Page 262 - Schaum's Outline of Theory and Problems of Advanced Calculus

CHAP. 10] LINE INTEGRALS, SURFACE INTEGRALS, AND INTEGRAL THEOREMS 253

Fig. 10-20

@A 1 @A 1
n k dS
½r ðA 1 iފ n dS ¼ n j ð1Þ
@z @y
If z ¼ f ðx; yÞ is taken as the equation of S,then the position vector to any point of S is r ¼ xi þ yj þ zk ¼
@r @z @ f @r
xi þ yj þ f ðx; yÞk so that @y ¼ j þ @y k ¼ j þ @y k. But @y is a vector tangent to S and thus perpendicular to
n,sothat
@r @z @z
n k ¼ 0 or n k
@y @y @y
n ¼ n j þ n j ¼
Substitute in (1)to obtain

@A 1 @A 1 @A 1 @z @A 1
n k dS
@z n j @y n k dS ¼ @z @y n k @y
or

@A 1 @A 1 @z
n k dS
@y @z @y
½r ðA 1 iފ n dS ¼ þ ð2Þ
@A 1 @A 1 @z @F
Now on S, A 1 ðx; y; zÞ¼ A 1 ½x; y; f ðx; yފ ¼ Fðx; yÞ;hence, þ ¼ and (2)becomes
@y @z @y @y
@F @F
dx dy
@y @y
½r ðA 1 iފ n dS ¼ n k dS ¼
Then
@F
ðð ðð
dx dy
@y
½r ðA 1 iފ n dS ¼
S r
where r is the projection of S on the xy plane. By Green’s theorem for the plane, the last integral equals
þ
Fdx where is the boundary of r.Since at each point ðx; yÞ of the value of F is the same as the value

of A 1 at each point ðx; y; zÞ of C, and since dx is the same for both curves, we must have
þ þ
A 1 dx
Fdx ¼
C