Calculus: Analytic Geometry and Calculus, with Vectors

13.4 Applications of double and iterated integrals^685

< a, and we can beinterested in situations in which e is small. Set up an integral
for the gravitationalforce which the rod on the right exerts upon the rod on the
left.
17 Figure 13.493 shows rods of lengths a and b that have constantlinear
density S, that lie on the x and y axes, and that arehinged at their ends at the

(0, b)

(0,y)

(x,0) (a,0) x

Figure 13.493

origin. A little segment of the horizontal rodin a neighborhood of (x,0) pro-

duces a little gravitational force u on alittle segment of the vertical rod in a

neighborhood of (O,y). This little force u has a little scalar horizontal compo-

nent ux which produces alittle torque (or first moment) yu= which tends to rotate

the vertical rod toward the horizontal one. There are hordes of little torques.

Set up an integral for the total torque.

18 Most people having serious interestin mathematics want to see and

perhaps study the nontrivial steps bywhich the important Euler gammaintegral

formula

(1) Z! = foo t=ea dt

(z > -1)

is derived from the definition of z!given in Problem 11 of Problems3.39. We

start with the fact that,when z is not a negative integer,z! is defined by the

formulas / nine

(2) z! = lim Fn(z), Fn\z)

_

• (z + 1) (z + 2) ... (z + n)

Expressing ((z + 1)(z + 2).. (z + n))-i as a sum of partial

fractions leads

to the formula

FF(z) = n=+i (-1)k (n k1)j_+ -k+

1.

k=0

To put this in a form that canbe simplified by use of the

binomial formula, we

use the fact that

(3)

I fofu=+kdu=J0 u=ukdu

when z > -1 and k = 0, 1, 2,. ...

Assuming henceforth that z> -1,

we find that

(4) FF(z) = n:+1 r 1 0 u=k=0(n k1) ln-k(-u)k du

and hence that

(5) FF(z) = ns+l f 01 us(1 -

n)n-1 du.