Chapter 12: Surface Area and Volume
Learning Objectives
- Compute the surface area of prisms, cylinders, pyramids, and cones using the correct formulas
- Distinguish between total surface area and lateral surface area
- Calculate volumes of prisms, cylinders, pyramids, and cones
- Apply Cavalieri's Principle to relate volumes of solids with equal cross-sections
- Find the surface area and volume of spheres and hemispheres
- Decompose composite solids and combine surface area and volume calculations
12.1 Surface Area of Prisms and Cylinders
A prism is a polyhedron with two congruent parallel bases connected by rectangular lateral faces. The surface area (SA) is the total area of all faces.
Prism Surface Area
For any prism with base area $B$, base perimeter $P$, and height $h$:
- Total SA $= 2B + Ph$
- Lateral area $= Ph$
- Cube (edge $s$): $\text{SA} = 6s^2$
- Rectangular prism ($l \times w \times h$): $\text{SA} = 2(lw + lh + wh)$
A cylinder has two circular bases. Unrolling the lateral surface gives a rectangle with width $2\pi r$ and height $h$.
Theorem: Surface Area of Prisms and Cylinders
- Any prism: $\text{SA} = 2B + Ph$, where $B$ = base area, $P$ = base perimeter, $h$ = height
- Cylinder: $\text{SA} = 2\pi r^2 + 2\pi r h = 2\pi r(r + h)$
- Cylinder lateral area: $= 2\pi r h$
Example 12.1 — Rectangular Prism
Find the surface area of a rectangular prism with dimensions $5 \times 4 \times 3$.
$\text{SA} = 2(lw + lh + wh) = 2(5 \cdot 4 + 5 \cdot 3 + 4 \cdot 3) = 2(20 + 15 + 12) = 2(47) = \mathbf{94 \text{ sq units}}$
Example 12.2 — Cylinder
A cylinder has radius $r = 6$ and height $h = 10$. Find the total surface area and lateral area.
Total SA: $2\pi(6)(6 + 10) = 12\pi(16) = 192\pi \approx \mathbf{603 \text{ sq units}}$
Lateral area: $2\pi(6)(10) = 120\pi \approx \mathbf{377 \text{ sq units}}$
A cube has edge length 7. Find the surface area and volume.
Show Answer
Volume: $V = s^3 = 7^3 = \mathbf{343 \text{ cu units}}$
Rectangular prism drawn with perspective — visible edges in purple, hidden edges dashed.
Figure 12.1 — Rectangular Prism: Surface Area Visualization
12.2 Surface Area of Pyramids and Cones
A regular pyramid has a regular polygon as its base, with its apex directly above the center. The slant height $l$ is the distance from the apex to the midpoint of a base edge (along the lateral face).
Pyramid and Cone Surface Area
- Any regular pyramid: $\text{SA} = B + \frac{1}{2}Pl$, where $B$ = base area, $P$ = base perimeter, $l$ = slant height
- Square pyramid (base $s$, slant height $l$): $\text{SA} = s^2 + 2sl$
- Cone: $\text{SA} = \pi r^2 + \pi r l$, where $l = \sqrt{r^2 + h^2}$ is the slant height
- Cone lateral area: $= \pi r l$
Theorem: Surface Area of Pyramids and Cones
- Regular pyramid: $\text{SA} = B + \frac{1}{2}Pl$
- Cone: $\text{SA} = \pi r^2 + \pi r l$ where slant height $l = \sqrt{r^2 + h^2}$
- Note: $l$ (slant height) is measured along the lateral face, not perpendicular to the base ($h$).
Example 12.3 — Square Pyramid
A square pyramid has base edge $s = 8$ and slant height $l = 10$. Find the total surface area.
$\text{SA} = s^2 + 2sl = 8^2 + 2(8)(10) = 64 + 160 = \mathbf{224 \text{ sq units}}$
Example 12.4 — Cone
A cone has radius $r = 5$ and height $h = 12$. Find the total surface area.
Slant height: $l = \sqrt{r^2 + h^2} = \sqrt{25 + 144} = \sqrt{169} = 13$
Total SA: $\pi(5)^2 + \pi(5)(13) = 25\pi + 65\pi = 90\pi \approx \mathbf{283 \text{ sq units}}$
A cone has $r = 3$ and slant height $l = 5$. Find the total surface area and the lateral area only (without the base).
Show Answer
Total SA: $\pi r^2 + \pi r l = \pi(9) + 15\pi = 24\pi \approx \mathbf{75.4 \text{ sq units}}$
AP Tip: Don't confuse slant height $l$ (measured along the lateral face from apex to base edge midpoint) with the actual height $h$ (perpendicular from the base to the apex). They are related by $l = \sqrt{r^2 + h^2}$ for a cone, but $l > h$ whenever $r > 0$.
