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@A3 dz dy dx @z
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f2 x;y
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! @A3 dz dy dx z f1 x;y @z
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 f2  A3 x; y; z  dy dx A3 x; y; f2 A3 x; y; f1 dy dx 
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For the upper portion S2 , dy dx cos
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2 dS2 k n2 dS2 since the normal n2 to S2 makes an acute angle
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2 with k. For the lower portion S1 , dy dx cos
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1 dS1 k n1 dS1 since the normal n1 to S1 makes an obtuse angle
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1 with k. Then
A3 x; y; f2 dy dx
A3 k n2 dS2 A3 k n1 dS1
A3 x; y; f1 dy dx
and A3 x; y; f2 dy dx
A3 x; y; f1 dy dx
A3 k n2 dS2
S S2
A3 k n1 dS1
A3 k n dS
LINE INTEGRALS, SURFACE INTEGRALS, AND INTEGRAL THEOREMS
[CHAP. 10
so that 1
@A3 dV @z
A3 k n dS
Similarly, by projecting S on the other coordinate planes, @A1 dV A1 i n dS 2 @x V S @A2 A2 j n dS dV 3 @y
Adding (1), (2), and (3), 
 @A1 @A2 @A3 A1 i A2 j A3 k n dS dV @x @y @z S r A dV A n dS
The theorem can be extended to surfaces which are such that lines parallel to the coordinate axes meet them in more than two points. To establish this extension, subdivide the region bounded by S into subregions whose surfaces do satisfy this condition. The procedure is analogous to that used in Green s theorem for the plane.
10.23. Verify the divergence theorem for A 2x z i x2 yj xz2 k taken over the region bounded by x 0; x 1; y 0; y 1; z 0; z 1.
We rst evaluate
A n dS where S is the surface of the cube in Fig. 10-19.
Face DEFG: n i; x 1. Then 1 1 A n dS f 2 z i j z2 kg i dy dz
DEFG 0 0
1 1
2 z dy dz 3=2
Face ABCO: n i; x 0. Then 1 1 A n dS zi i dy dz
ABCO 0 0
1 1
z dy dz 1=2 Fig. 10-19
Face ABEF: n j; y 1. Then 1 1 1 1 A n dS f 2x z i x2 j xz2 kg j dx dz x2 dx dz 1=3
ABEF 0 0 0 0
Face OGDC:
n j; y 0. Then 1 1 A n dS f 2x z i xz2 kg j dx dz 0
OGDC 0 0
CHAP. 10]
LINE INTEGRALS, SURFACE INTEGRALS, AND INTEGRAL THEOREMS
Face BCDE:
n k; z 1. Then 1 1 1 1 A n dS f 2x 1 i x2 yj xkg k dx dy x dx dy 1=2
BCDE 0 0 0 0
Face AFGO:
n k; z 0.
Then 1 1 A n dS f2xi x2 yjg k dx dy 0
AFGO 0 0
Adding,
A n dS 3 1 1 0 1 0 11 : 2 2 3 2 6 r A dV
Since
1 1 1
0 0 0
2 x2 2xz dx dy dz
11 6
the divergence theorem is veri ed in this case.
10.24. Evaluate
r n dS, where S is a closed surface.
By the divergence theorem, r n dS r r dV
 @ @ @ i j k xi yj zk dV @x @y @z V   @x @y @z dV 3 dV 3V @x @y @z

where V is the volume enclosed by S.
10.25. Evaluate xz2 dy dz x2 y z3 dz dx 2xy y2 z dx dy, where S is the entire surface of the p S hemispherical region bounded by z a2 x2 y2 and z 0 (a) by the divergence theorem (Green s theorem in space), (b) directly.
(a) Since dy dz dS cos ; dz dx dS cos ; dx dy dS cos
, the integral can be written fxz2 cos x2 y z3 cos 2xy y2 z cos
g dS A n dS
S 2 2 3 2 S
where A xz i x y z j 2xy y z k and n cos i cos j cos
k, the outward drawn unit normal. Then by the divergence theorem the integral equals ' & @ @ @ r A dV x2 y2 z2 dV xz2 x2 y z3 2xy y2 z dV @x @y @z
V V V
where V is the region bounded by the hemisphere and the xy plane. By use of spherical coordinates, as in Problem 9.19, 9, this integral is equal to =2 =2 2a5 r2 r2 sin  dr d d 4 5  0  0 r 0
LINE INTEGRALS, SURFACE INTEGRALS, AND INTEGRAL THEOREMS
[CHAP. 10
(b) If S1 is the convex surface of the hemispherical region and S2 is the base z 0 , then pa2 y2 q pa2 y2 a a q 2 2 2 2 2 xz dy dz z a y z dz dy z2 a2 y2 z2 dz dy x2 y z3 dz dx
S1 S1 y a z 0 y a z 0
pa2 x2
p fx2 a2 x2 z2 z3 g dz dx p f x2 a2 x2 z2 z3 g dz dx q y2 a2 x2 y2 g dy dx
x a x 0
2xy y2 z dx dy 1 xz2 dy dz 0; 2xy y2 z dx dy
S2 S2 S
a a
pa2 x2
p f2xy x a y a2 x2
x a z 0 pa2 x2
x2 y z3 dz dx 0;
f2xy y2 0 g dx dy
x a
pa2 x2
p y a2 x2
2xy dy dx 0
By addition of the above, we obtain p a a a2 y2 q z2 a2 y2 z2 dz dy 4 4
y 0 x 0
pa2 x2
p x2 a2 x2 z2 dz dx q y2 a2 x2 y2 dy dx
x 0 z 0 pa2 x2 x 0 y 0
Since by symmetry all these integrals are equal, the result is, on using polar coordinates, a pa2 x2 q =2 a p 2a5 12 y2 a2 x2 y2 dy dx 12 2 sin2  a2 2  d d 5 x 0 y 0  0  0
STOKES THEOREM 10.26. Prove Stokes theorem.
Let S be a surface which is such that its projections on the xy, yz, and xz planes are regions bounded by simple closed curves, as indicated in Fig. 10-20. Assume S to have representation z f x; y or x g y; z or y h x; z , where f ; g; h are single-valued, continuous, and di erentiable functions. We must show that r A n dS r A1 i A2 j A3 k n dS
A dr
where C is the boundary of S. Consider first r A1 i n dS:
  i   @ Since r A1 i   @x  A 1
 j k  @ @  @A1 @A  j 1 k; @y @z  @z @y  0 0
CHAP. 10]
LINE INTEGRALS, SURFACE INTEGRALS, AND INTEGRAL THEOREMS
Fig. 10-20  r A1 i n dS  @A1 @A n j 1 n k dS @z @y
If z f x; y is taken as the equation of S, then the position vector to any point of S is r xi yj zk @r @z @f @r xi yj f x; y k so that j k j k. But is a vector tangent to S and thus perpendicular to @y @y @y @y n, so that n @r @z n j n k 0 @y @y or n j @z n k @y
Substitute in (1) to obtain     @A1 @A @A @z @A n j 1 n k dS 1 n k 1 n k dS @z @y @z @y @y or  r A1 i n dS  @A1 @A1 @z n k dS @y @z @y @A1 @A1 @z @F and (2) becomes @y @z @y @y 2
Now on S, A1 x; y; z A1 x; y; f x; y F x; y ; hence, r A1 i n dS Then r A1 i n dS
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