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

Section 15.9 Change of Variables in Multiple Integrals 1057
¨
∫
å
0
y
0
r=a
a
¨=∫
r=a
∫
å
¨=∫
FIGURE 7
The polar coordinate transformation
R
S
¨=å
T
¨=å
b
r=b
r=b
r
x
As a first illustration of Theorem 9, we show that the formula for integration in polar
coordinates is just a special case. Here the transformation T from the r-plane to the
xy-plane is given by
x − tsr, d − r cos
y − hsr, d − r sin
and the geometry of the transformation is shown in Figure 7. T maps an ordinary rectangle
in the r-plane to a polar rectangle in the xy-plane. The Jacobian of T is
Z
−x −x
−sx, yd
−sr, d − −r −
2r sin
−y
r cos
Z − r cos 2 1 r sin 2 − r . 0
−r
Thus Theorem 9 gives
y
R
−y
−
Z
cos
− Z
sin
−sx, yd
y f sx, yd dx dy − yy f sr cos , r sin d Z
which is the same as Formula 15.3.2.
S
−sr, d
Z dr d
− y yb f sr cos , r sin d r dr d
a
ExamplE 2 Use the change of variables x − u 2 2 v 2 , y − 2uv to evaluate the integral
yy R
y dA, where R is the region bounded by the x-axis and the parabolas y 2 − 4 2 4x
and y 2 − 4 1 4x, y > 0.
SOLUtion The region R is pictured in Figure 2 (on page 1054). In Example 1 we
discovered that TsSd − R, where S is the square f0, 1g 3 f0, 1g. Indeed, the reason
for making the change of variables to evaluate the integral is that S is a much simpler
region than R. First we need to compute the Jacobian:
Z
−x −x
−sx, yd
−su, vd − −u −v 22v
−y −y
2u
Z − 4u 2 1 4v 2 . 0
Therefore, by Theorem 9,
y
R
−u
−v
Z
2u
− Z
2v
−sx, yd
y y dA − yy 2uv Z
−su, vd
Z dA − y 1
S
− 8 y 1
0 y1 0
s2uvd4su 2 1 v 2 d du dv
u−0 dv
0 y1 0 su3 v 1 uv 3 d du dv − 8 y 1
0 f 1 4 u4 v 1 1 2 u2 v 3 g u−1
− y 1
s2v 1 4v 3 d dv − fv 2 1 v 4 1
g 0
− 2 ■
0
note Example 2 was not a very difficult problem to solve because we were given
a suitable change of variables. If we are not supplied with a transformation, then the
first step is to think of an appropriate change of variables. If f sx, yd is difficult to inte-
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