3. Outline
Last Time
Double Integrals over General Regions
Again a Limit
Properties of Double Integrals
Iterated Integrals over Curved Regions
Regions of Type I
Regions of Type II
. . . . . .
4. Definition
The double integral of f over the rectangle R is
∫∫ ∑∑
m n
f(x, y) dA = lim f(x∗ , y∗ ) ∆A
ij ij
m,n→∞
R i=1 j=1
For continuous f this limit is the same regardless of method for
choosing the sample points.
. . . . . .
5. Fubini’s Theorem
Theorem (Fubini’s Theorem)
If f is continuous on R = [a, b] × [c, d], then
∫∫ ∫ b∫ d ∫ d∫ b
f(x, y) dA = f(x, y) dy dx = f(x, y) dx dy
a c c a
R
This is also true if f is bounded on R, f is discontinuous only on a finite
number of smooth curves, and the iterated integrals exist.
. . . . . .
6. Outline
Last Time
Double Integrals over General Regions
Again a Limit
Properties of Double Integrals
Iterated Integrals over Curved Regions
Regions of Type I
Regions of Type II
. . . . . .
7. Towards an integral over general regions
◮ Right now we can integrate over a rectangle
◮ Extend this to an integral over a union of rectangles possibly
overlapping:
∫∫ ∑ ∫∫
n
f(x, y) dA = f(x, y) dA
D1 ∪...∪Dn i=1 D
i
◮ Define the integral over a general region as
∫∫ ∫∫
f(x, y) dA = lim f(x, y) dA
R D1 ∪...∪Dn
where the limit is taken over all unions of rectangles
approximating R
. . . . . .
8. Properties of Double Integrals
∫∫ ∫∫ ∫∫
(a) [f(x, y) + g(x, y)] dA = f(x, y) dA + g(x, y) dA
∫∫
D ∫∫ D D
(b) cf(x, y) dA = c f(x, y) dA
D D
(c) If∫(x, y) ≥ g(x, y) ∫ ∫ all (x, y) ∈ D, then
f∫ for
f(x, y) dA ≥ g(x, y) dA.
D D
(d) If D = D1 ∪ D2 , where D1 and D2 do not overlap except possibly
on ∫
∫ their boundaries,∫
∫ then ∫∫
[f(x, y)] dA = f(x, y) dA + f(x, y) dA
∫∫
D D1 D2
(e) dA is the area of D, written A(D).
D
(f) If m ≤ f(x, y∫ ∫ M for all (x, y) ∈ D, then
)≤
m · A(D) ≤ f(x, y) dA ≤ M · A(D).
. . . . . .
9. Outline
Last Time
Double Integrals over General Regions
Again a Limit
Properties of Double Integrals
Iterated Integrals over Curved Regions
Regions of Type I
Regions of Type II
. . . . . .
10. Definition
A plane region D is said to be of Type I if it lies between the graphs
of two continuous functions of x:
D = { (x, y) | a ≤ x ≤ b, g1 (x) ≤ y ≤ g2 (x) }
. . . . . .
11. Definition
A plane region D is said to be of Type I if it lies between the graphs
of two continuous functions of x:
D = { (x, y) | a ≤ x ≤ b, g1 (x) ≤ y ≤ g2 (x) }
Question
What rectangular approximations for such a D would be good in
estimating an integral over D?
. . . . . .
12. Fact
If D is a region of Type I:
D = { (x, y) | a ≤ x ≤ b, g1 (x) ≤ y ≤ g2 (x) }
Then for any “mostly” continuous function f
∫∫ ∫ b∫ g2 (x)
f(x, y) dA = f(x, y) dy dx
D a g1 (x)
. . . . . .
13. Worksheet #1
Problem ∫
∫ 1 ey √
Evaluate x dx dy
0 y
. . . . . .
14. Worksheet #1
Problem ∫
∫ 1 ey √
Evaluate x dx dy
0 y
Answer
( )
4
−8 + 5e3/2
45
. . . . . .
15. Worksheet #2
Problem ∫
∫
2y √
Evaluate dA, where D = {(x, y)|0 ≤ x ≤ 1, 0 ≤ y ≤ x}.
x2 +1
D
. . . . . .
16. Worksheet #2
Problem ∫
∫
2y √
Evaluate dA, where D = {(x, y)|0 ≤ x ≤ 1, 0 ≤ y ≤ x}.
x2 +1
D
Answer
1
ln 2
2
. . . . . .
17. Definition
A plane region D is said to be of Type II if it lies between the graphs
of two continuous functions of y:
D = { (x, y) | c ≤ y ≤ d, h1 (y) ≤ x ≤ h2 (y) }
. . . . . .
18. Definition
A plane region D is said to be of Type II if it lies between the graphs
of two continuous functions of y:
D = { (x, y) | c ≤ y ≤ d, h1 (y) ≤ x ≤ h2 (y) }
Question
What rectangular approximations for such a D would be good in
estimating an integral over D?
. . . . . .
19. Fact
If D is a region of Type II:
D = { (x, y) | c ≤ x ≤ d, h1 (y) ≤ x ≤ h2 (y) }
Then for any “mostly” continuous function f
∫∫ ∫ d∫ h 2 (y )
f(x, y) dA = f(x, y) dx dy
D c h 1 (y )
. . . . . .
20. Worksheet #3
Problem ∫
∫
Evaluate xy2 dA, where D is bounded byy = x, and x = y2 − 2.
D
. . . . . .
21. Worksheet #3
Problem ∫
∫
Evaluate xy2 dA, where D is bounded byy = x, and x = y2 − 2.
D
Answer
9
7
. . . . . .
22. Worksheet #4
Problem
Find the volume of the solid under the surface z = xy and above the
triangle with vertices (1, 1), (4, 1), and (1, 2).
. . . . . .
23. Worksheet #4
Problem
Find the volume of the solid under the surface z = xy and above the
triangle with vertices (1, 1), (4, 1), and (1, 2).
Answer
∫ 4 ∫ 2−1/3(x−1)
31
xy dy dx =
1 1 8
. . . . . .
24. Worksheet #5
Problem ∫ 4∫
√
x
Sketch the region of integration for f(x, y) dy dx and change the
0 0
order of integration.
. . . . . .
25. Worksheet #5
Problem ∫ 4∫
√
x
Sketch the region of integration for f(x, y) dy dx and change the
0 0
order of integration.
Answer
The integral is equal to
∫ 2 ∫ y2
f(x, y) dx dy
0 0
. . . . . .
