We develop double integrals for measuring volume and iterated integrals for calculating double integrals.

1. Section 12.1–12.2
Double Integrals over Rectangles
Iterated Integrals
Math 21a
March 17, 2008
Announcements
◮ Office hours Tuesday, Wednesday 2–4pm SC 323
◮ Problem Sessions: Mon, 8:30; Thur, 7:30; SC 103b
. .
Image: Flickr user Cobalt123
. . . . . .

2. Announcements
◮ Office hours Tuesday, Wednesday 2–4pm SC 323
◮ Problem Sessions: Mon, 8:30; Thur, 7:30; SC 103b
. . . . . .

3. Outline
Last Time
Double Integrals over Rectangles
Recall the definite integral
Definite integrals in two dimensions
Iterated Integrals
Partial Integration
Fubini’s Theorem
Average value
. . . . . .

4. Outline
Last Time
Double Integrals over Rectangles
Recall the definite integral
Definite integrals in two dimensions
Iterated Integrals
Partial Integration
Fubini’s Theorem
Average value
. . . . . .

5. Cavalieri’s method
Let f be a positive function defined on the interval [a, b]. We want to
find the area between x = a, x = b, y = 0, and y = f(x).
For each positive integer n, divide up the interval into n pieces. Then
b−a
∆x = . For each i between 1 and n, let xi be the nth step
n
between a and b. So
x0 = a
b−a
x 1 = x 0 + ∆x = a +
n
b−a
x 2 = x 1 + ∆x = a + 2 ·
n
······
b−a
xi = a + i ·
n
x x x
.0 .1 .2 . i . n −1 . n
xx x ······
. . . . . . . .
.
a b−a
b
. xn = a + n · =b
. . n
. . . .

6. Forming Riemann sums
We have many choices of how to approximate the area:
Ln = f(x0 )∆x + f(x1 )∆x + · · · + f(xn−1 )∆x
Rn = f(x1 )∆x + f(x2 )∆x + · · · + f(xn )∆x
( ) ( ) ( )
x0 + x 1 x1 + x2 x n −1 + x n
Mn = f ∆x + f ∆x + · · · + f ∆x
2 2 2
. . . . . .

7. Forming Riemann sums
We have many choices of how to approximate the area:
Ln = f(x0 )∆x + f(x1 )∆x + · · · + f(xn−1 )∆x
Rn = f(x1 )∆x + f(x2 )∆x + · · · + f(xn )∆x
( ) ( ) ( )
x0 + x 1 x1 + x2 x n −1 + x n
Mn = f ∆x + f ∆x + · · · + f ∆x
2 2 2
In general, choose x∗ to be a point in the ith interval [xi−1 , xi ]. Form
i
the Riemann sum
Sn = f(x∗ )∆x + f(x∗ )∆x + · · · + f(x∗ )∆x
1 2 n
∑ n
= f(x∗ )∆x
i
i=1
. . . . . .

8. Definition
The definite integral of f from a to b is the limit
∫ b ∑
n
f(x) dx = lim f(x∗ )∆x
i
a n→∞
i=1
(The big deal is that for continuous functions this limit is the same no
matter how you choose the x∗ ).i
. . . . . .

9. The problem
Let R = [a, b] × [c, d] be a rectangle in the plane, f a positive function
defined on R, and
S = { (x, y, z) | a ≤ x ≤ b, c ≤ y ≤ d, 0 ≤ z ≤ f(x, y) }
Our goal is to find the volume of S
. . . . . .

10. The strategy: Divide and conquer
For each m and n, divide the interval [a, b] into m subintervals of
equal width, and the interval [c, d] into n subintervals. For each i and
j, form the subrectangles
Rij = [xi−1 , xi ] × [yj−1 , yj ]
Choose a sample point (x∗ , y∗ ) in each subrectangle and form the
ij ij
Riemann sum
∑∑m n
Smn = f(x∗ , y∗ ) ∆A
ij ij
i=1 j=1
where ∆A = ∆x ∆y.
. . . . . .

