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

1092 Chapter 16 Vector Calculus
Integrating (7) with respect to x, we obtain
9 f sx, yd − 3x 1 x 2 y 1 tsyd
Notice that the constant of integration is a constant with respect to x, that is, a function
of y, which we have called tsyd. Next we differentiate both sides of (9) with respect
to y:
10 f y sx, yd − x 2 1 t9syd
Comparing (8) and (10), we see that
Integrating with respect to y, we have
t9syd − 23y 2
tsyd − 2y 3 1 K
where K is a constant. Putting this in (9), we have
f sx, yd − 3x 1 x 2 y 2 y 3 1 K
as the desired potential function.
(b) To use Theorem 2 all we have to know are the initial and terminal points of C,
namely, rs0d − s0, 1d and rsd − s0, 2e d. In the expression for f sx, yd in part (a), any
value of the constant K will do, so let’s choose K − 0. Then we have
y C
F dr − y C
=f dr − f s0, 2e d 2 f s0, 1d − e 3 2 s21d − e 3 1 1
This method is much shorter than the straightforward method for evaluating line
integrals that we learned in Section 16.2.
■
A criterion for determining whether or not a vector field F on R 3 is conservative is
given in Section 16.5. Meanwhile, the next example shows that the technique for finding
the potential function is much the same as for vector fields on R 2 .
ExamplE 5 If Fsx, y, zd − y 2 i 1 s2xy 1 e 3z d j 1 3ye 3z k, find a function f such
that =f − F.
SOLUTION If there is such a function f, then
11 f x sx, y, zd − y 2
12
f y sx, y, zd − 2xy 1 e 3z
13 f z sx, y, zd − 3ye 3z
Integrating (11) with respect to x, we get
14 f sx, y, zd − xy 2 1 tsy, zd
where tsy, zd is a constant with respect to x. Then differentiating (14) with respect to y,
we have
f y sx, y, zd − 2xy 1 t y sy, zd
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