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Section 12.3Double Integrals over General Regions

Math 21a

March 19, 2008

Announcements

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

..Image: Flickr user Netream

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Announcements

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

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Outline

Last Time

Double Integrals over General RegionsAgain a LimitProperties of Double Integrals

Iterated Integrals over Curved RegionsRegions of Type IRegions of Type II

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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

For continuous f this limit is the same regardless of method forchoosing the sample points.

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Fubini’s Theorem

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.

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Outline

Last Time

Double Integrals over General RegionsAgain a LimitProperties of Double Integrals

Iterated Integrals over Curved RegionsRegions of Type IRegions of Type II

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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: ∫∫D1∪...∪Dn

f(x, y) dA =n∑

i=1

∫∫Di

f(x, y) dA

◮ Define the integral over a general region as∫∫R

f(x, y) dA = lim∫∫

D1∪...∪Dn

f(x, y) dA

where the limit is taken over all unions of rectanglesapproximating R

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Properties of Double Integrals(a)

∫∫D

[f(x, y) + g(x, y)] dA =

∫∫D

f(x, y) dA +

∫∫D

g(x, y) dA

(b)∫∫D

cf(x, y) dA = c∫∫D

f(x, y) dA

(c) If f(x, y) ≥ g(x, y) for all (x, y) ∈ D, then∫∫D

f(x, y) dA ≥∫∫D

g(x, y) dA.

(d) If D = D1 ∪ D2, where D1 and D2 do not overlap except possiblyon their boundaries, then∫∫

D

[f(x, y)] dA =

∫∫D1

f(x, y) dA +

∫∫D2

f(x, y) dA

(e)∫∫D

dA is the area of D, written A(D).

(f) If m ≤ f(x, y) ≤ M for all (x, y) ∈ D, then

m · A(D) ≤∫∫D

f(x, y) dA ≤ M · A(D).

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Outline

Last Time

Double Integrals over General RegionsAgain a LimitProperties of Double Integrals

Iterated Integrals over Curved RegionsRegions of Type IRegions of Type II

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DefinitionA plane region D is said to be of Type I if it lies between the graphsof two continuous functions of x:

D = { (x, y) | a ≤ x ≤ b, g1(x) ≤ y ≤ g2(x) }

QuestionWhat rectangular approximations for such a D would be good inestimating an integral over D?

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DefinitionA plane region D is said to be of Type I if it lies between the graphsof two continuous functions of x:

D = { (x, y) | a ≤ x ≤ b, g1(x) ≤ y ≤ g2(x) }

QuestionWhat rectangular approximations for such a D would be good inestimating an integral over D?

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FactIf 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∫∫D

f(x, y) dA =

∫ b

a

∫ g2(x)

g1(x)f(x, y) dy dx

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Worksheet #1

Problem

Evaluate∫ 1

0

∫ ey

y

√x dx dy

Answer445

(−8 + 5e3/2

)

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Worksheet #1

Problem

Evaluate∫ 1

0

∫ ey

y

√x dx dy

Answer445

(−8 + 5e3/2

)

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Worksheet #2

ProblemEvaluate

∫∫D

2yx2 + 1

dA, where D = {(x, y)|0 ≤ x ≤ 1, 0 ≤ y ≤√

x}.

Answer12

ln 2

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Worksheet #2

ProblemEvaluate

∫∫D

2yx2 + 1

dA, where D = {(x, y)|0 ≤ x ≤ 1, 0 ≤ y ≤√

x}.

Answer12

ln 2

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DefinitionA plane region D is said to be of Type II if it lies between the graphsof two continuous functions of y:

D = { (x, y) | c ≤ y ≤ d, h1(y) ≤ x ≤ h2(y) }

QuestionWhat rectangular approximations for such a D would be good inestimating an integral over D?

. . . . . .

DefinitionA plane region D is said to be of Type II if it lies between the graphsof two continuous functions of y:

D = { (x, y) | c ≤ y ≤ d, h1(y) ≤ x ≤ h2(y) }

QuestionWhat rectangular approximations for such a D would be good inestimating an integral over D?

. . . . . .

FactIf 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

f(x, y) dA =

∫ d

c

∫ h2(y)

h1(y)f(x, y) dx dy

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Worksheet #3

ProblemEvaluate

∫∫D

xy2 dA, where D is bounded byy = x, and x = y2 − 2.

Answer97

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Worksheet #3

ProblemEvaluate

∫∫D

xy2 dA, where D is bounded byy = x, and x = y2 − 2.

Answer97

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Worksheet #4

ProblemFind the volume of the solid under the surface z = xy and above thetriangle with vertices (1, 1), (4, 1), and (1, 2).

Answer ∫ 4

1

∫ 2−1/3(x−1)

1xy dy dx =

318

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Worksheet #4

ProblemFind the volume of the solid under the surface z = xy and above thetriangle with vertices (1, 1), (4, 1), and (1, 2).

Answer ∫ 4

1

∫ 2−1/3(x−1)

1xy dy dx =

318

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Worksheet #5

Problem

Sketch the region of integration for∫ 4

0

∫ √x

0f(x, y) dy dx and change the

order of integration.

AnswerThe integral is equal to ∫ 2

0

∫ y2

0f(x, y) dx dy

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Worksheet #5

Problem

Sketch the region of integration for∫ 4

0

∫ √x

0f(x, y) dy dx and change the

order of integration.

AnswerThe integral is equal to ∫ 2

0

∫ y2

0f(x, y) dx dy