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We develop double integrals for measuring volume and iterated integrals for calculating double integrals.
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. . . . . .
Section 12.1–12.2Double 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
. . . . . .
Announcements
◮ Office hours Tuesday, Wednesday 2–4pm SC 323◮ Problem Sessions: Mon, 8:30; Thur, 7:30; SC 103b
. . . . . .
Outline
Last Time
Double Integrals over RectanglesRecall the definite integralDefinite integrals in two dimensions
Iterated IntegralsPartial IntegrationFubini’s TheoremAverage value
. . . . . .
Outline
Last Time
Double Integrals over RectanglesRecall the definite integralDefinite integrals in two dimensions
Iterated IntegralsPartial IntegrationFubini’s TheoremAverage value
. . . . . .
Cavalieri’s methodLet f be a positive function defined on the interval [a, b]. We want tofind 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
∆x =b − a
n. For each i between 1 and n, let xi be the nth step
between a and b. So
..a .b. . . . . . .
.x0 .x1 .x2 .xi .xn−1.xn
x0 = a
x1 = x0 + ∆x = a +b − a
n
x2 = x1 + ∆x = a + 2 · b − an
· · · · · ·
xi = a + i · b − an
· · · · · ·
xn = a + n · b − an
= b
. . . . . .
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
Mn = f(
x0 + x1
2
)∆x + f
(x1 + x2
2
)∆x + · · · + f
(xn−1 + xn
2
)∆x
In general, choose x∗i to be a point in the ith interval [xi−1, xi]. Formthe Riemann sum
Sn = f(x∗1)∆x + f(x∗2)∆x + · · · + f(x∗n)∆x
=n∑
i=1
f(x∗i )∆x
. . . . . .
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
Mn = f(
x0 + x1
2
)∆x + f
(x1 + x2
2
)∆x + · · · + f
(xn−1 + xn
2
)∆x
In general, choose x∗i to be a point in the ith interval [xi−1, xi]. Formthe Riemann sum
Sn = f(x∗1)∆x + f(x∗2)∆x + · · · + f(x∗n)∆x
=n∑
i=1
f(x∗i )∆x
. . . . . .
DefinitionThe definite integral of f from a to b is the limit∫ b
af(x) dx = lim
n→∞
n∑i=1
f(x∗i )∆x
(The big deal is that for continuous functions this limit is the same nomatter how you choose the x∗i ).
. . . . . .
The problem
Let R = [a, b] × [c, d] be a rectangle in the plane, f a positive functiondefined 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
. . . . . .
The strategy: Divide and conquer
For each m and n, divide the interval [a, b] into m subintervals ofequal width, and the interval [c, d] into n subintervals. For each i andj, form the subrectangles
Rij = [xi−1, xi] × [yj−1, yj]
Choose a sample point (x∗ij , y∗ij ) in each subrectangle and form theRiemann sum
Smn =m∑
i=1
n∑j=1
f(x∗ij , y∗ij ) ∆A
where ∆A = ∆x ∆y.
. . . . . .
DefinitionThe double integral of f over the rectangle R is∫∫
R
f(x, y) dA = limm,n→∞
m∑i=1
n∑j=1
f(x∗ij , y∗ij )∆A
(Again, for continuous f this limit is the same regardless of methodfor choosing the sample points.)
. . . . . .
Worksheet #1
ProblemEstimate the volume of the solid that lies below the surface z = xy andabove the rectangle [0, 6] × [0, 4] in the xy-plane using a Riemann sumwith m = 3 and n = 2. Take the sample point to be the upper rightcorner of each rectangle.
Answer288
. . . . . .
Worksheet #1
ProblemEstimate the volume of the solid that lies below the surface z = xy andabove the rectangle [0, 6] × [0, 4] in the xy-plane using a Riemann sumwith m = 3 and n = 2. Take the sample point to be the upper rightcorner of each rectangle.
Answer288
. . . . . .
Theorem (Midpoint Rule)
∫∫R
f(x, y) dA ≈m∑
i=1
n∑j=1
f(xi, yj)∆A
where xi is the midpoint of [xi−1, xi] and yj is the midpoint of [yj−1, yj].
. . . . . .
Worksheet #2
ProblemUse the Midpoint Rule to evaluate the volume of the solid in Problem 1.
Answer144
. . . . . .
Worksheet #2
ProblemUse the Midpoint Rule to evaluate the volume of the solid in Problem 1.
Answer144
. . . . . .
Outline
Last Time
Double Integrals over RectanglesRecall the definite integralDefinite integrals in two dimensions
Iterated IntegralsPartial IntegrationFubini’s TheoremAverage value
. . . . . .
Partial Integration
Let f be a function on a rectangle R = [a, b] × [c, d]. Then for eachfixed x we have a number
A(x) =
∫ d
cf(x, y) dy
The is a function of x, and can be integrated itself. So we have aniterated integral∫ b
aA(x) dx =
∫ b
a
[∫ d
cf(x, y) dy
]dx
. . . . . .
Worksheet #3
ProblemCalculate∫ 3
1
∫ 1
0(1 + 4xy) dx dy and
∫ 1
0
∫ 3
1(1 + 4xy) dy dx.
. . . . . .
Fubini’s Theorem
Double integrals look hard. Iterated integrals look easy/easier. Thegood news is:
Theorem (Fubini’s Theorem)If f is continuous on R = [a, b] × [c, d], then∫∫
R
f(x, y) dA =
∫ b
a
∫ d
cf(x, y) dy dx =
∫ d
c
∫ b
af(x, y) dx dy
This is also true if f is bounded on R, f is discontinuous only on a finitenumber of smooth curves, and the iterated integrals exist.
. . . . . .
Worksheet #4
ProblemEvaluate the volume of the solid in Problem 1 by computing an iteratedintegral.
Answer144
. . . . . .
Worksheet #4
ProblemEvaluate the volume of the solid in Problem 1 by computing an iteratedintegral.
Answer144
. . . . . .
Meet the mathematician: Guido Fubini
◮ Italian, 1879–1943◮ graduated Pisa 1900◮ professor in Turin,
1908–1938◮ escaped to US and died
five years later
. . . . . .
Worksheet #5
ProblemCalculate ∫∫
R
xy2
x2 + 1dA
where R = [0, 1] × [−3, 3].
Answerln 512 = 9 ln 2
. . . . . .
Worksheet #5
ProblemCalculate ∫∫
R
xy2
x2 + 1dA
where R = [0, 1] × [−3, 3].
Answerln 512 = 9 ln 2
. . . . . .
Average value
◮ One variable: If f is a function defined on [a, b], then
fave =1
b − a
∫ b
af(x) dx
◮ Two variables: If f is a function defined on a rectangle R, then
fave =1
Area(R)
∫∫R
f(x, y) dA
. . . . . .
Worksheet #6
ProblemFind the average value of f(x, y) = x2y over the rectangleR = [−1, 1] × [0, 5].
Answer
110
∫ 5
0
∫ 1
−1x2y dx dy =
56
. . . . . .
Worksheet #6
ProblemFind the average value of f(x, y) = x2y over the rectangleR = [−1, 1] × [0, 5].
Answer
110
∫ 5
0
∫ 1
−1x2y dx dy =
56
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