Volume and surface area

Use the formula, or work back to a length

In 9 of 20 recent exams
2% of the marks
Paper 1 or 2
2 to 5 marks

How it is set in the exam

Both papers, 2–5 marks a part: 7 parts on Paper 1 (no calculator), 13 on Paper 2 (calculator). Common (9 of 20 exam versions). Usually the start of a longer question: coordinates give the lengths, the lengths give a volume or an area, and a later part asks for an angle (Triangles inside a solid) or an optimisation.

Spot it in the exam

Exam phrases to spot (8)

“find the volume of the cone”
“find the total surface area”
“find the curved surface area”
“the points A(…) and B(…) are the ends of a diameter of a sphere”
“the vertex of the cone is at V(…)”
“given that the volume is 250 cm³, find the height”
“the sector is folded to form a cone”
“show that the volume is … cm³, correct to three significant figures”

In disguise (4)

the solid is given only by coordinates, with no length written anywhere: the lengths come from the midpoint and distance formulas
a volume or surface area in terms of an angle, such as a cylinder whose rims touch a sphere: the radius and the height come from the right triangle in the cross-section (Triangles inside a solid first), and reaching the given form may need $\sin^2\theta + \cos^2\theta = 1$ (taught in The Pythagorean identity)
the tangent of an angle in the solid is given, not a length: the sine and cosine come from a right triangle, then the exact volume
two solids stuck together: add the volumes, and count only the outside surfaces

Variants (8)

the ends of a diameter → the centre, the radius, the volume of the sphere (Example 1.1)
a total surface area → the slant height → the vertical height → the exact volume of a cone (Example 1.2)
a cone and a hemisphere joined: a volume show-that, then the outside surface (Example 1.3)
a sector folded into a cone: the base radius, the height, the volume (Example 1.4)
a cone from coordinates: the base centre as a midpoint, the height as a distance (Your turn 1)
a pyramid's height from its volume (Your turn 2)
a sector in degrees folded into a cone: a show-that of the rounded radius, the volume, then a hemisphere glued on (Your turn 3)
the tangent of an angle given → an exact volume in a requested surd form (Your turn 4)

Looks like it, but isn't (4)

the unknown is an angle; or a length asked for itself, with no volume or area after it (an edge, a diagonal, a pyramid's height from its slant edge); or the radius or height of a solid inside a sphere or a hemisphere: Triangles inside a solid
the volume is fixed and a least surface area or a greatest volume is wanted: Optimisation: use the constraint, then optimise
a sector's arc, perimeter or area is asked, and nothing is folded: Arcs and sectors: arc length, perimeter and area
the height of a solid comes from an angle seen from outside it: Angles of elevation and depression first, then back here for the volume

Worked example

A sphere from the ends of a diameter

Paper 2 (calculator)

Question

The points $P(3, -2, 8)$ and $Q(-1, 6, -2)$ are the two ends of a diameter of a sphere. All lengths are in centimetres.
(a) Find the coordinates of the centre, $C$ , of the sphere. [2]
(b) Find the radius of the sphere. [2]
(c) Find the volume of the sphere. [2]
(d) The point $T$ has coordinates $(6, 5, 0)$ . Determine whether $T$ lies inside the sphere. [2]

Try it yourself first, then compare.

Now try one like it

Write your working on paper or on screen. Paper Zero marks it step by step, like the worked example, and shows where each mark went. Free, with no account.

Try one like it

Which method?

The formula is never the hard part: it is in the booklet, or it is half of one. What varies is where the missing length comes from.

the ends of a diameter as coordinates
The length you need
the centre and the radius
Tool
the midpoint; then the distance from the centre to an end, or half the diameter
the centre of the base and the apex as coordinates
The length you need
the height
Tool
the distance formula
two of $r$ , $h$ , $l$ (cone), or the height and the base (pyramid)
The length you need
the third
Tool
Pythagoras in the right triangle (The hidden right triangles)
an angle (or its tangent) and one length
The length you need
another length
Tool
right-angled trig; from a tangent, draw the right triangle to get the sine and cosine
a sector that is folded into a cone
The length you need
the base radius and the slant height
Tool
arc length = circumference of the base; the sector's radius = the slant height
the volume or an area
The length you need
the length in the formula
Tool
the formula set equal to the value, solved (positive root)

You know	The length you need	Tool
the ends of a diameter as coordinates	the centre and the radius	the midpoint; then the distance from the centre to an end, or half the diameter
the centre of the base and the apex as coordinates	the height	the distance formula
two of $r$ , $h$ , $l$ (cone), or the height and the base (pyramid)	the third	Pythagoras in the right triangle (The hidden right triangles)
an angle (or its tangent) and one length	another length	right-angled trig; from a tangent, draw the right triangle to get the sine and cosine
a sector that is folded into a cone	the base radius and the slant height	arc length = circumference of the base; the sector's radius = the slant height
the volume or an area	the length in the formula	the formula set equal to the value, solved (positive root)

Steps

Lengths first.
From coordinates (the centre is the midpoint of a diameter; the radius is centre to an end, or half the diameter; the height is base centre to apex), from Pythagoras ( $h$ and $l$ ), from an angle, or from a folded sector.
Name the solid, or solids, and the formula.
A hemisphere is half a sphere; a pyramid's $A$ is the area of its base.
Decide which surfaces count.
Total or curved; a solid hemisphere's flat face; no lid on an open container; where two solids join, the hidden faces don't count.
Substitute, or work back:
set the formula equal to the given value and solve for the length.
Answer in the right form.
: exact, with $\pi$ and surds (and the requested form, e.g. $p\pi\sqrt{10}$ ). : write the expression with its values, then the value to 3 s.f., with units. A show-that of a rounded value needs the unrounded (or exact) value written before it. GDC: keep full-accuracy values: carry the full value, not a rounded one.

Common mistakes

The slant height in the volume,
or the vertical height in $\pi rl$ . The volume uses $h$ , the curved surface uses $l$ : markschemes give no method mark for $\pi rl$ with the vertical height.
A "total" surface area without the base
(or without the flat face of a solid hemisphere). The curved surface alone earns the method mark at most.
$4\pi r^2$ for a hemisphere.
Its curved surface is half that, $2\pi r^2$ ; add $\pi r^2$ only for a solid hemisphere's flat face.

More mistakes (7)

Counting hidden faces
where two solids are joined, or a lid on an open container.
The diameter used as the radius
when the two ends are given as coordinates: halve it.
Rounding before the end
keep the full value ( $\sqrt{45}$ , or the stored value) until the final answer. Markschemes accept a later answer worked from an earlier part's correct 3 s.f. answer (in Example 1.1, $6.71$ gives $1270$ , accepted as "from 3 s.f."), but not one worked from a value rounded to fewer figures, or rounded partway through a part. The full value always scores.
A numerical show-that that starts at the rounded value
write the unrounded (or exact) value first, then the given one.
Not in the requested form
"in the form $p\pi\sqrt{10}$ " needs the denominator rationalised; $\frac{64\pi}{10\sqrt{10}}$ loses the last mark.
A folded sector
its radius becomes the cone's slant height, and its arc the base's circumference. If the sector's angle is in degrees, the arc is $\frac{\theta}{360} \times 2\pi R$ : $R\theta$ needs radians.
Units
cm³ for a volume, cm² for an area.

Easy to confuse with

Next typeTriangles inside a solid

Now try one like it

Keep going

Easy to confuse with