Calculus: Analytic Geometry and Calculus, with Vectors

4.8 Moments and centroids in E2 and E3 275

dl < d < d2 and the magnitude of the force is p11, where p is wd, the pressure

at depth d. Our little lesson in hydrostatics is ended, and we can now formulate
our problem. We confine our attention to forces upon one side of a plane region R
which lies, as in Figure 4.891, beneath the sur-
face of a liquid and in a vertical plane. The x 1Top surface of liquid
axis is taken to be horizontal and in the top
surface of the liquid. The y axis is taken to a-------
be vertical with y positive measured down- A/yk-i
ward; this means that the point (x,y) lies y
below the x axis when y > 0. It is supposed
that the region R lies between the lines having bk-
the equations y = a and y = b and that, when
a < y < b, the line having the equation
y y* intersects R in an interval (or collec-

Figure 4.891

tion of intervals) having length (or total length) f(y*). It is assumed that the

region R and the function f are bounded. It is not necessary that f be con-

tinuous, but we do assume that R has area IRI and that IRI = fa

b

f(x) dx. Our

problem is to set up an integral for the magnitude of the force which the liquid

exerts upon one side of the region R. The procedure should now be completely

familiar. We make a partition of the interval a 5 y < b into subintervals and

choose yk such that yk-i <- y* < yk. The number

(1) f(Yk) QYk

is taken as an approximation to the area of the part of R that lies in a strip parallel

to the surface of the liquid. Multiplying this by wyk gives an approximation to

the magnitude of the force of the part of R. Since the forces on the parts of R

all have the same direction, the sum

(2) wZYaf(Yk) AYk

gives an approximation to the magnitude IFI of the force on the whole region R,

and the approximation should be good when the norm of the partition is small-

Thus it should be true that

(3) IFI = lim wEykf(Yk) DYk

Since the right member of (3) exists and is a Riemann integral, our work motivates

the definitions where IFI is defined by the formula

(4) IFI = w f ab Yf(y) dy.

In order to make numerical calculations, we must know or be able to compute

a, b, and f(y). The really interesting thing about our result is that it can be put

in the form

(5) IFI = wyA,

where 4 is the area of the region R and 9 is the depth of its centroid. Thus the

magnitude of the force on the plane region R is the product of the pressure at the een-

troid and the area of the region. Many problems can be solved very quickly by

use of this fact.