Square pyramid — lateral edges in purple, base in blue, dashed height line from apex to center of base.
Figure 12.2 — Square Pyramid: Slant Height vs. True Height
12.3 Volume of Prisms, Cylinders, Pyramids, and Cones
Volume measures the amount of three-dimensional space enclosed by a solid. The fundamental principle: volume = (area of base) × height for prisms and cylinders.
Volume Formulas: Prisms and Cylinders
- Any prism: $V = Bh$ (base area × height)
- Rectangular prism: $V = lwh$
- Cube: $V = s^3$
- Cylinder: $V = \pi r^2 h$
Pyramids and cones hold exactly $\frac{1}{3}$ the volume of the corresponding prism or cylinder with the same base and height.
Volume Formulas: Pyramids and Cones
- Any pyramid: $V = \frac{1}{3}Bh$
- Cone: $V = \frac{1}{3}\pi r^2 h$
Cavalieri's Principle
If two solids have the same height and equal cross-sectional areas at every level, then they have equal volumes. This justifies using $V = Bh$ for oblique prisms and $V = \frac{1}{3}Bh$ for oblique pyramids.
| Solid | Volume Formula | Surface Area Formula |
|---|---|---|
| Rectangular prism | $V = lwh$ | $\text{SA} = 2(lw+lh+wh)$ |
| Cube | $V = s^3$ | $\text{SA} = 6s^2$ |
| Cylinder | $V = \pi r^2 h$ | $\text{SA} = 2\pi r(r+h)$ |
| Square pyramid | $V = \frac{1}{3}s^2 h$ | $\text{SA} = s^2 + 2sl$ |
| General pyramid | $V = \frac{1}{3}Bh$ | $\text{SA} = B + \frac{1}{2}Pl$ |
| Cone | $V = \frac{1}{3}\pi r^2 h$ | $\text{SA} = \pi r^2 + \pi rl$ |
Example 12.5 — Cylinder Volume
A cylinder has $r = 4$ and $h = 9$. Find the volume.
$V = \pi r^2 h = \pi(16)(9) = 144\pi \approx \mathbf{452 \text{ cu units}}$
Example 12.6 — Square Pyramid Volume
A square pyramid has base $s = 6$ and height $h = 8$. Find the volume.
$B = s^2 = 36$. $\quad V = \frac{1}{3}Bh = \frac{1}{3}(36)(8) = \mathbf{96 \text{ cu units}}$
Example 12.7 — Cone vs. Cylinder Comparison
A cone and a cylinder have the same radius $r$ and height $h$. Compare their volumes.
$V_{\text{cylinder}} = \pi r^2 h$, $\quad V_{\text{cone}} = \frac{1}{3}\pi r^2 h$
The cone's volume is exactly $\frac{1}{3}$ that of the cylinder. Three congruent cones would fill the cylinder completely.
A triangular prism has a right triangle base with legs $3$ and $4$, and the prism has height $10$. Find the volume.
Show Answer
$V = Bh = 6 \times 10 = \mathbf{60 \text{ cu units}}$
Cylinder (left, purple) and its equal-volume comparison cone (right, blue) — same radius and height, but the cone holds ⅓ the volume.
Figure 12.3 — Cylinder and Cone: Same Base and Height, Volumes in Ratio 3:1
12.4 Spheres and Composite Solids
A sphere is the set of all points in space equidistant from a center point. Its surface area and volume depend only on the radius $r$.
Sphere and Hemisphere Formulas
- Sphere SA: $\text{SA} = 4\pi r^2$
- Sphere Volume: $V = \dfrac{4}{3}\pi r^3$
- Hemisphere SA (curved surface + flat circular base): $\text{SA} = 3\pi r^2$
- Hemisphere Volume: $V = \dfrac{2}{3}\pi r^3$
Example 12.8 — Sphere
A sphere has radius $r = 6$. Find the surface area and volume.
SA: $4\pi(6)^2 = 4\pi(36) = 144\pi \approx \mathbf{452 \text{ sq units}}$
Volume: $\dfrac{4}{3}\pi(6)^3 = \dfrac{4}{3}\pi(216) = 288\pi \approx \mathbf{905 \text{ cu units}}$
Example 12.9 — Composite Solid (Ice Cream Cone)
An ice cream treat consists of a cone ($r = 3$, $h = 10$) with a hemisphere ($r = 3$) on top. Find the total volume.
