Calculus: Analytic Geometry and Calculus, with Vectors

13.3 Double integrals 673

Be sure to realize that this is correct because, for each fixedy, Q(x,y) is a function

of x whose derivative with respect to x is the integrand Qx(x,y). We must now

look at our figure. As y increases fromc to d, the point with coordinates ($2(y), y)

traverses, in the positive direction, the part Cl of C which lies to the right of S.

It is a consequence of the definition of curve integrals given in Problem15 of

Section 7.2 that

(5) 1d Q(g2(Y), Y) dY =

fc.Q(x,Y) dy,

the parameter now being y instead oft. The last term in (4) is more troublesome.

As y increases from c to d, the point with coordinates (g,(y), y)traverses, in the

negative direction, the part C2 of C which lies to the left ofS. This reversal of

direction introduces a change in sign so that

(6) -fed Q(gl(y),y) dy = f Q(x,y) dy.

Show that combining these results gives the formula

(7) ff, ax dS =fc Q dy.

Tell why

b s fs(x)

(8) f = fa dx fl(M) ) Pv(x,Y) dy = fab dx [P(x,Y)]

v°fl(x)

and

=L6P(x, f2(x)) dx- fbP(x, f, (x)) dx

• ffsay dS =fcP dy.

Note that combining (7) and (8) gives the formula

(10) ffs (aQ - ay) dS = fC (P dx Q dy).

This formula appeared in the works of George Green (1793-1841), a pioneer in

applied mathematics who originated the term potential function, and the formula

is called a (or, sometimes, the), Green formula.

The Green formula and its extensions have very important applications, some

of which involve vectors. To be broad-minded about the matter, let

(11) V(x,y,z) = P(x,y,z)i + Q(x,y,z)j + R(x,y,z)k,

where V is a vector function having scalar components P, Q, R that are con-
tinuous and have continuous partial derivatives over a part of E3 in which our
sets and curves are supposed to lie. The curl of the vector function Y is a vector
function which is written V X V, which is read "the curl of V" or "del cross V"
and is defined by the formula

(12) V x V =

I j k

a a a

axay az

P Q R