Part 3 of 6 Note that N dS = VG(x, y, z) dA, and recall that VG(x, y, z) = (curl F). NdS = (2yi + (y − 2x)j – zk). || 2 x 2 X 49 - ² 49-² [-z) dA X 9-1² 49 49 - +² X 49 - x² + k dA dA. + k and z= 49 x
Part 3 of 6 Note that N dS = VG(x, y, z) dA, and recall that VG(x, y, z) = (curl F). NdS = (2yi + (y − 2x)j – zk). || 2 x 2 X 49 - ² 49-² [-z) dA X 9-1² 49 49 - +² X 49 - x² + k dA dA. + k and z= 49 x
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Question
9- 15-8

Transcribed Image Text:Part 3 of 6
Note that N ds = VG(x, y, z) dA, and recall that VG(x, y, z) =
(curl F). NdS = (2yi + (y − 2x)j – zk).
11
2 x
2 X
49 - ²
49
[-z) dA
X
Submit Skip (you cannot come back)
49
T
x
49 -
+²
X
49 - x²
+ k dA
dA.
+ k and z=
49x2. Therefore,

Transcribed Image Text:Use Stokes's Theorem to evaluate
F(x, y, z) = yzi + (2 − 6y)j + (x² + y²)k, x² + y² ≤ 49
S: the first-octant portion of x² + z² = 49 over x² + y² = 49
Part 1 of 6
The surface S is described by x² + z² = 49 over x² + y² = 49.
To evaluate a line integral of F over this surface, write S as z = g(x, y), where
g(x, y) = √√49x².
Since S is orientable, it can be described by the function G(x, y, z), where
G(x, y, z) = z
g(x, y) = z-V 49
Therefore,
VG(x, y, z) =
curl F =
ScF F. dr. Use a computer algebra system to verify your results. In each case, C is oriented counterclockwise as viewed from above.
=
ə
X
= i(||
G(x, y, z)i +G(x, y, z)j + G(x, y, z)k
дz
?х
Part 2 of 6
It is given that F(x, y, z) = yzi + (2 − 6y)j + (x² + y²)k. Therefore,
i
j
k
ə
Ə
əx ду
əz
yz 2-6y x² + y²
49
49 -².
X + k.
−(x² + y²) - 0₂ (2 − 6y)) - ³(2x (x² + y²) − 0₂y²) + k( 0 (2 − 6y) - 8yz)
y²)
əy
дz
i + (y-
- 0) - j(2x - y) + k(0 - z)
)j - zk.
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