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.
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