Surface Area and Volume of 3D Solids

Surface Area and Volume of 3D Solids

Surface area is the total area of all outer surfaces. Volume is the total space enclosed. For pyramids, cones and spheres the formulae involve a third of the base area or a cube of the radius — these must be memorised (they are given on the AQA formula sheet, but working with them fluently requires familiarity).

Formulae

SolidVolumeSurface Area
Sphere\( \frac{4}{3}\pi r^3 \)\( 4\pi r^2 \)
Cone (right circular)\( \frac{1}{3}\pi r^2 h \)\( \pi r l + \pi r^2 \) where \( l=\sqrt{r^2+h^2} \)
Pyramid (any base)\( \frac{1}{3} \times \text{base area} \times h \)base area + sum of lateral face areas
Cylinder\( \pi r^2 h \)\( 2\pi r h + 2\pi r^2 \)

Slant Height of a Cone

The slant height \( l \) connects the apex to the circumference of the base. It is found using Pythagoras:

\[ l = \sqrt{r^2 + h^2} \]

The curved surface area uses \( l \), not \( h \). Do not confuse them.

Composite Solids

For solids built from standard parts (e.g. cylinder with hemispherical cap, cone on a cylinder), find each volume separately and add. For surface area, identify which faces are exposed — internal joins are not counted.

Worked Examples

Find the volume and surface area of a sphere with radius 6 cm.
\[ V = \frac{4}{3}\pi\times216 = 288\pi\approx905 \text{ cm}^3 \] \[ SA = 4\pi\times36 = 144\pi\approx452 \text{ cm}^2 \]
A cone has base radius 5 cm and perpendicular height 12 cm. Find the volume and total surface area.
\[ l = \sqrt{25+144} = \sqrt{169} = 13 \text{ cm} \] \[ V = \frac{1}{3}\pi\times25\times12 = 100\pi\approx314 \text{ cm}^3 \] \[ SA = \pi\times5\times13 + \pi\times25 = 65\pi + 25\pi = 90\pi\approx283 \text{ cm}^2 \]
A solid consists of a cylinder (radius 4 cm, height 10 cm) with a hemisphere (radius 4 cm) on top. Find the total volume.
\[ V_{\text{cyl}} = \pi\times16\times10 = 160\pi \] \[ V_{\text{hemi}} = \frac{1}{2}\times\frac{4}{3}\pi\times64 = \frac{128\pi}{3} \] \[ V_{\text{total}} = 160\pi + \frac{128\pi}{3} = \frac{480\pi+128\pi}{3} = \frac{608\pi}{3}\approx636 \text{ cm}^3 \]

 Key Takeaways

  • Sphere: \( V=\frac{4}{3}\pi r^3 \), \( SA=4\pi r^2 \).
  • Cone: \( V=\frac{1}{3}\pi r^2 h \), \( SA=\pi r l+\pi r^2 \), \( l=\sqrt{r^2+h^2} \).
  • Pyramid: \( V=\frac{1}{3}\times\text{base area}\times h \).
  • Composite solids: volumes add; surface areas require identifying exposed faces only.
  • Slant height \( l \neq h \) — always use Pythagoras to find \( l \) from \( r \) and \( h \).