684 Iterated and multiple integrals
line x = xa. Ans.:
M(p)
xa
dx
yi+m2(x-xl)
x-xo ixi yi+m.a(x-xt)
X3 y,+m,(x-x,)
+ fSa dx yx+'ma(x-x2) (xx - x dy.
14 Develop computational skill by simplifying the answer to Problem 13
for the case in which S(x,y) = 1 and p = 0 so the answer is numerically equal to
the area of the lamina. Ans.:
[(x,y2 - x2Yi) - (xIy2 - xaYi) + (xzya - x3Y2)] or.
Yi 1
Y2 1
Y3 1
15 Figure 13.492 shows two parallel rods of lengths a and b. The rod on
the left has linear density (mass per unit length) 81(t) at distance t from its lower
(x - xo)PS(x,y) dy
xl
x2
x3
Y
b
a
pt
------------
u
01 _D X
Figure 13.492
end, and the rod on the right has linear density 62(u) at distance u from its lower
end. We undertake to learn about the total or resultant gravitational force F on
the left-hand rod that is produced by the right-hand rod. Assume that particles
of mass m I and m2 at points P, and P2 attract each other with a force of magnitude
Gm1m2/IP1P212, where G is a gravitational constant, and the actual force pulling
the particle at P, toward P2 is obtained by multiplying this magnitude by P,P2/
lP,P21, the unit vector which has its tail at Pl and points toward P2. Derive
the formula
51(t) At 62(U) Au
G
[D2 + (u - t)2] [Di + (u - t)J]
for the force which an element (or subset) of the rod on the right exerts upon an
element of the rod on the left. Then derive the formula
G51(t) At b a2(u)[Di + Cu- t)9]du
fo [D2 + (u - t)2]%
for the force which the whole rod on the right exerts upon the element of the rod
on the left. Then derive the formula
F = G f a Si(t) 'it fb 12(u)[Di + (U- t)j]du.
0 o [D2 + (u - t)2]%
16 Two identical slender rods having constant linear density S occupy the
intervals -a < x 5 -e and e <x S a of an x axis. It is supposed that 0 < e