12 1. Various Ways of Representing Surfaces and Examples

To obtain the surface of revolution, we replace x with the distance

from the z-axis by making the substitution x → x2 + y2, and obtain

T2

= (x, y, z) ∈

R3

( x2 + y2 −

2)2

+

z2

− 1 = 0 .

At first glance, then, setting F (x, y, z) = ( x2 + y2 −

2)2

+

z2

− 1

gives our desired solution. However, this suffers from the defect that

F is not differentiable along the z-axis; we can overcome this fairly

easily with a little algebra. Expanding the equation, isolating the

square root, and squaring both sides, we obtain

x2

+

y2

+ 4 − 4 x2 + y2 +

z2

− 1 = 0,

x2

+

y2

+

z2

+ 3 = 4 x2 + y2,

(x2

+

y2

+

z2

+

3)2

=

16(x2

+

y2),

and hence consider the function F defined by

(1.5) F (x, y, z) =

(x2

+

y2

+

z2

+

3)2

−

16(x2

+

y2).

It is easy to check that the new choice of F from (1.5) does not

introduce any extraneous points to the solution set, and now F is

differentiable on all of

R3.

Exercise 1.5. Prove that a sphere with m ≥ 2 handles cannot be

represented as a surface of revolution.

Due to the result in Exercise 1.5, this argument cannot be applied

directly to find an equation whose set of solutions look like a sphere

with m ≥ 2 handles, but we can reverse engineer the result to find

a general method. Instead of beginning with a vertical plane, we

consider the intersection of the torus and the horizontal xy-plane,

which is given by two concentric circles. F (x, y, 0) is negative between

the circles, hence F (x, y, z) = F (x, y, 0) +

z2

= 0 has two solutions

for those values of x and y, leading to the torus shape. By beginning

with three or more circles (no longer concentric) we may use this idea

to represent a sphere with any number of handles.

Exercise 1.6. Represent a sphere with two handles as the set of

solutions of the equation F (x, y, z) = 0, where F is a differentiable

function, and none of its critical points satisfy this equation.