Calculus: Analytic Geometry and Calculus, with Vectors

7.2 Lengths and integrals 425

11 Let units of measurement be so employed that the potential at Po(xo,yo,zo)
due to a particle of mass m concentrated at a point P(x,y,z) is m/IPoF. Let
x, y, z have continuousderivatives over a < u 5 b, and let P(x,y,z) traverse a
curve C as u increasesfrom a to b. Let C have linear density 3. Let Po(xo,yo,zo)
be a point not on C. Remembering that potentials are scalars, and remembering
or learning quicklythat the potential due to a collection of particles is equal to the
sum of the potentials due to theindividual particles, set up an integral which is
equal to the potential at Po due to C.
12 Set up an integral equal to the moment of inertia (second moment)
about the z axis of the curve C of the preceding problem.
13 When a particle has fallen (or slid without friction) from height G(b) to
a lower height G(u),its speed is 2g[G(b) - G(u)], where g is the acceleration of
gravity. We all know that if we travel a short distance with a constant speed,
we get the time required for the trip by dividing the distance by the speed.
With this basic information, we are prepared to attack a problem. Let F and
G be functions having positive continuous derivatives over 0 =< u 5 b and let
the point P(F(u), G(u)) traverse a curve C from the origin to a point B in the
first quadrant of an xy plane as u increases from 0 to b. Sketch a schematic
figure showing the curve C. Set up an integral which gives the time T required
for a particle starting from rest at B to slide without friction down the curve C
to the origin.
14 The reflecting surface of a headlight is a part of a paraboloid, of depth
4 inches and diameter 12 inches, obtained by rotating a part of a parabola about
its axis. Find its area. Solution: This is not an easy problem, since a basic
difficulty lurks in the fact that areas of such surfaces have not been defined.
We start by so determining k that the parabola having the equation y = kx2
contains (or passes through) the point
(6,4). This gives y = $x2 for the equa-
tion of the generating parabola. With ---------------
the possibility of setting a = 0, b = 6,

y

and f(x) = gx2, we suppose that f has I &xk

a continuous derivative over an interval

a < x < b. Let S be the surface gen-

erated by rotating, about the y axis, the

part of the graph G of y = f (x) for which

a 5 x < b. Expecting to use some intui-

tive ideas about areas, we make a parti-

tion P of the interval a < x :5 b and

xk x

Figure 7.293

consider one particular subinterval xk_1 < x < X. As an approximation to

the length Lk of the segment Gk of the graph G containing points x for which

xk_1 S x 5 xk, we may use the number

2

(1) V xk +

ay!

=^1

+ 'Nxk

= 1 'f U (xk )12 k,

where Axk = xk- xk-1, Ayk = f(xk) - f(xk-1), and xk is a number for which

xk_1 < xk < xk. Figure 7.293 is helpful. Even though areas are not yet defined.

we can have a feeling that when the segment Ck is rotated about the y axis, its