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...

How they get the boundaries!?

Extracted 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