MATH 209—
Calculus,
III
Volker Runde
Double
integrals over
general
regions
MATH 209—Calculus, III
Volker Runde
University of Alberta
Edmonton, Fall 2011
Examples, I
MATH 209—
Calculus,
III
Volker Runde
Example
Let
Double
integrals over
general
regions
n
o
p
D := (x, y ) ∈ R2 : −1 ≤ x ≤ 1, 0 ≤ y ≤ 1 − x 2 .
RR
What is D x 2 y dA?
Set R := [−1, 1] × [0, 1], and define F : R → R as
2
x y , (x, y ) ∈ D,
F (x, y ) :=
0,
otherwise.
Set
ZZ
D
x 2 y dA :=
ZZ
F (x, y ) dA.
R
Examples, II
MATH 209—
Calculus,
III
Volker Runde
Double
integrals over
general
regions
Example (continued)
Then:
ZZ
Z
ZZ
2
R
Z
1
√
Z
=
=
−1
F (x, y ) dy
−1
dx
0
!
1−x 2
2
x y dy
−1
1
1
Z
F (x, y ) dA =
x y dA =
D
Z
1
dx
0
y =√1−x 2 !
Z 1 2
x 2 y 2 x − x4
dx =
dx
2 y =0
2
−1
x=1 !
1 x 3 x 5 2
=
− = .
2
3
5 x=−1
15
Integrals over general regions
MATH 209—
Calculus,
III
Volker Runde
Definition
Double
integrals over
general
regions
Let D ⊂ [a, b] × [c, d] =: R, and let
F : R → R as
f (x, y ),
F (x, y ) :=
0,
RR
If R F (x, y ) dA exists, define
ZZ
ZZ
f (x, y ) dA :=
D
f : D → R. Define
R
(x, y ) ∈ D,
otherwise.
F (x, y ) dA.
Regions of type I
MATH 209—
Calculus,
III
Volker Runde
Definition
We say that D is of type I if there are continuous functions
g1 , g2 : [a, b] → R with g1 ≤ g2 such that
Double
integrals over
general
regions
D := {(x, y ) ∈ R2 : x ∈ [a, b], g1 (x) ≤ y ≤ g2 (x)}.
Integration over regions of type I
If D is of type I:
ZZ
ZZ
f (x, y ) dA :=
D
Z
F (x, y ) dA
R
b
Z
=
d
F (x, y ) dy
a
c
Z
b
Z
!
g2 (x)
dx =
f (x, y ) dy
a
g1 (x)
dx.
Examples, III
MATH 209—
Calculus,
III
Volker Runde
Double
integrals over
general
regions
Example
RR
What is D x + 2y dA, where D is the region bounded by the
parabolas y = 2x 2 and y = 1 + x 2 .
We have
D = {(x, y ) ∈ R2 : x ∈ [−1, 1], 2x 2 ≤ y ≤ 1 + x 2 }.
Hence, D is of type I.
Examples, IV
MATH 209—
Calculus,
III
Volker Runde
Double
integrals over
general
regions
Example (continued)
Then:
ZZ
x + 2y dA
D
Z
1
Z
!
1+x 2
=
x + 2y dy
2x 2
−1
Z
Z
1
dx =
−1
y =1+x 2
dx
xy + y 2 2
y =2x
1
=
(x(1 + x 2 ) + (1 + x 2 )2 − x(2x 2 ) − (2x 2 )2 ) dx
−1
Z
1
=
−3x 4 − x 3 + 2x 2 + x + 1 dx
−1
x=1
x5 x4 2 3 x2
32
= −3 −
+ x +
+ x = .
5
4
3
2
15
x=−1
Regions of type II
MATH 209—
Calculus,
III
Volker Runde
Double
integrals over
general
regions
Definition
We say that D is of type II if there are continuous functions
h1 , h2 : [c, d] → R with h1 ≤ h2 such that
D := {(x, y ) ∈ R2 : y ∈ [c, d], h1 (y ) ≤ x ≤ h2 (y )}.
Integration over regions of type II
If D is of type II:
ZZ
Z
d
Z
f (x, y ) dA =
D
!
h2 (y )
f (x, y ) dx
c
h1 (y )
dy .
Examples, V
MATH 209—
Calculus,
III
Volker Runde
Double
integrals over
general
regions
Example
RR
Evaluate D xy dA, where D is the region bounded by the line
y = x − 1 and the parabola y 2 = 2x + 6.
Then D is both of type II:
1 2
2
D = (x, y ) ∈ R : y ∈ [−2, 4], y − 3 ≤ x ≤ y + 1 .
2
Examples, VI
MATH 209—
Calculus,
III
Volker Runde
Double
integrals over
general
regions
Example (continued)
Thus:
!
x=y +1
x 2 xy dA =
xy dx dy =
y
dx
1 2
D
−2
y −3
−2 2
x= 21 y 2 −3
2
2 !
