0580 · 5.4Surface area and volume

Surface area, volume and cuboids

0:00A storage box needs enough material for its outside and enough space for everything inside. Those are two different measurements, even though they describe the same object. Surface area measures its covering, while volume measures the space it occupies. By the end, you will understand how these measurements work for cuboids, prisms, cylinders, pyramids, cones and spheres. Surface area is measured in square units because we are covering surfaces with small squares.

0:24Volume is measured in cubic units because we are filling space with small cubes. A length in centimetres therefore gives area in square centimetres and volume in cubic centimetres. Make all lengths use the same unit before substituting them into a formula. You need the areas of rectangles, triangles and circles, together with the circumference of a circle. A circle has area pi times radius squared and circumference 2 × π times radius.

0:49The radius runs from the centre to the circumference. If a question gives the diameter, divide it by two before using a radius formula. A cuboid has six rectangular faces, with opposite faces equal in area. Its total surface area comes from adding all six face areas. There are two faces with area length times width, two with area length times height, and two with area width times height. Pairing equal faces makes the calculation shorter without leaving anything out.

1:15Its volume has a different explanation. One horizontal layer contains length times width unit cubes, and the height tells us how many layers fit. So volume equals length times width times height. This is not adding the outside faces; it is counting the space throughout the solid. Calculate the volume and total surface area of a cuboid measuring five centimetres by three centimetres by two centimetres. For volume, multiply five by three by two.

1:40This gives thirty cubic centimetres. For surface area, begin by calculating one face from each opposite pair. The length times width gives fifteen square centimetres. Multiplying length by height produces ten square centimetres. The remaining face area is 3 × 2, giving six square centimetres. Add fifteen, ten and six, then double the total of thirty-one. The total surface area is sixty-two square centimetres.

Prisms and cylinders

2:06A prism is a solid with a uniform cross-section. That means a slice taken perpendicular to its length has the same shape and size wherever you take it. Imagine sliding one flat shape forwards without changing it. The space swept out gives volume equal to cross-sectional area multiplied by length. The cross-section does not have to be a rectangle or triangle. In this syllabus, the term includes any solid with a uniform cross-section, such as a cylindrical sector.

2:31For that shape, find the sector area and multiply by the solid’s length. Choosing the correct cross-section matters more than recognising a familiar-looking name. For a straight prism with polygonal ends, the remaining faces are rectangles. Each rectangle has the prism’s length as one dimension and a cross-section edge as the other. Adding their areas gives cross-section perimeter multiplied by length. Total surface area is therefore two end areas plus the areas of the connecting faces.

2:59Calculate the volume and total surface area of the triangular prism shown. Its cross-section is a right-angled triangle, with perpendicular sides of three and four centimetres. Half times 3 × 4 gives a cross-sectional area of six square centimetres. Multiply six by the prism length of ten to obtain sixty cubic centimetres. The cross-section perimeter is 3 + 4 + 5, giving twelve centimetres. Multiplying twelve by ten gives one hundred and twenty square centimetres for the connecting faces.

3:27The two triangular ends contribute twice six, which is twelve square centimetres. Adding these contributions gives a total surface area of one hundred and thirty-two square centimetres. A cylinder follows the same volume principle because its circular cross-section stays uniform. Its volume is pi times radius squared times height. To understand its curved surface, imagine cutting vertically down the side and unrolling it. The resulting rectangle has height equal to the cylinder’s height and width equal to the circular circumference.

3:56Multiplying that circumference by height gives curved surface area equal to 2 × π times radius times height. A closed cylinder also has two circular ends. Its total surface area is the curved area plus 2 × π times radius squared. Curved surface area and total surface area are therefore different quantities. A closed cylinder has radius three centimetres and height eight centimetres. Calculate its volume and total surface area, giving exact answers in terms of pi.

4:25Substitute into the volume formula to get π × 3² × 8. Since 3² is nine, the volume becomes 72π cubic centimetres. The curved area is 2 × π × 3 × 8, giving 48π square centimetres. The two ends have combined area 2 × π × 3², which is 18π square centimetres. Adding 48π and 18π gives 66π square centimetres. Keeping pi preserves the exact answer requested. For an idealised container without a lid, only one circular end would be included.

4:57Read the description before deciding which surfaces to count. A missing lid changes the surface area but not the space available inside. Surface area calculations must match the actual surfaces described, rather than automatically using a closed-solid formula.

Pyramid volume and surface area

5:12A pyramid has a polygonal base and triangular faces meeting at an apex. Unlike a prism, its cross-sections become smaller towards the top. Its volume is one third times base area times perpendicular height. Perpendicular height means the distance straight from the apex to the plane of the base, measured at a right angle. The factor of one third accounts for the narrowing throughout the solid. A pyramid has one third the volume of a prism with the same base area and perpendicular height.

