Geometry: An Interactive Journey to Mastery

The Volume and Surface Area of a Sphere

Appendix to Lesson 23

*UHHNVFKRODU$UFKLPHGHVRI6\UDFXVHͼ௘FD±%&(௘ͽSURYHGWKDWWKHYROXPHRIDKHPLVSKHUHRIUDGLXVr is
given by the formula 32 Sr^3 ͼ௘DQGWKXVWKHYROXPHRIDIXOOVSKHUHLV^43 Sr^3 ௘ͽ
He did this by enclosing the hemisphere in a circular cylinder of radius r and height r and comparing the space
EHWZHHQWKHWZRZLWKWKHYROXPHRIDFLUFXODUFRQHHQFORVHGLQWKHVDPHF\OLQGHUͼ௘6HH)LJXUH௘ͽ

More precisely, Archimedes compared cross-sectional slices of this space with cross-sectional slices at matching
heights x.
On the right, the cross section is a circle. By noting that we have isosceles right triangles, we see that the radius
of this circle is x, so this cross section has area ʌ[^2.
On the left, the cross section is a ring made of a large circle of radius r and a small circle of radius rx^22 .
Its area is SSrrxx^2222  2 S.
Because these areas are the same, Archimedes argued, like Cavalieri, that the volume outside the sphere equals
the volume inside the cone. Now,
23
23

cylinder.

outside hemisphere cone^1133.

Vrrr

VVrrr

SS

SS

˜

˜

Thus,

33

3

hemisphere cylinder outside hemisphere

(^13)
(^23).

VVV

rr

r

SS

S





The volume of the entire sphere is indeed^43 Sr^3.

r

x r

r

x x r

Figure 23.20