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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
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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9- 15-8

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: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,
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.
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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