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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