Example 7-29. Evaluate the surface integral
∫∫
R
F·dS, where Sis the surface
of the cube bounded by the planes
x= 0, x = 1, y = 0, y = 1 , z = 0, z = 1
and F is the vector field F= (x^2 +z)ˆe 1 + (xy −z)ˆe 2 + (x+y)ˆe 3
Solution The given surface is illustrated in figure 7-21.
Figure 7-21. Surface of a cube.
The given surface is piecewise continuous and thus the surface integral can be
broken up and written as the sum of the surface integrals over each face of the cube.
The following calculations illustrates the mechanics involved in evaluating this type
of surface integral.
(i) On face ABCD the unit normal to the surface is the vector n =ˆe 1 and xhas the
value 1 everywhere so that
∫∫
R
F·dS=
∫ 1
0
∫ 1
0
F·n dS =
∫ 1
0
∫ 1
0
(1 + z)dydz =
∫ 1
0
(1 + z)dz =
3
2
(ii) On face EFG0 the unit normal to the surface is the vector n =−ˆe 1 and xhas
the value 0 everywhere so that
∫∫
R
F·dS=
∫ 1
0
∫ 1
0
F·n dS =
∫ 1
0
∫ 1
0
−z dydz =
∫ 1
0
−z dz =−^1
2
