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@F dx dy @y
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where r is the projection of S on the xy plane. By Green s theorem for the plane, the last integral equals F dx where is the boundary of r. Since at each point x; y of the value of F is the same as the value of A1 at each point x; y; z of C, and since dx is the same for both curves, we must have F dx
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LINE INTEGRALS, SURFACE INTEGRALS, AND INTEGRAL THEOREMS
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[CHAP. 10
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or r A1 i n dS
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Similarly, by projections on the other coordinate planes, r A2 j n dS A2 dy; r A3 k n dS A3 dz
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S C S C
Thus, by addition, r A n dS
A dr
The theorem is also valid for surfaces S which may not satisfy the restrictions imposed above. For assume that S can be subdivided into surfaces S1 ; S2 ; . . . ; Sk with boundaries C1 ; C2 ; . . . ; Ck which do satisfy the restrictions. Then Stokes theorem holds for each such surface. Adding these surface integrals, the total surface integral over S is obtained. Adding the corresponding line integrals over C1 ; C2 ; . . . ; Ck , the line integral over C is obtained.
10.27. Verify Stoke s theorem for A 3yi xzj yz2 k, where S is the surface of the paraboloid 2z x2 y2 bounded by z 2 and C is its boundary. See Fig. 10-21.
The boundary C of S is a circle with equations x2 y2 4; z 2 and parametric equations x 2 cos t; y 2 sin t; z 2, where 0 @ t < 2. Then A dr 3y dx xz dy yz2 dz
3 2 sin t 2 sin t dt 2 cos t 2 2 cos t dt 12 sin2 t 8 cos2 t dt 20       z2 x i z 3 k   yz2  k @ z r x2 y2 2z xi yj k p : jr x2 y2 2z j x2 y2 1 r A n dS
2
Also,
  i j   @ @  r A   @x @y   3y xz
Fig. 10-21
and Then
r A n
dx dy jn kj
xz2 x2 z 3 dx dy
8 9 !2 < = x2 y2 x2 y2 3 dx dy x x2 : ; 2 2
In polar coordinates this becomes 2 2 f  cos  4 =2 2 cos2  2 =2 3g  d d 20
 0  0
CHAP. 10]
LINE INTEGRALS, SURFACE INTEGRALS, AND INTEGRAL THEOREMS
10.28. Prove that a necessary and su cient condition that r A 0 identically.
Su ciency. Suppose r A 0.
A dr 0 for every closed curve C is that
Then by Stokes theorem A dr r A n dS 0
Necessity. Suppose
A dr 0 around every closed path C, and assume r A 6 0 at some point P.
Then
assuming r A is continuous, there will be a region with P as an interior point, where r A 6 0. Let S be a surface contained in this region whose normal n at each point has the same direction as r A, i.e., r A n where is a positive constant. Let C be the boundary of S. Then by Stokes theorem A dr r A n dS n n dS > 0
which contradicts the hypothesis that
A dr 0 and shows that r A 0.
It follows that r A 0 is also a necessary and su cient condition for a line integral independent of the path joining points P1 and P2 .
P2
A dr to be
10.29. Prove that a necessary and su cient condition that r A 0 is that A r.
Su ciency. If A r, then r A r r 0 by Problem 7.80, Chap. 7, Page 179. Necessity. If r A 0, then by Problem 10.28, A dr 0 around every closed path and A dr is independent of the path joining two points which we take as a; b; c and x; y; z . Let us de ne x;y;z x;y;z A dr A1 dx A2 dy A3 dz  x; y; z
a;b;c a;b;c C
Then x x;y;z  x x; y; z  x; y; z
x;y;z
A1 dx A2 dy A3 dz
Since the last integral is independent of the path joining x; y; z and x x; y; z , we can choose the path to be a straight line joining these points so that dy and dz are zero. Then  x x; y; z  x; y; z 1 x x;y;z A1 dx A1 x  x; y; z 0<<1 x x x;y;z where we have applied the law of the mean for integrals. Taking the limit of both sides as x ! 0 gives @=@x A1 . Similarly, we can show that @=@y A2 ; @=@z A3 : Thus, A A1 i A2 j A3 k @ @ @ i j k r: @x @y @z
10.30. (a) Prove that a necessary and su cient condition that A1 dx A2 dy A3 dz d, an exact di erential, is that r A 0 where A A1 i A2 j A3 k. (b) Show that in such case, x2 ;y2 ;z2 x2 ;y2 ;z2 A1 dx A2 dy A3 dz d  x2 ; y2 ; z2  x1 ; y1 ; z1
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