Surface area is the total area of a solid's outside, the paint it would take to cover it. Unfold the solid into a flat net, as in Area and Surface Area Problems, and add the areas of the pieces. The height and slant height from Volume of Pyramids, Cones and Spheres play different roles here.
Why it works
Cylinder: two circles plus a rectangle whose length is the circumference, $2\pi r$, so $S = 2\pi r^2 + 2\pi rh$. Cone: the curved part unrolls into a sector of a circle with radius $\ell$ (the slant height) and arc length $2\pi r$. Its area is $\frac{1}{2} \cdot \ell \cdot 2\pi r = \pi r \ell$, like the "half the apothem times the perimeter" idea. So $S = \pi r^2 + \pi r \ell$. Pyramid: the base plus triangles whose heights are the slant height $\ell$. Sphere: $S = 4\pi r^2$, exactly four times the area of a great circle (Archimedes showed this; a full proof needs calculus).
| Solid | Surface area |
|---|---|
| rectangular prism $l \times w \times h$ | $2(lw + lh + wh)$ |
| cylinder | $2\pi r^2 + 2\pi rh$ |
| square pyramid, base edge $s$, slant height $\ell$ | $s^2 + 2s\ell$ |
| cone | $\pi r^2 + \pi r \ell$ |
| sphere | $4\pi r^2$ |
Key idea
Surface area adds the areas of every face. The slanted faces of pyramids and cones use the slant height $\ell$, not the height $h$; they are related by the Pythagorean theorem.
Worked example: A cylinder
Find the surface area of a cylinder with radius $3$ and height $8$.
- Bases: $2 \cdot \pi \cdot 9 = 18\pi$. Side: $2\pi \cdot 3 \cdot 8 = 48\pi$.
- $S = 66\pi \approx 207.3$.
Answer: $66\pi$
Worked example: A pyramid from its height
A square pyramid has base edge $10$ and height $12$. Find its surface area.
- The slant height is the hypotenuse of a right triangle with legs $12$ (the height) and $5$ (half the base edge): $\ell = \sqrt{144 + 25} = 13$.
- Base: $100$. Four triangles: $4 \cdot \frac{1}{2} \cdot 10 \cdot 13 = 260$.
- $S = 360$.
Answer: $360$ square units
Worked example: A cone
A cone has radius $6$ and height $8$. Find its surface area.
- $\ell = \sqrt{6^2 + 8^2} = 10$.
- $S = \pi \cdot 36 + \pi \cdot 6 \cdot 10$ $= 36\pi + 60\pi = 96\pi \approx 301.6$.
Answer: $96\pi$
Common mistake
Using the height instead of the slant height for the lateral area of a cone or pyramid. The triangles and the cone's curved surface slope, so their "height" is the slant height. Check: the slant height is longer than the height, so the correct lateral area is the larger one.