Begin2.DVI

in the y−direction from y= 0 to the circle y=

√

4 −x^2 to form a slab. The slab is

then summed in the x−direction from x= 0 to x= 2.

Figure 8-4. Integration over closed surface area defined by S 1 ∪S 2 ∪S 3 ∪S 4.

The resulting volume integral is then represented

∫∫∫

V

div F dV
=

∫x=2

x=0

∫y=√ 4 −x 2

y=0

∫z=4

z=x^2 +y^2

3 dzdy dx

=

∫ 2

0

∫√ 4 −x 2

0

3[4 −(x^2 +y^2 )]dy dx

=

∫ 2

0

3(4y−x^2 y−^13 y^3 )

√ 4 −x 2

0

dx

=

∫ 2

0

(8 − 2 x^2 )

√

4 −x^2 dx = 6π.

For the surface integral part of Gauss’ divergence theorem, observe that the surface

enclosing the volume is composed of four sections which can be labeled S 1 , S 2 , S 3 , S 4

as illustrated in the figure 8-4. The surface integral can then be broken up and

written as a summation of surface integrals. One can write

∫∫

S

F
·dS
=

∫∫

S 1

F
·dS
+

∫∫

S 2

F
·dS
+

∫∫

S 3

F
·dS
+

∫∫

S 4

F
·dS .

Each surface integral can be evaluated as follows.