Calculus: Analytic Geometry and Calculus, with Vectors

10.3 Areas and integrals involving polar coordinates 547

is called the witch of Agnesi because Maria Gaetena Agnesi (1718-1799) discovered
a spooky ruler-and-compass method for constructing points on it. Let C be
the circle of diameter 1 with center at (0, 4). Let The a point on the line tangent
to the circle at the point (0, 1), and let Q be the point, different from the origin 0,
at which the line OT intersects the circle. Show that the vertical line through T
and the horizontal line through Q intersect at a point on the witch.

10.3 Areas and integrals involving polar coordinates Problems

involving "areas in polar coordinates" can be formulated in different

ways. To begin, we suppose that f is a nonnegative integrable function

of 0 over some interval a 5 0 < 0 for which 0 < a + 27r. Let S be

the set of points having polar coordinates p, 0 for which a < 0 S 0

and 0 < p =< f(c). In case f is continuous, S may be described as the

set S bounded by the polar graphs of the equations 0 = a, 0 = 0, and

Figure 10.31

p = f(O). The schematic Figure 10.31 should not be misleading. To

find the area ISI of S, we make a partition of the interval a < 0 < 0

into subintervals of which a representative one of length Dq5k contains

a particular 0k. When there is a necessity for being precise about this

matter, we set OqSk = Ok - 4k-i and choose qsk such that 4k-i < 0k

Ok. The area of the subset of S containing points for which 4,4-i 5 0 5

Ok can now be approximated by the area -[f(.0*)]2 ilk of the sector of

radius f(4k) having central angle Acyk. When we are hurried, we need

not make the usual remarks about the way in which the approximation

depends upon the choice of 0k, but in any case we should know why we

are using the area of the sector instead of its perimeter. An application

of fundamental ideas about estimating, summing, and taking limits to

set up integrals then gives

(10.32) ISj = lim I.[f(4k )]2 Q4)k= J 1 s [f(0)12 d4,,

where the integral is a Riemann integral. When we are not required

to write detailed explanations of the reasons for doing what we do, we

replace (10.32) by the simpler calculation

(10.33) ISI = lim I ..[f(0)J2 0(A = 1 f ' [f(4,)]2 do