Calculus 2nd Edition Rogawski

904 C H A P T E R 16 MULTIPLE INTEGRATION
y
P ij = (r ij , θ ij )
y
r = r 2 ( θ)
θ
R ij
θ 1
x
θ r 1 r
r 2 2
FIGURE 4 Decomposition of a polar
rectangle into subrectangles.
θ
2
r = r 1 ( θ)
θ 1
FIGURE 5 General polar region.
x
THEOREM 1 Double Integral in Polar Coordinates
the domain
For a continuous function f on
Eq. (4) is summarized in the symbolic
expression for the “area element” dA in
polar coordinates:
dA = rdrdθ
∫∫
f (x, y) dA =
D
D : θ 1 ≤ θ ≤ θ 2 , r 1 (θ) ≤ r ≤ r 2 (θ)
∫ θ2
∫ r2 (θ)
θ 1
r=r 1 (θ)
f (r cos θ,rsin θ)rdrdθ 4
π
θ =
2
∫∫
EXAMPLE 1 Compute (x + y)dA, where D is the quarter annulus in Figure 6.
D
Solution
Step 1. Describe D and f in polar coordinates.
The quarter annulus D is defined by the inequalities (Figure 6)
D : 0 ≤ θ ≤ π 2 , 2 ≤ r ≤ 4
2 4
θ = 0
FIGURE 6 Quarter annulus 0 ≤ θ ≤ π 2 ,
2 ≤ r ≤ 4.
In polar coordinates,
f (x, y) = x + y = r cos θ + r sin θ = r(cos θ + sin θ)
Step 2. Change variables and evaluate.
To write the integral in polar coordinates, we replace dA by rdrdθ:
The inner integral is
∫ 4
r=2
∫∫
∫ π/2 ∫ 4
(x + y)dA = r(cos θ + sin θ)rdrdθ
D
0 r=2
( 4
(cos θ + sin θ)r 2 3
)
dr = (cos θ + sin θ)
3 − 23
3
= 56 (cos θ + sin θ)
3
and
∫∫
(x + y)dA = 56
D
3
∫ π/2
0
(cos θ + sin θ)dθ = 56
3 (sin θ − cos θ) ∣ ∣∣∣
π/2
0
= 112
3