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.

The total surface integral is the summation of the surface integrals over each

section of the surface and produces the result 6 π which agrees with our previous

result.

Sometimes it is convenient to change the variables in a surface or volume inte-

gral. For example, the integral over the surface S 4 is not an integral which is easily