Surface areas of basic solids
R
Cube of side a
Elliptical
crosssection
A  6a 2
Sphere of radius r
A  4 r 2
Cuboid
h
A  2  ab  bc  ca 
Torus, with a circular
cross section of radius r
c
Cylinder, with radius r and height h
A  2 r  2 R  4 2 rR
A  2 r 2  2 rh
b
a
P
a
a
h
p
a
s
h
Frustum, (a truncated pyramid)
of base perimeter P, top area
perimeter p and slant height s.
a and a are the respective base
and top areas
r
s
Assume all
slants are the
same length
a
P
A  a  a  12 s  p  P 
A  2a  Ph
Otherwise we have
an oblique frustum or
pyramid and these
formulae will not hold
P
Prism, of length h with
uniform cross sectional
area a with perimeter P
Cone, with base radius r
and slant height s
A   rs
s
s
Pyramid, with base perimeter P and
slant height s
A  12 Ps
This is the curved surface area
i.e. an ‘open’ cone
Mathematics topic handout: Areas Dr Andrew French. www.eclecticon.info PAGE 1
Calculating the surface area of
a (right) pyramid
n0
Therefore the total exterior of the
pyramid is the sum of the exterior
areas of the N laminae
slant height
i.e. all slants are
the same length
Otherwise we have
an oblique pyramid
and this formula does
not hold
Pn 
s
s
N
b
Now the sum of the first N integers is
1  2  ....  N  12 N ( N  1)
Hence:
a
P
A pyramid of perpendicular slant height s and base
perimeter P can be thought of being composed of an
(infinitely) large number of similar laminae.
Let there be N laminae of thickness s/N. Since all
laminae are similar, the perimeters scale as n/N where n
is the lamina number from the apex. The exterior area of
each lamina is:
A  12 Ps
Ps 1
2 N ( N  1)
N2
N2  N
AN  12 Ps
N2
AN  12 Ps 1  N1 
AN 
Area of a right cone
A   r 2  12 2 rs
A   r 2   rs
s
s
Add the base area a to make the
total surface area
r
i.e. “the total area of a pyramid is the base plus
half the product of the base perimeter and the
slant height”
Note we can find the curved area of a right cone by noting how it can be constructed from a circle
with a sector missing

 2 s  2 r
Arc length
o
s
N
n
s
P
N
N
Ps
An  2 n
N
a
Therefore as N becomes infinite, the area
tend towards
The enlargement factor of each lamina is in direct
proportion to the slant distance from the apex of the
pyramid. This is true since the sides of the pyramid are
straight lines with a constant gradient.
An 
A  ab  s  a  b 
s
s
n
P
N
An  Pn 
A  ab  12 s  2a  2b 
Ps
Ps
Ps
AN  2 1  2 2  ....  2 N
N
N
N
Ps
AN  2 1  2  ....  N 
N
nN
perimeter
Area of a right rectangular based pyramid
360
r
A
s
s

360o

 s
2


360
o

r
s
r
 A    s2
s
 A   rs
s
Mathematics topic handout: Areas Dr Andrew French. www.eclecticon.info PAGE 2
Calculating the area of a sphere without using calculus
So we can conclude that the strip area, in the limit
of small h, does not depend on angle .
i.e. each strip on the hemisphere ‘maps’ (i.e. can be
projected onto) a cylindrical strip of (small) width h.
This means the curved surface of the hemisphere
must equal that of the enclosing cylinder.
B
x 

h
A
r
r

r
r cos
Hence the curved surface area of the hemisphere
must be:
2 r  r  2 r 2
Therefore the surface area of a sphere is:
A  4 r 2
Consider the tangent to a circular cross section of a hemisphere
at point A, which is at elevation angle  from the base of a hemisphere of radius r.
Point B is a distance away x along the tangent such that x cos  h , where h is the projected
width of the hemisphere on a cylinder of radius r. (The cylinder width between points at
the heights of positions A and B).
Now if we shrink h, then position B will move closer to the circle, and hence in this limit
we would expect the arc length between A and B to become:
x
h
cos
The area of a circular strip with width x between A and B on the hemisphere is approximately:
A  2 r cos  x
A  2 r cos
Take an orange.
Draw round it four times.
Peel the orange.
The peel should fit
almost exactly into
the four circles, since
the area of each circle is  r 2
h
 2 rh
cos
Mathematics topic handout: Areas Dr Andrew French. www.eclecticon.info PAGE 3
A frustum is a
truncated pyramid
Curved area of entire pyramid (assume NOT oblique)
Calculating the area of a sphere using calculus
Note all angles are in radians i.e.  radians = 180o.
Aentire  12 P(s  l )
Since cap is similar to entire pyramid,
Aremoved 
curved of removed ‘cap’ pyramid is:
1
2
rd
r
pl
Divide up the surface of a hemisphere
into circular ribbons
The radius of each ribbon is r sin 
and the width is the arc length rd
l
l
Therefore frustum curved area is A  Aenitre  Aremoved
A  P  s  l   pl
1
2
1
2
a
r
r
s
a
P
s
r
1

2
A  2  2 r 2  sin  d
0
A  4 r   cos  0
A  P  s  l   pl
2
1
2
1

2
A  4 r 2   cos 12      cos(0) 
A  12 Ps  12 l  P  p 
A  4 r 2 0   1
ps
P  p
P p
A  12 s ( P  p )
A  12 Ps  12
Therefore total surface area of
the frustum is:
dA  2  r sin   rd  2 r 2 sin  d

A  2   dA
Hence:
1
2
The area of each ribbon is
The total area of the sphere is therefore
Now since the top and base of the frustum are similar
P sl

p
l
P s
 1
p l
P
s
1 
p
l
P p s

p
l
ps
l
P p
p
r sin 
A  4 r 2
p
a
s
A  a  a  12 s  p  P 
a
This is only true for non
oblique frustums!
P
Mathematics topic handout: Areas Dr Andrew French. www.eclecticon.info PAGE 4
Archimedes’ method for calculating the area of a sphere
A volume of revolution of the octagon is the volume of two cones
formed by the rotation of triangle ABH and two frustums formed
by the rotation of trapezium BHGC
Firstly this involves finding a formula for the surface area of a
volume of revolution of a regular octagon about the x axis. Generalizing
for regular even-number sided polygons of increasing large numbers
of sides, we anticipate the resulting surface area to be that of a sphere.
The total surface area is therefore:
A  2 as  2  12 s  2 a  2 b 
C

B
s 
a
A

a
D
s
b
a


Noting the parallel lines
connecting vertices AB
and HC, and by application of
the ‘Z-angle’ theorem, one can
show that the green right angled
triangles highlighted are similar
s
s

E
x
Hence:
a
s
A  2 k  2a   
A   k   4a  2 
a
A   k  AE
Now we would expect the same result for any even polygon, but as the
number of sides increase, we expect k to approach AE as s shrinks
s
For a circle AE = 2r. Therefore if in this scenario k = AE
G
Since we can circumscribe the
octagon with a circle, we can use
the ‘right angle diameter theorem’
to show the orange circle is also
similar to the green triangles.
s
s
B
D

k
s 
A   2r  2r
A  4 r 2
s
a
Define k = BE

a
3a  2
E
a
s
Using the results above
a
F
s



A  2 s  2a  b 
s

H
A
b
A  2 as  2 s  a  b 
s

H
F
s
s
x
k b
 b
s 
k a
   a a
s a

k
s
k
s
100 sided polygon of
revolution – this
looks very much like
a sphere!
20 sided
polygon of
revolution
G
Mathematics topic handout: Areas Dr Andrew French. www.eclecticon.info PAGE 5