Z
1
1 4
dy
=
y (y + 1)2 −
y2 − 3
2 −2
2
5
Z
1 4
y
3
2
=
y − + 4y + 2y − 8y dy
2 −2
4
x=1 !
3
1
y6
y
=
− + y 4 + 2 − 4y 2 = 36.
2
24
3
x=−1
ZZ
Z
4
Z
y +1
Z
4
Examples, VII
MATH 209—
Calculus,
III
Volker Runde
Double
integrals over
general
regions
Example
pπ pπ
Let p
D be
the triangle
with the vertices (0, 0),
2,
2 , and
RR
0, π2 . What is D sin(y 2 ) dA?
D is of type I:
r
r π
π
2
D = (x, y ) ∈ R : 0 ≤ x ≤
,x ≤y ≤
.
2
2
Hence:
ZZ
sin(y 2 ) dA =
D
Z √π Z √π
2
0
2
x
sin(y 2 ) dy dx = ???
Examples, VIII
MATH 209—
Calculus,
III
Example (continued)
Volker Runde
Double
integrals over
general
regions
D is also of type II:
r
π
2
D = (x, y ) ∈ R : 0 ≤ y ≤
,0≤x ≤y .
2
Thus:
Z √π Z
ZZ
sin(y 2 ) dA =
D
Z √π
2
=
0
2
0
0
y
sin(y 2 ) dx dy
y =√ π
2
1
1
2 y sin(y ) dy = − cos(y )
= .
2
2
y =0
2
Properties of the integral, I
MATH 209—
Calculus,
III
Volker Runde
Double
integrals over
general
regions
Proposition
ZZ
1
[f (x, y ) + g (x, y )] dA =
Z ZD
ZZ
f (x, y ) dA +
g (x, y ) dA;
D
2
D
for c constant:
ZZ
ZZ
c f (x, y ) dA = c
f (x, y ) dA;
D
3
D
for f (x, y ) ≥ g (x, y ) on D:
ZZ
ZZ
f (x, y ) dA ≥
g (x, y ) dA.
D
D
Properties of the integral, II
MATH 209—
Calculus,
III
Volker Runde
Double
integrals over
general
regions
Proposition
Suppose that D1 and D2 have only boundary points in
common. Then
ZZ
ZZ
ZZ
f (x, y ) dA =
f (x, y ) dA +
f (x, y ) dA.
D1 ∪D2
D1
D2
Examples, IX
MATH 209—
Calculus,
III
Example
Volker Runde
Let
Double
integrals over
general
regions
D1 := {(x, y ) ∈ R2 : x ∈ [−1, 0], y ∈ [x + 1, 1]},
D2 := {(x, y ) ∈ R2 : x ∈ [0, 1], y ∈ [1 − x, 1]},
D3 := {(x, y ) ∈ R2 : x ∈ [−1, 0], y ∈ [−1, −1 − x]},
D4 := {(x, y ) ∈ R2 : x ∈ [0, 1], y ∈ [−1, x − 1]},
(All of type I.)
Set
D = D1 ∪ D2 ∪ D3 ∪ D4 ,
and evaluate
RR
D
x 2 dA.
Examples, X
MATH 209—
Calculus,
III
Example (continued)
Volker Runde
Thus:
Double
integrals over
general
regions
ZZ
4 ZZ
X
2
x dA =
D
Dj
j=1
0 Z 1
Z
x 2 dy dx +
=
Z
x 2 dA
−1 x+1
0 Z −1−x
0
x 2 dy dx +
+
−1
Z
3
Z
−x dx +
=
−1
Z
−1
0
1Z 1
Z
1
3
Z
1−x
1 Z x−1
x 2 dy dx
−1
0
0
3
Z
−x dx +
x dx +
0
x 2 dy dx
−1
1
x 3 dx
0
Z
=4
0
1
x=1
x 3 dx = x 3 = 1.
x=0
Properties of the integral, III
MATH 209—
Calculus,
III
Volker Runde
Double
integrals over
general
regions
Notation
For a region D ⊂ R2 , we denote its area by A(D).
Proposition
Suppose that
m ≤ f (x, y ) ≤ M
on D. Then
ZZ
f (x, y ) dA ≤ M A(D).
m A(D) ≤
D
Examples, IX
MATH 209—
Calculus,
III
Volker Runde
Double
integrals over
general
regions
Example
For
D = {(x, y ) ∈ R2 : x 2 + y 2 ≤ 4},
RR
estimate D e (sin x)(cos y ) dA.
We know that A(D) = 4π and
1
≤ e (sin x)(cos y ) ≤ e.
e
Hence,
4π
≤
e
ZZ
D
e (sin x)(cos y ) dA ≤ 4πe.