5:39This relationship can be demonstrated by comparing how much the two solids hold. It applies to different base shapes, not just square-based pyramids. Surface area needs a separate calculation because the triangular faces slope. Their triangle heights lie along the faces, rather than through the inside of the pyramid. For a right square-based pyramid, these equal face heights are called slant heights. Add the base area and all four triangular face areas to obtain the total.

6:04Calculate both measurements for the pyramid shown, using the given perpendicular and slant heights. The square base has area 6², which is thirty-six square centimetres. Volume is ⅓ × 36 × 4, giving forty-eight cubic centimetres. The four centimetres measures perpendicular height, so it belongs in the volume calculation. One triangular face has area half times 6 × 5, giving fifteen square centimetres. Multiply fifteen by four to obtain sixty square centimetres for the triangular faces.

6:34Add the base area of thirty-six to get ninety-six square centimetres overall. The five centimetres belongs here because it measures height within each triangular face. It fits Pythagoras’ theorem, since half the base edge is three, and 3² + 4² is 5².

Cone formulas and calculations

6:48A cone narrows from a circular base to an apex. Its volume is ⅓ × π times radius squared times perpendicular height. This is one third of the corresponding cylinder’s volume, just as a pyramid relates to a prism. Its curved surface area uses slant height instead, because that measurement follows the surface being covered. A cone’s curved surface unrolls into a sector with radius equal to the cone’s slant height.

7:13Its arc length equals the circumference of the cone’s base. A full circle of this radius has circumference 2 × π times slant height. The sector is therefore the fraction base radius over slant height of that full circle. Apply that fraction to the full circle’s area, which is pi times slant height squared. One slant-height factor cancels, leaving pi times radius times slant height. That is the cone’s curved surface area.

7:36For total surface area, add the circular base area of pi times radius squared. Calculate the cone’s volume and total surface area, giving exact answers in terms of pi. Its volume is ⅓ × π × 3² × 4. Replacing 3² with nine gives 12π cubic centimetres. Use the perpendicular height of four, not the slant height of five. For curved area, π × 3 × 5 gives 15π square centimetres.

8:04The base area is π × 3², giving 9π square centimetres. Adding these areas produces 24π square centimetres in total. The radius, perpendicular height and slant height form a right-angled triangle. When one of them is missing, find it with Pythagoras’ theorem.

Spheres and exam technique

8:20A sphere consists of all points in space at a fixed distance from its centre. That distance is its radius, and its surface has no flat faces. Surface area equals 4 × π times radius squared. Volume equals 4/3 × π times radius cubed. The surface formula gives exactly four times the area of a circle with the same radius. That comparison helps interpret the formula, although it is not a derivation.

8:43The squared radius belongs to area, while the cubed radius belongs to volume. A sphere cannot be unfolded into a flat net without stretching or cutting its surface. An exam question gives a sphere’s diameter as twelve centimetres and asks for its volume to three significant figures. Divide twelve by two to obtain radius six centimetres. Cubing six gives two hundred and sixteen. Substitute this into the volume formula to obtain 288π cubic centimetres.

9:10Using the calculator pi button gives approximately nine hundred and four point seven seven eight cubic centimetres. Rounded to three significant figures, the answer is nine hundred and five cubic centimetres. Do not round during the calculation, because early rounding can change the final answer. Here a decimal answer is needed, rather than an answer left in terms of pi. Paper Two is non-calculator, so its answers are often left in terms of pi, while Paper Four allows a calculator.

9:37The exam formula list gives the curved surface areas of a cylinder and a cone, and the surface area of a sphere. It also gives the volumes of a prism, pyramid, cylinder, cone and sphere. You must know cuboid volume and surface area, circle area and circumference, and how to add ends and bases. You still decide which dimensions and surfaces count. Questions may ask you to calculate a measurement or determine a missing dimension.

10:01For example, reverse the prism formula by dividing volume by cross-sectional area to find length. Show your substitution, intermediate working and final units clearly. Unless instructed otherwise, round answers that are not exact to three significant figures. Common mistakes include using diameter as radius, confusing perpendicular height with slant height, and omitting a base. Diameter as radius makes squared and cubed calculations much too large. The wrong height measures a different part of the solid, so the formula no longer represents the required quantity.

10:31A missing face means the calculated covering is incomplete. The central distinction is between covering the outside and measuring the space inside. Uniform cross-sections give volume through area multiplied by length. Pyramids and cones introduce a factor of one third because they narrow towards an apex. For every problem, identify the shape, label the correct dimensions, count the required surfaces and finish with the appropriate units.