(^662) Iterated and multiple integrals
and we have shown that, in the special cases being considered,our
iterated integral is the volume of the block H.
Our work involving Figure 13.21 is only half finished because wemust
investigate the integral J defined by
(13.25) J = fol dyfa2f(x,y) dx
and must assimilate some new ideas. For each y in the interval 0 5
y < 1, the integrand in the integral
(13.251) foe f(x,y) dx
has values that depend upon x. As in Figure 13.21 we mark a point y
on the y axis between 0 and 1 and draw a line through this point parallel
to the x axis. A part of this line is in S3 where f(x,y) = C, another part
is in S2 where f(x,y) = B, a third part is in Si where f(x,y) = A4, and,
moreover, the end points of these parts depend upon the fixed y. We
must examine Figure 13.21 carefully enough to see that
f(x,y) = C when 0 < x < 2y,
f (x,y) = B when 2y < x < 2 -N/Y-1
f(x,y) = 1I when 2 < x S 2,
and that
(13.252)
so
(13.253)
fat f(x,y) dx = f7 C dx +J2yB dx+ f2 A dx,
f 2f(x,y) dx = 214(1 - V) +2B(-- y) + 2Cy.
Substituting this in (13.25) gives
J=2fol[4(1-/Y)+B(/3-y)+Cy]dy
and hence
(13.26) J=14+B+C.
Comparison of (13.225) and (13.26) shows that I and j are equal.
When 1I, B, C are nonnegative, I is the volume of the block H and it
follows that J must also be this volume. We need the experience gained
by using the slab method to prove that J, like I, is the volume of H so we
will have another and more informative proof that J = I. Let P be a
partition of the interval 0 < y < 1 into subintervals of lengths Ayl,
oy2, ... , Ay and let yk be a point in the kth subinterval such that
yx-1 < yk < yk. While we should acquire the ability to do our chores