Example 7-28. Find the surface area of that part of the plane
φ(x, y, z ) = 2x+ 2y+z−12 = 0
which lies in the first octant.
Figure 7-20. Surface area of plane in first octant.
Solution The given plane is sketched as in figure 7-20.
The unit normal to the plane is
ˆen= grad φ
|grad φ|
=^2
3
ˆe 1 +^2
3
ˆe 2 +^1
3
ˆe 3.
The projection of the surface element dS onto the z= 0 plane produces
dS = dx dy
|ˆe 3 ·eˆn|
= 3dx dy
By summing dx dy over the region where x > 0 , y > 0 ,and x+y≤ 6 ,one obtains the
limits of integration for the surface area. The surface area is determined from the
integral
S=
∫x=6
x=0
∫y=6−x
y=0
3 dx dy =
∫ 6
0
3(6 −x)dx =−^3
2
(6 −x)^2
6
0
= 54.
If the element of surface area is projected onto the plane y= 0,there results
dS = dx dz
|ˆe 2 ·ˆen|
=^3
2
dx dz
