Calculus: Analytic Geometry and Calculus, with Vectors

13.2 Iterated integrals and volumes 663

without benefit of elegant figures, we
can use Figure 13.27 to help us see
what we are doing.
For each k, the number

(13.271) foe f(x,yk) dx

is the area of the intersection of the
plane y = yk and the body H. De-

pending upon the choice of yk, the
number

(13.272) t yk fo f(x)yk) dx

is exactly equal to or is an approxima-

Figure 13.27

Lion to the volume of the slab of H between the planesy = yk_1 and
y = yk. The sum in the formula

(13.273) Y =

Jim

k=1Il Ayk f 2 f(x,yk) dx

is then exactly equal to the volume V of H or is an approximation to Y,

and in any case the limit as IPI -4 0 is Y. Thus

(13.274) Y= J = fo1 dyfo2 f(x,1') dx.

For the special case being considered, equality of the last two members

of the formula

(13.275) Y = fox dx folf(x,y) dy= fol dyf 2f(x,y) dx

is a consequence of the fact that each member is equal to the volume Y

of a solid body H.

Some special applications of these ideas and formulas are particularly

worthy of notice. In case R = C = 0 and B > 0, the number V is the

volume of a body which rests upon the base S2 and (13.275) reduces to

the formula

(13.276) V =

J2

dx f : f (x,y) dy = fol dy f2y'f (x,y) dx.

In case 11 = B = C = 1, the solid H has unit height over the whole

rectangular set R and the volume of H must be the same as the area (R1

of R. We could therefore be sure that there would be a mistake in our

work if it were not true that the numbers I and J in (13.225) and (13.26)

reduce to 2 when A = B = C = 1. In case A = C = 0 and B = 1, the

numbers Y, I, and j reduce to the area IS21 of the set S2.

Among other things, the above example leads us to the following idea.

Suppose we want to find the volume Y of a solid body K which rests upon