Surface Area and Volume of Solids
The surface area of a solid is the total area of its faces or curved surfaces, and its volume is the amount of space it occupies; both can be found with formulas based on the shape of the base and the height. In brief: a cuboid with edges a, b, c has volume abc and surface area 2(ab + bc + ca). For a 3 by 4 by 5 box, V = 60 and S = 2(12 + 20 + 15) = 94. Volume is measured in cubic units, and surface area in square units.
Surface area tells how much material covers a solid (paint, wrapping paper, sheet metal) and volume tells how much it holds (water, concrete, air). This guide collects the main formulas for prisms, pyramids, cylinders, cones, spheres and frustums, introduces Cavalieri’s principle and the scaling rules, and shows how to choose and use formulas, with a complete table and worked examples.
What are surface area and volume?
Surface area measures the outside of a solid. For a polyhedron it is the sum of the areas of all faces, which can be seen from a net (the unfolded faces). For curved solids, such as a cylinder, the lateral surface unrolls to a rectangle or a sector. Volume measures the space inside. The unit cube with an edge of 1 unit has a volume of 1 cubic unit, and the volume of a box counts how many unit cubes it holds.
For prisms and cylinders, V = (area of base) · height. For pyramids and cones, V = (1/3) · (area of base) · height. Special formulas apply to spheres.
What is Cavalieri’s principle?
If two solids lie between two parallel planes and every plane parallel to them cuts both solids in cross-sections of equal area, then the solids have equal volumes. For example, an oblique prism has the same volume as a right prism with the same base and height, because every cross-section is congruent to the base. A stack of coins that is pushed to a slant has the same volume as the straight stack. The principle justifies the formula V = B · h for oblique prisms and cylinders, and the factor 1/3 for all pyramids and cones.
What are the formulas for frustums?
A frustum is the part of a cone or a pyramid that remains after a smaller, similar cone or pyramid is cut off by a plane parallel to the base.
- Frustum of a cone with radii R and r and height h: V = (1/3)πh(R2 + Rr + r2). The slant height is s = √(h2 + (R − r)2), the lateral area is π(R + r)s, and the total area is π(R + r)s + πR2 + πr2.
- Frustum of a pyramid with base areas B1 and B2 and height h: V = (1/3)h(B1 + √(B1B2) + B2).
Example 1: a bucket
A bucket is a frustum with the top radius 15 cm, the bottom radius 10 cm and height 24 cm. V = (1/3)π · 24 · (225 + 150 + 100) = 8π · 475 = 3800π ≈ 11,938 cm3 (about 11.9 litres). The slant height is √(576 + 25) = √601 ≈ 24.52, and the lateral area is π · 25 · 24.52 ≈ 1,925.6 cm2. With the bottom only: 1925.6 + π · 100 ≈ 2,239.7 cm2 of metal.
Example 2: a pyramid frustum
Square bases with sides 8 and 4 and height 6. B1 = 64, B2 = 16, √(64 · 16) = 32. V = (1/3) · 6 · (64 + 32 + 16) = 224.
How do surface area and volume scale?
If all lengths of a solid are multiplied by k, then all areas are multiplied by k2 and all volumes by k3. This follows from similarity. A cube with edge 2 has volume 8 and area 24; with edge 6 (k = 3) the volume is 216 (27 times 8) and the area is 216 (9 times 24). The ratio S/V decreases as the solid grows: small objects have a large surface compared to their volume, which is why small animals lose heat faster.
Worked examples
Example 3: a swimming pool
A pool is 25 m long, 10 m wide and 2 m deep. V = 500 m3 = 500,000 litres. The surface to tile (floor and four walls): 250 + 2 · (25 · 2) + 2 · (10 · 2) = 250 + 100 + 40 = 390 m2.
Example 4: a sphere in a cube
The largest sphere inside a cube of edge 10 has radius 5. Its volume is (4/3)π · 125 ≈ 523.6, which is 52.4% of the cube’s 1,000.
Example 5: a cone-shaped pile
A heap of sand forms a cone with the base radius 3 m and height 2 m. V = (1/3)π · 9 · 2 = 6π ≈ 18.85 m3. With sand density about 1.6 tonnes per cubic meter, the mass is about 30 tonnes (an approximate value).
Example 6: unknown edge
A cube has a surface area of 150 cm2. Then 6a2 = 150, so a = 5 and V = 125 cm3.
Example 7: a cylinder cut in half
A half-cylinder (trough) has radius 10 cm and length 100 cm. V = (1/2)π · 100 · 100 = 5000π ≈ 15,708 cm3. The surface: curved part π · 10 · 100 = 3141.6, the flat rectangle 20 · 100 = 2000 and two half discs 2 · (1/2)π · 100 = 314.2, which totals 5,455.8 cm2.
Master table of formulas
| Solid | Volume | Surface area |
|---|---|---|
| Cube (edge a) | a3 | 6a2 |
| Cuboid (a, b, c) | abc | 2(ab + bc + ca) |
| Prism (base B, perimeter P, height h) | Bh | 2B + Ph |
| Pyramid (base B, perimeter P, slant s) | (1/3)Bh | B + (1/2)Ps (regular) |
| Cylinder (r, h) | πr2h | 2πr(r + h) |
| Cone (r, h, s) | (1/3)πr2h | πr(r + s) |
| Sphere (r) | (4/3)πr3 | 4πr2 |
| Cone frustum (R, r, h) | (1/3)πh(R2 + Rr + r2) | π(R + r)s + πR2 + πr2 |
Where do these formulas come from?
Egyptian and Babylonian scribes computed volumes of boxes, cylinders and frustums for granaries and building works, and the Moscow Papyrus gives a correct formula for the volume of a square frustum. Greek mathematicians proved the formulas rigorously: Eudoxus and Euclid for pyramids and cones, Archimedes for the sphere. Bonaventura Cavalieri formulated his principle in the seventeenth century, a forerunner of integral calculus.
Where are they used today?
Architects compute volumes of concrete, painters compute areas of walls and engineers calculate the capacity of tanks. Packaging uses the surface area to estimate the cardboard needed. In medicine, the volume of an organ is estimated from images, and in science the ratio of surface area to volume determines how fast a body exchanges heat.
Which mistakes are common?
Mixing up units
1 m3 = 1000 litres, and 1 m2 = 10,000 cm2. Convert before computing.
Using the wrong height
Volume formulas use the perpendicular height. Surface formulas for cones and pyramids use the slant height.
Forgetting a face
An open box (without a lid) has five faces. Count the faces needed for the question.
Scaling volume linearly
Doubling all lengths multiplies the volume by 8, not 2.