Example 7-31. Evaluate the surface integral
∫∫RF×dS, where Sis the plane
2 x+ 2 y+z−1 = 0 in the first octant and F =xˆe 1 +yˆe 2 +zeˆ 3.
Solution Here ∫∫
RF×dS=∫∫RF׈endSand from the previous example
ˆen=2
3 ˆe^1 +2
3 ˆe^2 +1
3 ˆe^3.As in the previous example, the element of surface area is projected upon the xy
plane to obtain dS = 3dx dy. Therefore,
F׈en=∣∣
∣∣
∣∣ˆe 1 ˆe 2 ˆe 3
x y z
2
32
31
3∣∣
∣∣
∣∣=1
3(y− 2 z)ˆe 1 −^1
3(x− 2 z)eˆ 2 +^2
3(x−y)ˆe 3and the surface integral is
∫∫RF×dS=ˆe 1∫∫R(y− 2 z)dx dy −ˆe 2∫∫R(x− 2 z)dx dy + 2 ˆe 3∫∫R(x−y)dx dy.Here the element of surface area has been projected upon the xy plane and all
integrations are with respect to xand y. Consequently, one must express zin terms
of xand y. From the equation of the plane, the value of zon the surface is given by
z= 1 − 2 x− 2 yand the surface integral becomes
∫∫RF×dS= ˆe 1∫ (^12)
0
∫ (^12) −x
0
(5 y+ 4 x−2) dy dx
−ˆe 2
∫^12
0
∫^12 −x
0
(5 x+ 4 y−2) dy dx
- 2 ˆe 3
∫ (^12)
0
∫ (^12) −x
0
(x−y)dy dx.
These integrals are easily evaluated and the final result is
∫∫RF×dS=−^1
16 ˆe^1 +1
16 ˆe^2 +1
48 ˆe^3.