$V_{\text{cone}} = \dfrac{1}{3}\pi(3)^2(10) = \dfrac{1}{3}\pi(9)(10) = 30\pi$
$V_{\text{hemisphere}} = \dfrac{2}{3}\pi(3)^3 = \dfrac{2}{3}\pi(27) = 18\pi$
Total: $30\pi + 18\pi = 48\pi \approx \mathbf{150.8 \text{ cu units}}$
A ball (sphere) has diameter $14$ cm. Find the surface area and volume.
Show Answer
SA: $4\pi(7)^2 = 4\pi(49) = 196\pi \approx \mathbf{616 \text{ sq cm}}$
Volume: $\dfrac{4}{3}\pi(7)^3 = \dfrac{4}{3}\pi(343) \approx \mathbf{1437 \text{ cu cm}}$
AP Tip: Sphere formulas ($\text{SA} = 4\pi r^2$ and $V = \frac{4}{3}\pi r^3$) must be memorized — they are not provided on most formula sheets, including the SAT and many state geometry exams.
Practice Problems
A rectangular prism has dimensions $8 \times 5 \times 4$. Find the surface area and volume.
Show Solution
$V = 8 \times 5 \times 4 = \mathbf{160 \text{ cu units}}$
A cylinder has $r = 7$ and $h = 15$. Find the surface area and volume. Leave answers in terms of $\pi$.
Show Solution
$V = \pi(7)^2(15) = \mathbf{735\pi \text{ cu units}}$
A square pyramid has $s = 12$ and slant height $l = 10$. Find the total surface area and the lateral area.
Show Solution
Lateral area: $\frac{1}{2}Pl = \frac{1}{2}(48)(10) = \mathbf{240 \text{ sq units}}$
A cone has $r = 6$ and $h = 8$. Find the slant height, total surface area, and volume. Leave in terms of $\pi$.
Show Solution
Total SA: $\pi(6)^2 + \pi(6)(10) = 36\pi + 60\pi = \mathbf{96\pi \text{ sq units}}$
Volume: $\frac{1}{3}\pi(6)^2(8) = \frac{1}{3}(36)(8)\pi = \mathbf{96\pi \text{ cu units}}$
A storage silo is a cylinder topped with a hemisphere. The cylinder has $r = 10$ ft and $h = 30$ ft. Find the total volume.
Show Solution
$V_{\text{hemi}} = \frac{2}{3}\pi(10)^3 = \frac{2000}{3}\pi$
Total: $3000\pi + \frac{2000}{3}\pi = \frac{11000}{3}\pi \approx \mathbf{11{,}519 \text{ cu ft}}$
A sphere has surface area $100\pi$. Find the radius and volume.
Show Solution
$V = \frac{4}{3}\pi(5)^3 = \frac{4}{3}\pi(125) = \frac{500}{3}\pi \approx \mathbf{524 \text{ cu units}}$
Two similar cones have radii in ratio $3:5$. Find the ratio of their volumes.
Show Solution
$\dfrac{V_1}{V_2} = \left(\dfrac{3}{5}\right)^3 = \dfrac{27}{125}$
Volume ratio: $27 : 125$
AP Problem: A sculpture is a cube ($s = 6$ cm) with a cone ($r = 3$, $h = 4$) on top. The cone's base sits on the top face of the cube. Find the total surface area of the composite solid.
Show Solution
Cube has 5 full faces + 1 top face minus circle: $5(36) + (36 - 9\pi) = 180 + 36 - 9\pi$
Cone lateral area: $\pi(3)(5) = 15\pi$ (no base — it's attached to cube)
Total SA: $216 - 9\pi + 15\pi = 216 + 6\pi \approx \mathbf{234.9 \text{ sq cm}}$
📋 Chapter Summary
Surface Area Formulas
$SA = 2B + Ph$ where $B$ = base area, $P$ = base perimeter, $h$ = height.
$SA = 2\pi r^2 + 2\pi rh$ (two circular bases plus lateral surface).
$SA = B + \frac{1}{2}P\ell$ where $\ell$ = slant height and $P$ = perimeter of base.
$SA = \pi r^2 + \pi r\ell$ where $\ell = \sqrt{r^2 + h^2}$ is the slant height.
Volume Formulas
$V = Bh$ — base area times height. Cylinder: $V = \pi r^2 h$.
$V = \frac{1}{3}Bh$ — one-third of base area times height. Cone: $V = \frac{1}{3}\pi r^2 h$.
$V = \frac{4}{3}\pi r^3$ $SA = 4\pi r^2$. Both depend only on the radius $r$.
If linear scale factor is $k$: areas scale by $k^2$, volumes scale by $k^3$.