Calculus: Analytic Geometry and Calculus, with Vectors

674 Iterated and multiple integrals

in which the right member is a determinant. The expanded form of (12) is

(13) VXV=(ay-a.)i+(8 -ax)j+(a-aylk.

For the special case to which (10) applies, S is a patch of surface in the xy plane

and k is the unit normal to S which lies in the direction of the thumb on a right

hand when the fingers point in the direction in which C is oriented. If we denote

this unit normal by n, then in the special case (13) reduces to

(14) (V TX ay

Therefore, the left member of (10) is, in the special case, the left member of the

formula

(15) f fs(V frV.dr.

Our next step is to show that the right of (10) is, in the special case, the right

member of (15). This is quite easy. Let functions x(t), y(t), z(t) be such that

the point P(t) having coordinates x(t), y(t), z(t) traverses C once in the positive

direction as t increases from tl to t2 and let

(16) r(t) = x(t)i + y(t)j + z(1)k.

For the case in which these functions have piecewise continuous derivatives, dif-

ferentiation gives

(17) r'(t) = x'(t)i + y'(t)j + z'(t)k.

Then

(18) Px'(t) + Qy'(t) + Rz'(t),

where V stands for V(x(t), y(t), z(t)), P stands for P(x(t), y(t), z(t)), and so on.

Hence

(19) Jti=V.r'(t) dt= f=i' [Px'(t) + Qy'(t) + Rz'(t)] dt

and these integrals are, bydefinitionsof curve integrals, respectively equal to

those in the formula

(20) fC V.dr= fC (P dx + Q dy + R dz).

In the special cases where C lies in the xy plane, we have z(t) = 0 for each t, so

the right member of (10) is equal to the right members of (19) and (20) and hence

is equal to the left member of (20), which is the right member of (15).

The formula (15), that is,

(21) f fs (VX dS =

which reduces to the Green formula (10) when S and C lie in the xy plane, is

known as the Stokes formula. So far, (21) has been proved only when S is a

set in the xy plane which satisfies the heavy restriction given at the beginning