Double Integrals - Math 212bfitzpat/teaching/212s20/... · 2020-04-29 · Double Integrals Iterated Integrals Example Suppose density throughout D = [1;5] [0;2] is f(x;y) = xy2 kg=m2

Double IntegralsMath 212

Brian D. Fitzpatrick

Duke University

March 2, 2020

MATH

Overview

Double IntegralsMotivationIterated IntegralsNonrectangular Regions

Example Ix-slicingy -slicing

Example IIx-slicingy -slicing

Double IntegralsMotivation

Example

A sheet of material conforms to the shape of a domain D in R2.

P1

f (P1)=7 kg/m2 P2

f (P2)=17 kg/m2

P3

f (P3)=3 kg/m2

P4

f (P4)=21 kg/m2

Suppose f ∈ C (R2) measures density (kg/m2) at every point of D.

DefinitionThe double integral of f over D is∫∫

Df dA = mass of D (in kg)

Double IntegralsMotivation

ObservationTracking units allows us to interpret double integrals.∫∫

Df

mass units

area unit

dA

area unit

= mass of D

Double IntegralsIterated Integrals

QuestionHow can we calculate a double integral?

AnswerIntegrate variable-by-variable!

Double IntegralsIterated Integrals

DefinitionSuppose f ∈ C (R2) measures density throughout a rectangle D.

a b

c

d

x

density =∫ dc f (x , y) dy

The iterated integral with x-slicing computes mass with∫∫Df dA =

∫ b

a

∫ d

cf (x , y) dy dx

Double IntegralsIterated Integrals

Example

Suppose density throughout D = [1, 5]× [0, 2] is

f (x , y) = xy2 kg/m2

We may compute the mass of D using x-slicing.∫∫Df dA =

∫ 5

1

∫ 2

0xy2 dy

x constant

dx

=

∫ 5

1

{1

3xy3}y=2

y=0

dx

=

∫ 5

1

8

3x dx

= 32 kg

Double IntegralsIterated Integrals

DefinitionSuppose f ∈ C (R2) measures density throughout a rectangle D.

a b

c

d

y

density =∫ ba f (x , y) dx

The iterated integral with y -slicing computes mass with∫∫Df dA =

∫ d

c

∫ b

af (x , y) dx dy

Double IntegralsIterated Integrals

Example

Suppose again density throughout D = [1, 5]× [0, 2] is

f (x , y) = xy2 kg/m2

We may compute the mass of D using y -slicing.∫∫Df dA =

∫ 2

0

∫ 5

1xy2 dx

y constant

dy

=

∫ 2

0

{1

2x2y2

}x=5

x=1

dy

=

∫ 2

012 y2 dy

= 32 kg

Double IntegralsIterated Integrals

Example

Suppose density throughout D = [1, 2]× [0, π] is

f (x , y) = x sin (xy) kg/m2

We may compute the mass of D using x-slicing.∫∫Df dA =

∫ 2

1

∫ π

0x sin (xy) dy dx

=

∫ 2

1{− cos (xy) }y=πy=0 dx

=

∫ 2

1− cos (πx) + 1 dx

=

{−sin(πx)

π+ x

}x=2

x=1

= 1

Double IntegralsNonrectangular Regions

QuestionHow do we compute

∫∫D f dA if D is not rectangular?

AnswerOur slicing method will depend on the shape of D.

Double IntegralsNonrectangular Regions

DefinitionSuppose D is enclosed between the graphs of y1(x) and y2(x).

a bx

density =

∫ y2(x)

y1(x)f (x , y) dy

y2(x)

y1(x)

The iterated integral with x-slicing computes mass with∫∫Df dA =

∫ b

a

∫ y2(x)

y1(x)f (x , y) dy dx

Double IntegralsNonrectangular Regions

Example

Consider the region D bounded by y = x2 and y = x + 2.

