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13.5 Curl and Divergence
Copyright © Cengage Learning. All rights reserved.
: closed surface
13.5 Curl and Divergence
Copyright © Cengage Learning. All rights reserved.
Vector Forms of Green’s Theorem
Its line integral is
and, regarding F as a vector field on ¡ 3 with third component 0, we have
F = P i + Q j + 0 k.
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Vector Forms of Green’s Theorem
Then, from equation 16.3 (3)
we have
F = P i + Q j + 0 k.
N
M
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The Divergence Theorem
We write Green’s Theorem in a vector version as
where C is the positively oriented boundary curve of the plane region D.
If we were seeking to extend this theorem to vector fields on
we might
make the guess that
where S is the boundary surface of the solid region E.
It turns out that Equation 1 is true, under appropriate hypotheses, and is
called the Divergence Theorem.
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13.9 The Divergence Theorem
Copyright © Cengage Learning. All rights reserved.
The Divergence Theorem
The boundary of E is a closed surface, and we use the convention(?), that the
positive orientation is outward; that is, the unit normal vector n is directed
outward from E.
Thus the Divergence Theorem states that, under the given conditions, the flux
of F across the boundary surface of E is equal to the triple integral of the
divergence of F over E.
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Example 4 (in 13.7)
Find the flux of the vector field F(x, y, z) = z i + y j + x k over the unit sphere x 2 +
y 2 + z 2 =1.
Solution:
We use the parametric representation
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Example 1(We use the Divergence Theorem)
Find the flux of the vector field F(x, y, z) = z i + y j + x k over the unit sphere x 2 +
y 2 + z 2 =1.
Solution:
First we compute the divergence of F:
The unit sphere S is the boundary of the unit ball B given by
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Example 2
Solution:
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Example 2
Solution:
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The Divergence Theorem
Let’s consider the region E that lies between the closed surfaces S1 and S2, where
S1 lies inside S2.
Let n1 and n2 be outward normals of S1 and S2.
Then the boundary surface of E is S = S1 ∪ S2 and its normal n is given by n = −n1
on S1 and n = n2 on S2. (See Figure 3.)
Applying the Divergence Theorem to S, we get
Figure 3
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#2
with upward orientation.
Solution:
( )
(-)
with downward orientation.
D
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# 37
Solution:
?
= 20
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Final exam
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Type I
=y
=y
Solution:
Type II
Solution: The region R can be described as
R = {(x, y) | x, y  0, 1  x 2 + y 2  4}
It is the half-ring shown in Fig.,
and in polar coordinates it is given
by 1  r  2, 0     /4.
Type_1
위
아래
Figure 9
Type 1; 위/아래
#. Evaluate the integral by changing to spherical coordinates.
#.
25.
Solution:
R
#.
Solution:
Example 3
Solution:
Using Equation 2, we have
f (1,1,2) - f (0,0,0) = 8
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Recall (Green’s Theorem)
C:+
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Example 1
Solution:
n = <0, 1, 1> / 2
curl F ≠ 0
:+
choice!
r (x, y) = < x, y, g(x, y) > = < x, y, 2 - y >
Figure 3
rx × ry =
= < 0, 1, 1 >
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Example 1
Solution:
curl F ≠ 0
:+
choice!
Figure 3
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①
Example 2 (의도/의미?)
Solution:
뿔룩한 곡면
: + : counterclockwise
(+ orientation)
②
Note that in Example 2 we computed a surface integral simply by knowing the values of F on the
boundary curve C. This means that if we have another oriented surface with the same boundary
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curve C, then we get exactly the same value for the surface integral!
②
Example 2 (의도/의미?)
Solution:
: + : counterclockwise
(+ orientation)
n = < 0, 0, 1>
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Example 3
: counterclockwise
(+ orientation)
Solution:
n(=k)= < 0, 0, 1>
-n
:+
< 0, 0, -1>
(-1)
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Recall (Surface Integrals, Flux)
③
+
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13.7. #31
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