James Stewart-Calculus_ Early Transcendentals-Cengage Learning (2015)

1142 Chapter 16 Vector Calculus
equations:
2 y
S
3 y
S
4 y
S
y P i n dS − y y
E
y Q j n dS − y y
E
y R k n dS − y y
y −P
−x dV
y −Q
−y dV
E
y −R
−z dV
To prove Equation 4 we use the fact that E is a type 1 region:
E − hsx, y, zd | sx, yd [ D, u 1sx, yd < z < u 2 sx, ydj
where D is the projection of E onto the xy-plane. By Equation 15.6.6, we have
y yy −R
E
−z dV − y
D
u2sx, yd
y Fy
u1sx, yd
and therefore, by the Fundamental Theorem of Calculus,
5 y yy −R
E
−R
−z sx, y, zd dz G dA
−z dV − yy fR(x, y, u 2 sx, yd) 2 R(x, y, u 1 sx, yd)g dA
D
z
S£
0
x
FIGURE 1
E
D
S {z=u(x, y)}
S¡ {z=u¡(x, y)}
y
The boundary surface S consists of three pieces: the bottom surface S 1 , the top
surface S 2 , and possibly a vertical surface S 3 , which lies above the boundary curve of D.
(See Figure 1. It might happen that S 3 doesn’t appear, as in the case of a sphere.) Notice
that on S 3 we have k n − 0, because k is vertical and n is horizontal, and so
yy R k n dS − yy 0 dS − 0
S 3 S 3
Thus, regardless of whether there is a vertical surface, we can write
6 yy R k n dS − yy R k n dS 1 yy R k n dS
S
S 1 S 2
The equation of S 2 is z − u 2 sx, yd, sx, yd [ D, and the outward normal n points
upward, so from Equation 16.7.10 (with F replaced by R k) we have
yy R k n dS − y
S 2 D
y Rsx, y, u 2 sx, ydd dA
On S 1 we have z − u 1 sx, yd, but here the outward normal n points downward, so we
multiply by 21:
yy R k n dS − 2y
S 1 D
y Rsx, y, u 1 sx, ydd dA
Therefore Equation 6 gives
y
S
y R k n dS − y
D
y fRsx, y, u 2 sx, ydd 2 Rsx, y, u 1 sx, yddg dA
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