−1 2

0 = x2 − (x + 2)

= (x + 1)(x − 2)

y2(x) = x + 2y1(x) = x2

The mass of this region can be calculated using x-slicing.∫∫Df dA =

∫ 2

−1

∫ x+2

x2f dy dx

Double IntegralsNonrectangular Regions

Example

Suppose density is f (x , y) = x + 2 y A/m2 in the previous example.∫∫Df dA =

∫ 2

−1

∫ x+2

x2x + 2 y dy dx

=

∫ 2

−1

{xy + y2

}y=x+2

y=x2dx

=

∫ 2

−1{x · (x + 2) + (x + 2)2} − {x · (x2) + (x2)2} dx

=

∫ 2

−1−x4 − x3 + 2 x2 + 6 x + 4 dx

=333

20A

Double IntegralsNonrectangular Regions

DefinitionSuppose D is enclosed between the graphs of x1(y) and x2(y).

y

c

d

density =

∫ x2(y)

x1(y)f (x , y) dx

x1(y)x2(y)

The iterated integral with y -slicing computes mass with∫∫Df dA =

∫ d

c

∫ x2(y)

x1(y)f (x , y) dx dy

Double IntegralsNonrectangular Regions

Example

Consider the region D bounded by y =√x

x = y2

and y = 2− xx = 2− y.

1x1(y) = y2

x2(y) = 2− y

The mass of this region can be calculated using y -slicing.∫∫Df dA =

∫ 1

0

∫ 2−y

y2

f dx dy

Double IntegralsNonrectangular Regions

Example

Suppose density is f (x , y) = y $/m2 in the previous example.∫∫Df dA =

∫ 1

0

∫ 2−y

y2

y dx dy

=

∫ 1

0{xy}x=2−y

x=y2 dy

=

∫ 1

0

{(2− y)y − (y2)y

}dy

=

∫ 1

0

{2 y − y2 − y3

}dy

=

{y2 − y3

3− y4

4

}y=1

y=0

=5

12$

Example Ix-slicing

Example

Consider a direct calculation of∫ 21

∫ x2

1x/y dy dx .

∫ 2

1

∫ x2

1

x

ydy dx =

∫ 2

1x log (y) |y=x2

y=1 dx

=

∫ 2

1x log(x2) dx

u = x2 u(2) = 4(1/2) du = x dx u(1) = 1

=1

2

∫ 4

1log(u) du

w = log(u) dw = 1/u dudv = du v = u

=1

2

{u log(u)|u=4

u=1 −∫ 4

1du

}=

1

2{4 log(4)− 3}

= 2 log(4)− 3

2

Note that this method uses x-slicing.

Example Iy -slicing

Example

Calculating∫ 21

∫ x2

1x/y dy dx with y -slicing gives

∫ 2

1

∫ x2

1

x

ydy dx =

∫ 4

1

∫ 2

√y

x

ydx dy

=1

2

∫ 4

1

x2

y

∣∣∣∣x=2

x=√y

dy

=1

2

∫ 4

1

4

y− 1 dy

=1

2{4 log(y)− y}y=4

y=1

=1

2{(4 log(4)− 4)− (4 log(1)− 1)}

= 2 log(4)− 3

2

y = 1

x = 2y = x2

Here we avoid the u-substitution and the integration by parts!

Example IIx-slicing

Example

Consider the iterated integral∫ 1

0

∫ 1

yex

2dx

no elementary antiderivative!

dy

Example IIy -slicing

Example

Calculating∫ 10

∫ 1y ex

2dx dy using x-slicing gives

∫ 1

0

∫ 1

yex

2dx dy =

∫ 1

0

∫ x

0ex

2dy dx

=

∫ 1

0y ex

2∣∣∣y=x

y=0dx

=

∫ 1

0x ex

2dx

u = x2 u(1) = 1(1/2) du = x dx u(0) = 0

=1

2

∫ 1

0eu du

=e − 1

2

y = 0

x = 1x=y