IB Analysis and Approaches HL · Geometry in space · Section 7 of 7

Volumes and surface areas of solids

3 worked examples · By Álvaro López

Key idea

Solid problems ask for two things: how much it holds (volume) and how much surface it has (area), and often also a length or an angle hidden inside the body. With a few formulas, Pythagoras' theorem, trigonometry and the vectors from this topic, almost everything can be solved. In the exam a formula booklet is provided, so what matters is knowing what each letter measures and recognising the right-angled triangle you need to draw.

Prisms and cylinders

A prism has two equal, parallel bases joined by lateral faces that are parallelograms. The cylinder is the same with a circular base. When the lateral faces are perpendicular to the bases, the prism or cylinder is called right, and then the height $h$ is the length of a lateral edge.

The idea behind the volume is simple: the prism is made of identical layers of the base, stacked up to height $h$ . The lateral area of the cylinder is a rectangle rolled up, with width $2 π r$ and height $h$ .

Pyramids and cones

A pyramid has a polygonal base and a vertex (the apex) to which all the lateral faces, which are triangles, meet. The cone is the version with a circular base. Here you need to tell apart three lengths that are easy to confuse:

the height $h$ , which is perpendicular to the base and reaches the apex;
the apothem of a regular pyramid, which is the height of a lateral face (it runs from the apex to the midpoint of an edge of the base);
the slant height $ℓ$ of a cone, which runs from the apex to a point on the edge of the base.

Regular pyramid with a square base, showing the height VO, the apothem VM and the segment OM, half of a side of the base

The triangle $V O M$ in the figure is right-angled at $O$ , and from it comes the relation between height, apothem and side: $a^{2} = h^{2} + (\frac{ℓ}{2})^{2}$ for a square base of side $ℓ$ . In a cone, the height, the radius and the slant height satisfy $ℓ^{2} = h^{2} + r^{2}$ .

If the lateral surface of a cone is unrolled onto a plane, it becomes a circular sector of radius $ℓ$ and arc length $2 π r$ ; its area is $\frac{1}{2} \cdot ℓ \cdot 2 π r = π r ℓ$ . It is the same formula as for the pyramid, taking the circumference of the base as the "perimeter".

Sphere

Composite solids and similarity

A composite solid is built by joining or removing simple pieces: a silo (a cylinder with a hemisphere on top), a pencil (a prism and a cone), a piece with a hole in it.

There is also a useful rule when two bodies are similar (one is an enlargement or reduction of the other). If lengths are multiplied by $k$ , areas are multiplied by $k^{2}$ and volumes by $k^{3}$ . This applies, for example, to the small cone cut off the tip of a large cone, or to the volume of a balloon when its radius is doubled (it is multiplied by $8$ ).

Lengths and angles inside a solid

Almost everything asked for inside a body (a slanted edge, a diagonal, an angle) comes from right-angled triangles. The key is to choose the right triangle and work out its sides with Pythagoras.

Diagonal of a cuboid. In a cuboid with edges $a$ , $b$ , $c$ , the diagonal of the base is $a^{2} + b^{2}$ . That diagonal and the vertical edge $c$ form a right-angled triangle whose hypotenuse is the diagonal of the solid:

Angle between a line and a plane. The angle that an edge (or a diagonal) makes with the base is the angle it makes with its orthogonal projection onto the base: the right-angled triangle has as one leg the height of the endpoint and as the other leg the projection. In a cube of edge $2$ , the diagonal of the solid is $23$ , its projection is the diagonal of the base, $22$ , and the height is $2$ , so the angle with the base satisfies $tan θ = \frac{2}{2 2} = \frac{1}{2}$ , that is, $θ \approx 35. 3^{\circ}$ .

Height of a pyramid. In a regular pyramid, the foot of the height is the centre of the base. With that, the height comes from a right-angled triangle: apex, centre of the base and vertex (for the lateral edge) or midpoint of an edge (for the apothem).

With the help of vectors

When the solid is given by the coordinates of its vertices, vectors replace the triangles in the drawing:

a length is the magnitude $∣ A B ∣$ ;
the angle between two edges comes from the scalar (dot) product, $cos θ = \frac{u \cdot v}{∣ u ∣ ∣ v ∣}$ ;
the area of a triangular face $A B C$ is $\frac{1}{2} ∣ A B \times A C ∣$ , and the volume of the tetrahedron (a pyramid with a triangular base) is $\frac{1}{6} ∣[A B, A C, A D]∣$ , as in the vector (cross) product;
the height from a vertex to the opposite face is the distance from a point to a plane (see angles and distances); it can also be found from $V = \frac{1}{3} A h$ if the volume and the area of that face are already known.

Worked examples

Example 1 Basic

A silo with a hemispherical cap

A silo has the shape of a right cylinder of radius $3$ m and height $8$ m, topped by a hemisphere of the same radius. The bottom base is flat and is closed by a slab.

a) Find the volume of the silo, in exact form.

b) Find the area of the exterior surface of the silo (including the bottom base), in exact form.

Solution

a) Add the volumes of the two pieces:

$V = π r^{2} h + \frac{2}{3} π r^{3} = π \cdot 9 \cdot 8 + \frac{2}{3} π \cdot 27 = 72 π + 18 π = 90 π m^{3}$

b) The exterior surface is made up of the bottom base, the lateral wall of the cylinder and the hemispherical dome. The circular joining face does not count.

