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Part 3 of 6 Note that N ds = VG(x, y, z) dA, and recall that VG(x, y, z)= = 2 2 (curl F). N dS = (2yi + (y-2x)j - zk). xy 49 - x² xy 49-² xy xy -SA( (curl F). N ds - z dA Part 4 of 6 Recall that according to Stokes's Theorem, Ic F. dr = (curl F) - N ds. Using the above value for (curl F) N dS, we, therefore, have [ F. dr - Sl = 2xy 49X² 49 49x² 49 - .x² dA 49 - ·(√²²=²++) ₂ dA 49 - 49 xy + k and z = % dA. 49x2. Therefore, x2 dydx.

Part 3 of 6 Note that N ds = VG(x, y, z) dA, and recall that VG(x, y, z)= = 2 2 (curl F). N dS = (2yi + (y-2x)j - zk). xy 49 - x² xy 49-² xy xy -SA( (curl F). N ds - z dA Part 4 of 6 Recall that according to Stokes's Theorem, Ic F. dr = (curl F) - N ds. Using the above value for (curl F) N dS, we, therefore, have [ F. dr - Sl = 2xy 49X² 49 49x² 49 - .x² dA 49 - ·(√²²=²++) ₂ dA 49 - 49 xy + k and z = % dA. 49x2. Therefore, x2 dydx.

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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Part 3 of 6 Note that N dS = VG(x, y, z) dA, and recall that VG(x, y, z)= 2 2 IF F. dr (curl F). N dS = (2yi + (y - 2x)j – zk). F. dr = xy xy 49 49 x² Part 4 of 6 Recall that according to Stokes's Theorem, ¹ r = 16₂₁ (curl F) . N ds. Using the above value for (curl F). N ds, we, therefore, have I.F. - 16 = -6² - 16(5 xy xy (curl F) . N dS T 2xy 49 - x² - z dA 49 49 - 49-² dA 49 xy X 49 X 49 - i+k and z= √√49 - x². Therefore, x² x² i + k dA dA. x² dydx.
Transcribed Image Text:Part 3 of 6 Note that N dS = VG(x, y, z) dA, and recall that VG(x, y, z)= 2 2 IF F. dr (curl F). N dS = (2yi + (y - 2x)j – zk). F. dr = xy xy 49 49 x² Part 4 of 6 Recall that according to Stokes's Theorem, ¹ r = 16₂₁ (curl F) . N ds. Using the above value for (curl F). N ds, we, therefore, have I.F. - 16 = -6² - 16(5 xy xy (curl F) . N dS T 2xy 49 - x² - z dA 49 49 - 49-² dA 49 xy X 49 X 49 - i+k and z= √√49 - x². Therefore, x² x² i + k dA dA. x² dydx.
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