On S 1 , z = 4, eˆn=ˆe 3 , dS =dx dy and
∫∫
S 1
F·dS=
∫ 2
0
∫√ 4 −x 2
0
16 dy dx = 16
∫ 2
0
√
4 −x^2 dx = 16 π.
On S 2 , y = 0, ˆen=−ˆe 2 , dS =dx dz and
∫∫
S 2
F·dS=
∫∫
− 3 y dx dz = 0.
On S 3 , x = 0, ˆen=−ˆe 1 , dS =dy dz and
∫∫
S 3
F·dS=
∫∫
− 2 x dy dz = 0.
On S 4 the surface is defined by φ=x^2 +y^2 −z= 0,and the normal is determined
from
eˆn= grad φ
|grad φ|=
(^2) √xˆe 1 + 2yˆe 2 −ˆe 3
4(x^2 +y^2 ) + 1
2 xˆe (^1) √+ 2 yeˆ 2 −ˆe 3
4 z+ 1
,
and consequently the element of surface area can be represented by
dS =
dx dy
|ˆen·ˆe 3 |=
√
4 z+ 1 dx dy.
One can then write
∫∫
S 4
F·dS=
∫∫
S 4
F·ˆendS =
∫∫
S 4
(4 x^2 − 6 y^2 − 4 z)dx dy
=
∫∫
S 4
(4 x^2 − 6 y^2 −4(x^2 +y^2 )) dx dy
=
∫ 2
0
∫√ 4 −x 2
0
− 10 y^2 dy dx =−
∫ 2
0
10
3 y
3
√ 4 −x 2
0
dx
=−
∫ 2
0
10
3
√
(4 −x^2 )^3 dx =− 10 π.