11. 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
(Again, for continuous f this limit is the same regardless of method
for choosing the sample points.)
. . . . . .

12. Worksheet #1
Problem
Estimate the volume of the solid that lies below the surface z = xy and
above the rectangle [0, 6] × [0, 4] in the xy-plane using a Riemann sum
with m = 3 and n = 2. Take the sample point to be the upper right
corner of each rectangle.
. . . . . .

13. Worksheet #1
Problem
Estimate the volume of the solid that lies below the surface z = xy and
above the rectangle [0, 6] × [0, 4] in the xy-plane using a Riemann sum
with m = 3 and n = 2. Take the sample point to be the upper right
corner of each rectangle.
Answer
288
. . . . . .

14. Theorem (Midpoint Rule)
∫∫ ∑∑
m n
f(x, y) dA ≈ f(¯i , ¯j ) ∆A
x y
R i=1 j=1
where ¯i is the midpoint of [xi−1 , xi ] and ¯j is the midpoint of [yj−1 , yj ].
x y
. . . . . .

15. Worksheet #2
Problem
Use the Midpoint Rule to evaluate the volume of the solid in Problem 1.
. . . . . .

16. Worksheet #2
Problem
Use the Midpoint Rule to evaluate the volume of the solid in Problem 1.
Answer
144
. . . . . .

17. Outline
Last Time
Double Integrals over Rectangles
Recall the definite integral
Definite integrals in two dimensions
Iterated Integrals
Partial Integration
Fubini’s Theorem
Average value
. . . . . .

18. Partial Integration
Let f be a function on a rectangle R = [a, b] × [c, d]. Then for each
fixed x we have a number
∫ d
A(x) = f(x, y) dy
c
The is a function of x, and can be integrated itself. So we have an
iterated integral
∫ b ∫ b [∫ d ]
A(x) dx = f(x, y) dy dx
a a c
. . . . . .

19. Worksheet #3
Problem
Calculate
∫ 3∫ 1 ∫ 1∫ 3
(1 + 4xy) dx dy and (1 + 4xy) dy dx.
1 0 0 1
. . . . . .

20. Fubini’s Theorem
Double integrals look hard. Iterated integrals look easy/easier. The
good news is:
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.
. . . . . .

21. Worksheet #4
Problem
Evaluate the volume of the solid in Problem 1 by computing an iterated
integral.
. . . . . .

22. Worksheet #4
Problem
Evaluate the volume of the solid in Problem 1 by computing an iterated
integral.
Answer
144
. . . . . .

23. Meet the mathematician: Guido Fubini
◮ Italian, 1879–1943
◮ graduated Pisa 1900
◮ professor in Turin,
1908–1938
◮ escaped to US and died
five years later
. . . . . .

24. Worksheet #5
Problem
Calculate ∫∫
xy2
dA
x2 + 1
R
where R = [0, 1] × [−3, 3].
. . . . . .

25. Worksheet #5
Problem
Calculate ∫∫
xy2
dA
x2 + 1
R
where R = [0, 1] × [−3, 3].
Answer
ln 512 = 9 ln 2
. . . . . .

26. Average value
◮ One variable: If f is a function defined on [a, b], then
∫ b
1
fave = f(x) dx
b−a a
◮ Two variables: If f is a function defined on a rectangle R, then
∫∫
1
fave = f(x, y) dA
Area(R)
R
. . . . . .

27. Worksheet #6
Problem
Find the average value of f(x, y) = x2 y over the rectangle
R = [−1, 1] × [0, 5].
. . . . . .

28. Worksheet #6
Problem
Find the average value of f(x, y) = x2 y over the rectangle
R = [−1, 1] × [0, 5].
Answer
∫ 5∫ 1
1 5
x2 y dx dy =
10 0 −1 6
. . . . . .