Lesson18 Double Integrals Over Rectangles Slides

. . . . . .

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