Calculus: Analytic Geometry and Calculus, with Vectors

552 Polar, cylindrical, and spherical coordinates

Q1PQ2 is a point of S. No interior point Q of the shaded sector of Figure 10.392

can be a point of S. Otherwise, the point P, being an inner point of the inside

of the triangle Q1QQ2, would be a point of S and so would also each point of some

Figure 10.392

circular disk having its center at P. There would then be a number po greater

than f (B) for which S contains each point having polar coordinates (p,o) for which

0 5 p 5 po, and this is impossible. We now know that each point inside the

circle with center at the origin and radius S is a point of S, that each point inside

the triangle Q1PQ2 is a point of S, and that each point Q interior to the shaded

sector is not a point of S. It follows from this that f (¢) is the one and only

positive number such that the point P having polar coordinates (f (q5), 0) is a

point on the boundary B of S Remark: Without going into details we observe

that this proof provides supplementary information that enables us to relate

f(O + h) to f(O) It can be shown that f is continuous and hence that the

boundary B becomes a simple closed curve C when its points are so ordered that

the point having polar coordinates (f(01), 01) precedes the point having polar

coordinates (f(¢2), 02) when 0 =< (P1 < 02 5 2a. The objection that curves

were defined in terms of rectangular coordinates is overcome by the observation

that if f is continuous, then so also are the functions g1 and g2 defined by

x=g1(0)=f(4)cos¢

y = g2(4) = f(O) sin ¢.

It can be proved that the curve C has finite length L.