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 )