Begin2.DVI

The element of surface area dS is projected upon the xy plane giving

dS = dx dy

|ˆe 3 ·eˆn|

= 3 dx dy

Figure 7-22. Plane 2 x+ 2y+z= 1 in first octant.

On the surface z= 1 − 2 x− 2 y,and therefore the surface integral can be represented

in terms of only xand y. One finds

∫∫

R

f(x, y, z )dS
=

∫∫

R

xyz eˆndS

=

∫x=^12

x=0

∫y=^12 −x

y=0

xy (1 − 2 x− 2 y)

[

2

3

ˆe 1 +^2

3

ˆe 2 +^1

3

ˆe 3

]

3 dx dy

= (2 ˆe 1 + 2 ˆe 2 +ˆe 3 )

∫ (^12)
0
∫ (^12) −x
0
(xy − 2 x^2 y− 2 xy^2 )dx dy

Integrate with respect to yand show

∫∫

R

f(x, y, z )dS
= (2 ˆe 1 + 2 ˆe 2 +ˆe 3 )

∫ (^12)
0
[
1
2 x(
1
2 −x)
(^2) −x (^2) (^1
2 −x)
(^2) −^2
3 x(
1
2 −x)
3
]
dx

Now integrate with respect to xand simplify the result to obtain

∫∫

R

f(x, y, z )dS
=^1

1920

(2 ˆe 1 + 2 ˆe 2 +eˆ 3 )