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