Page 414 - Advanced engineering mathematics

394 CHAPTER 12 Vector Integral Calculus

Then
2 4 2
N(u,v) =u [sin(v)cos(2v) − cos(v)sin(2v)]
4
2
+ u [sin(v)sin(2v) + cos(v)cos(2v)] + u 2
2
2
=u (1 + u ),
so
√
N(u,v) = u 1 + u .
2
The surface integral is
1 2
√
2
xyzdσ = [u cos(v)][u sin(v)] u sin(2v) u 1 + u dA
D 2
π 2 √
2
= cos(v)sin(v)sin(2v)dv u 5 1 + u du
0 1
π 100 √ 11 √
= 21 − 2 .
4 21 105
EXAMPLE 12.21

We will evaluate zdσ over the part of the plane x + y + z = 4 lying above the rectangle

D : 0 ≤ x ≤ 2,0 ≤ 1 ≤ 1. This surface is shown in Figure 12.14.
With z = S(x, y) = 4 − x − y we have

2
2
zdσ = z 1 + (−1) + (−1) dy dx
D
2
√ 1
= 3 (4 − x − y)dy dx.
0 0
First compute
1 1
1
(4 − x − y)dy = (4 − x)y − y 2
2
0 0
1 7
= 4 − x − = − x.
2 2
z

(0, 0, 4) (0, 1, 3)

Part of the plane
x + y + z = 4
(2, 0, 2)
(0, 1, 0)
y
(2, 1, 1) (0, 4, 0)
(4, 0, 0)
(2, 1, 0)
x
FIGURE 12.14 Part of the plane x + y + z = 4.

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October 14, 2010 14:53 THM/NEIL Page-394 27410_12_ch12_p367-424