$S = π r^{2} + 2 π r h + 2 π r^{2} = 9 π + 48 π + 18 π = 75 π m^{2}$

Example 2 Medium

Regular pyramid: volume, area and angles

A regular pyramid has a square base of side $8$ cm and its height is $3$ cm.

a) Find the volume.

b) Find the total surface area.

c) Find, in degrees, the angle that a lateral edge makes with the base.

d) Find, in degrees, the angle that a lateral face makes with the base.

Solution

a) The base area is $64$ , and

$V = \frac{1}{3} \cdot 64 \cdot 3 = 64 cm^{3}$

b) The apothem $a$ is needed. The triangle formed by the height, half of the side and the apothem is right-angled:

$a = 3^{2} + 4^{2} = 5 cm$

Each lateral face has area $\frac{1}{2} \cdot 8 \cdot 5 = 20$ , so

$S = 64 + 4 \cdot 20 = 144 cm^{2}$

c) The projection of a lateral edge onto the base is half of the diagonal of the base: $\frac{1}{2} \cdot 82 = 42$ . In the right-angled triangle with height $3$ :

$tan θ = \frac{3}{4 2} ⟹ θ = arctan \frac{3 2}{8} \approx 27. 9^{\circ}$

d) The angle between a face and the base is measured in the triangle formed by the height, half of the side and the apothem (the two segments $O M$ and $V M$ are perpendicular to the edge of the base):

$tan φ = \frac{3}{4} ⟹ φ = arctan \frac{3}{4} \approx 36. 9^{\circ}$

Example 3 Hard

A pyramid given by coordinates

A pyramid with a triangular base has vertices $A (0, 0, 0)$ , $B (4, 0, 0)$ , $C (0, 3, 0)$ and $V (0, 0, 6)$ (units in metres).

a) Calculate its volume.

b) Find the area of the face $V B C$ .

c) Find the total surface area of the pyramid.

d) Find the distance from the vertex $A$ to the plane of the face $V B C$ .

Solution

a) The base $A B C$ is a right-angled triangle with legs $4$ and $3$ , so its area is $6$ . The edge $A V$ is perpendicular to it, so the height is $6$ :

$V = \frac{1}{3} \cdot 6 \cdot 6 = 12 m^{3}$

(With the scalar triple product: $A B = (4, 0, 0)$ , $A C = (0, 3, 0)$ , $A V = (0, 0, 6)$ , determinant $72$ , and $\frac{1}{6} \cdot 72 = 12$ .)

b) Take $V B = (4, 0, - 6)$ and $V C = (0, 3, - 6)$ :

$V B \times V C = 0 \cdot (- 6) - (- 6) \cdot 3 (- 6) \cdot 0 - 4 \cdot (- 6) 4 \cdot 3 - 0 \cdot 0 = 182412$

The magnitude is $324 + 576 + 144 = 1044 = 629$ , so the area is

$\frac{1}{2} \cdot 629 = 329 m^{2}$

c) The other three faces are right-angled triangles: $A B C$ with area $6$ , $V A B$ with $\frac{1}{2} \cdot 4 \cdot 6 = 12$ and $V A C$ with $\frac{1}{2} \cdot 3 \cdot 6 = 9$ . Therefore

$S = 6 + 12 + 9 + 329 = 27 + 329 m^{2}$

d) The distance from $A$ to the face $V B C$ is the height of the pyramid when $V B C$ is taken as the base. Since $V = \frac{1}{3} A_{V B C} h$ :

$h = \frac{3 V}{A _{V B C}} = \frac{36}{3 29} = \frac{12}{29} = \frac{12 29}{29} m$

Check using the plane: the vector $(18, 24, 12)$ is proportional to $(3, 4, 2)$ , and the plane through $B$ is $3 x + 4 y + 2 z = 12$ ( $C$ and $V$ also satisfy it). The distance from the origin is $\frac{∣ - 12 ∣}{9 + 16 + 4} = \frac{12}{29}$ , which coincides.

In the exam: when you calculate an area with the vector product, write down the two vectors you use and do not forget the factor $\frac{1}{2}$ . For the height, make the method clear (solving for $h$ from $V = \frac{1}{3} A h$ with the chosen face as base, or the point-to-plane formula), because the method is marked as well as the value.

Common mistakes

Summary

Prism and cylinder: $V = A h$ ; total area $2 A + P h$ . Cylinder: $V = π r^{2} h$ , $S = 2 π r^{2} + 2 π r h$ .
Pyramid and cone: $V = \frac{1}{3} A h$ . Regular pyramid: $S = A + \frac{1}{2} P a$ . Cone: $S = π r^{2} + π r ℓ$ , with $ℓ^{2} = h^{2} + r^{2}$ .
Sphere: $S = 4 π r^{2}$ , $V = \frac{4}{3} π r^{3}$ . A formula booklet is provided in the exam; what you must know is what each letter measures.
Composite solids: add and subtract volumes; for surface area, count only the exterior faces. Similar solids with ratio $k$ : areas $\times k^{2}$ , volumes $\times k^{3}$ .
Diagonal of a cuboid: $a^{2} + b^{2} + c^{2}$ ; the angle of an edge with the base is measured using its projection onto the base.
With coordinates: magnitude for lengths, scalar product for angles, vector product for areas, scalar triple product for volumes, and point-to-plane distance for the height.

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