Surface Integral Evaluate (3) when F = [x", 0, 3y21 and S is the portion of the plane x + y + z = 1 in the first octant (Fig. 246). Solution. Writing x =u and y = v, we have z =1-x-y =1-u-v. Hence we can represent the plane x +y+=1 in the form r(u, v) = [u, v, 1- u-v). We obtain the first-octant portion S of this plane by restricting x u and y = v to the projection R of S in the xy-plane. R is the triangle bounded by the two coordinate axes and the straight line x +y = 1, obtained from x+y + z =1 by setting z = 0. Thus 05XS1-y, 0 y 1. | %3D
Surface Integral Evaluate (3) when F = [x", 0, 3y21 and S is the portion of the plane x + y + z = 1 in the first octant (Fig. 246). Solution. Writing x =u and y = v, we have z =1-x-y =1-u-v. Hence we can represent the plane x +y+=1 in the form r(u, v) = [u, v, 1- u-v). We obtain the first-octant portion S of this plane by restricting x u and y = v to the projection R of S in the xy-plane. R is the triangle bounded by the two coordinate axes and the straight line x +y = 1, obtained from x+y + z =1 by setting z = 0. Thus 05XS1-y, 0 y 1. | %3D
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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Transcribed Image Text:Surface Integral
Evaluate (3) when F = [x", 0, 3y21 and S is the portion of the plane x + y + z = 1 in the first octant (Fig. 246).
Solution. Writing x =u and y = v, we have z =1-x-y =1-u-v. Hence we can represent the
plane x +y+=1 in the form r(u, v) = [u, v, 1- u-v). We obtain the first-octant portion S of this plane
by restricting x u and y = v to the projection R of S in the xy-plane. R is the triangle bounded by the two
coordinate axes and the straight line x +y = 1, obtained from x+y + z =1 by setting z = 0. Thus
05XS1-y, 0 y 1.
|
%3D
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