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

Section 15.3 Double Integrals in Polar Coordinates 1011
¨=¨j
¨=¨j-1
¨=∫
r=b
R ij
(r i
*, ¨j*)
R
Ψ
r=a
¨=å
r=r i
O
∫
å
O
r=r i-1
FIGURE 3 Polar rectangle
FIGURE 4 Dividing R into polar subrectangles
The “center” of the polar subrectangle
has polar coordinates
R ij − hsr, d | r i21 < r < r i , j21 < < j j
r i * − 1 2 sr i21 1 r i d
j * − 1 2 s j21 1 j d
We compute the area of R ij using the fact that the area of a sector of a circle with radius
r and central angle is 1 2 r 2 . Subtracting the areas of two such sectors, each of which
has central angle D − j 2 j21 , we find that the area of R ij is
DA i − 1 2 r i 2 D 2 1 2 r 2
i21 D − 1 2 sr i 2 2 r 2 i21 d D
− 1 2 sr i 1 r i21 dsr i 2 r i21 d D − r i * Dr D
Although we have defined the double integral yy R
f sx, yd dA in terms of ordinary rectangles,
it can be shown that, for continuous functions f, we always obtain the same
answer using polar rectangles. The rectangular coordinates of the center of R ij are
sr i * cos j *, r i * sin j *d, so a typical Riemann sum is
1 o m
i−1
o n
f sr i * cos j *, r i * sin j *d DA i − o m
o n
f sr i * cos j *, r i * sin j *d r i * Dr D
j−1
i−1 j−1
If we write tsr, d − r f sr cos , r sin d, then the Riemann sum in Equation 1 can be
written as
o m
o n
tsr i *, j *d Dr D
i−1 j−1
which is a Riemann sum for the double integral
Therefore we have
y f sx, yd dA −
R
−
y yb tsr, d dr d
a
lim o n
m, nl ` om f sr i * cos j *, r i * sin j *d DA i
i−1 j−1
lim o n
m, nl ` om tsr i *, j *d Dr D − y
i−1 j−1
yb tsr, d dr d
a
− y yb f sr cos , r sin d r dr d
a
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