text{m}^3, \text{in}^3, \text{ft}^3$).

Under Cavalieri's Principle, any right solid whose parallel cross-sections are congruent to its base polygon from bottom to top has a volume equal to the area of its base ($B$) multiplied by its perpendicular height ($h$):

$$V = B \cdot h$$

Right Rectangular Prisms (Including Fractional Edges)

Because the base is a rectangle ($B = l \cdot w$):

$$V = l \cdot w \cdot h$$

Under Florida B.E.S.T. benchmark MA.6.GR.2.3, students must compute volumes of rectangular prisms with fractional edge lengths using two equivalent conceptual approaches:

1. Direct Fractional Multiplication: For a prism with $l = \frac{5}{2}\text{ in}$, $w = \frac{4}{3}\text{ in}$, and $h = \frac{3}{5}\text{ in}$: $$V = \frac{5}{2} \times \frac{4}{3} \times \frac{3}{5} = \frac{60}{30} = 2\text{ in}^3$$
2. Unit Cube Packing Model: Packing the solid with fractional unit cubes of side length $\frac{1}{6}\text{ in}$. The volume of each miniature cube is $(\frac{1}{6})^3 = \frac{1}{216}\text{ in}^3$. Converting each dimension to sixths: $l = \frac{15}{6}$, $w = \frac{8}{6}$, $h = \frac{6}{6}$. Total cubes = $15 \times 8 \times 6 = 720$ cubes. Multiplying count by individual cube volume: $720 \times \frac{1}{216} = \frac{720}{216} = \frac{10}{3} = 3\frac{1}{3}\text{ in}^3$.

Right Triangular Prisms

The base is a triangle ($B = \frac{1}{2} b_{\text{tri}} h_{\text{tri}}$):

$$V = B \cdot H_{\text{prism}} = \left(\frac{1}{2} b_{\text{tri}} h_{\text{tri}}\right) \cdot H_{\text{prism}}$$

Right Circular Cylinders

The base is a circle ($B = \pi r^2$):

$$V = B \cdot h = \pi r^2 h$$

Worked Example: A cylinder has radius $r = 4\text{ in}$ and height $h = 9\text{ in}$.

• Exact Volume: $V = \pi(4^2)(9) = 144\pi\text{ in}^3$.
• Approximate Volume ($\pi \approx 3.14$): $V \approx 144 \times 3.14 = 452.16\text{ in}^3$.

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Tapered Solids and Spheres: Pyramids, Cones, and Spheres

The One-Third Principle for Tapered Solids

Any solid that tapers uniformly from a base of area $B$ to a single apex point at height $h$ holds exactly one-third the volume of a prism or cylinder possessing the identical base and height:

$$V_{\text{tapered}} = \frac{1}{3} B h$$

1. Pyramids:
   • Square Pyramid (base side $s$): $V = \frac{1}{3} s^2 h$
   • Rectangular Pyramid: $V = \frac{1}{3} l w h$
2. Cones: Because the base is a circle ($B = \pi r^2$): $$V_{\text{cone}} = \frac{1}{3} \pi r^2 h$$
   • Physical Relationship: If a cone and a cylinder have the same radius and height, filling the cone with water and pouring it into the cylinder will fill exactly $\frac{1}{3}$ of the cylinder's volume. It requires exactly $3$ full cones to fill $1$ cylinder.

Spheres and Hemispheres

A sphere is completely defined by its radius $r$:

• Volume of a Sphere: $$V_{\text{sphere}} = \frac{4}{3} \pi r^3$$
• Surface Area of a Sphere: $$SA_{\text{sphere}} = 4\pi r^2$$
• Hemisphere (Half-Sphere):
  • Volume: $V_{\text{hemi}} = \frac{1}{2} \left(\frac{4}{3}\pi r^3\right) = \frac{2}{3} \pi r^3$
  • Curved Dome Area: $2\pi r^2$
  • Total Surface Area of Solid Hemisphere: Curved dome plus the flat circular base: $2\pi r^2 + \pi r^2 = 3\pi r^2$.

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Dimensional Scaling in Three Dimensions: $1\text{D} \to 2\text{D} \to 3\text{D}$

When all linear dimensions of a three-dimensional figure are dilated by a uniform positive scale factor $k$:

1. Linear Dimensions (radius, height, perimeter, edge length) scale by $k^1$: $$L_{\text{new}} = k \cdot L_{\text{original}}$$
2. Surface Areas (lateral area, base area, total surface area) scale by $k^2$: $$SA_{\text{new}} = k^2 \cdot SA_{\text{original}}$$
3. Volumes (internal capacity, displacement) scale by $k^3$: $$V_{\text{new}} = k^3 \cdot V_{\text{original}}$$

Worked Problem: A manufacturer produces a standard spherical ball with radius $3\text{ cm}$. They design an oversized display model with radius $6\text{ cm}$ (scale factor $k = 2$).

• Surface area increases by $k^2 = 2^2 = 4\times$: $$SA_{\text{orig}} = 4\pi(3^2) = 36\pi \implies SA_{\text{new}} = 4\pi(6^2) = 144\pi = 4 \times 36\pi$$
• Volume increases by $k^3 = 2^3 = 8\times$: $$V_{\text{orig}} = \frac{4}{3}\pi(3^3) = 36\pi \implies V_{\text{new}} = \frac{4}{3}\pi(6^3) = 288\pi = 8 \times 36\pi$$

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Common Exam Traps & Misconceptions