Math 275 Notes
Topic 4.1: Introduction to Double Integrals in
Cartesian Coordinates
Textbook Sections: 15.1, 15.2
From the Toolbox (what you need from previous classes):
Calc I/II: Be able to evaluate definite integrals of functions of a single
variable.
Algebra: Be able to find points of intersection of the graphs of two
functions, and to solve an equation in x and y in terms of either variable:
y = g(x) or x = h(y ).
Learning Objectives (New Skills) & Important Concepts
Learning Objectives (New Skills):
Set up and evaluate double integrals. This includes finding limits of
integration, and determining whether a given region of integration
requires more than one double integral.
Use the limits of integration for a given double integral to sketch
the region of integration, and change the order of integration.
Use double integrals to compute area, volume, and mass.
Important Concepts:
The region of integration for a double integral is a two-dimensional
region in the plane.
Double integrals are evaluated by evaluating two single integrals, using the techniques from Calc I/II, while treating the second variable
as a constant (more Jedi mind tricks, analogous to those used to
compute partial derivatives).
The area element dA is the area of an infinitesimal (very very small)
rectangle in the domain of integration. In Cartesian coordinates,
dA = dx dy = dy dx. The way dA is written indicates the order of
integration.
Applications of the double integral include:
? Measuring the area AD of a planar region: AD =
˜
D
dA.
? Measuring the volume V of a region between a planar region D
and the
˜ graph of a non-negative function f (x, y ):
V = D f (x, y ) dA, where f (x, y ) ≥ 0 over D.
? Computing
˜ the mass mD of a 2-dimensional flat object:
mD = D σ(x, y ) dA, where σ(x, y ) ≥ 0 is a function giving
the density per unit area at each point in D (a surface density).
The Big Picture
A double integral is an integral where the region of integration is a 2-dimensional
region in the plane R2 , and the integrand is a function of two variables.
There are many similarities between evaluating double integrals, and evaluating
single integrals. In fact, double integrals are generally evaluated by evaluating
two single integrals, using the techniques from in Calc I/II. When evaluating
the first of the two single integrals, the second variable is treated as a constant
(more Jedi mind tricks, analogous to those used to compute partial derivatives).
One of the major differences in evaluating double integrals lies in finding the
limits of integration. In Calc I and Calc II, the domain of integration was an
interval in R, so the limits of integration were just constants — the endpoints
of the interval. For a double integral, however, the region of integration is a
two-dimensional region in the plane R2 , and the limits of integration are given
by the equations of the curves bounding the region. So in fact, the “new”
thing to be learned for double integrals is finding limits of integration.
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How to “Read” a Double Integral
Double integrals over planar regions, in Cartesian coordinates:
¨
f (x, y ) dA
D
D is the region (or domain) of integration. D is a planar region: D ⊆ R2 .
f (x, y ) is called the integrand.
dA is the area element. dA represents the area of an infinitesimal (very very
small) rectangle in R2 . In Cartesian coordinates:
dA = dx dy = dy dx
Evaluating Double Integrals
A double integral is evaluated by evaluating two single integrals, using the
methods already learned in Calc I and II. The boundary curves of the region D
determine the limits of integration.
#
¨
ˆ b ˆ g2 (x)
ˆ b " ˆ g2 (x)
f (x, y ) dA =
f (x, y ) dy dx =
f (x, y ) dy dx
D
a
¨
ˆ
g1 (x)
d ˆ
or:
h2 (y )
f (x, y ) dA =
D
a
ˆ
g1 (x)
d
"ˆ
f (x, y ) dx dy =
c
h1 (y )
#
h2 (y )
f (x, y ) dx
c
dy
h1 (y )
◦ Evaluate the “inside” integral first, with respect to the variable indicated
by the differential. Then evaluate the “outside” integral, with respect to
the other variable.
◦ Jedi Mind Tricks:
? If the differential is dx, integrate with respect to x and treat y as a
constant.
? If the differential is dy , integrate with respect to y and treat x as a
constant.
◦ Note: the limits of integration of the “outside” integral are always constant.
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More Details
◦ If f (x, y ) is continuous,
and the boundary curves of D are continuous,
˜
the double integral D f (x, y ) dA can be evaluated as two single integrals
(sometimes called iterated integrals):
? If the order of integration is dA = dy dx, and the boundary curves
are y = g1 (x) and y = g2 (x) with endpoints a ≤ x ≤ b, then:
#
¨
ˆ ˆ
ˆ "ˆ
g2 (x)
b
b
f (x, y ) dA =
D
g2 (x)
f (x, y ) dy dx =
a
g1 (x)
f (x, y ) dy
a
dx
g1 (x)
? If the order of integration is dA = dx dy , and the boundary curves
are x = h1 (y ) and x = h2 (y ) with endpoints c ≤ y ≤ d, then:
#
ˆ "ˆ
ˆ ˆ
¨
d
h2 (y )
f (x, y ) dx
f (x, y ) dx dy =
c
D
d
h2 (y )
f (x, y ) dA =
c
h1 (y )
h1 (y )
◦ If f (x, y ) is continuous, and the boundary curves of D are continuous,
the order of integration does not matter:
ˆ d ˆ h2 (y )
ˆ b ˆ g2 (x)
¨
f (x, y ) dx dy
f (x, y ) dy dx =
f (x, y ) dA =
D
a
c
g1 (x)
h1 (y )
◦ Some Applications of Double Integrals All applications of integrals
begin with the idea that an integral works by “chopping up” and “adding”.
For double integrals in particular, the region of integration is chopped up
into infinitely many infinitesimal (very very small) rectangles of area dA.
The “adding up” is accomplished by integrating over the region D.
? Double Integrals and Area Chop up the region D into infinitesimal
rectangles, each having area dA. To find the total area of the region
D, add up the areas of all of the rectangles by integrating dA over
D:
¨
AreaD =
dA
D
4
dy
? Double Integrals and Volume Suppose f (x, y ) ≥ 0 over D. Chop
up the region D into infinitesimal rectangles, each having area dA.
Then f (x, y ) dA is the volume of a box with height f (x, y ) and (an
infinitesimal) base of area dA. To find the total volume between the
surface generated by the graph of f and the xy -plane over the region
D, add up the volume of all of the boxes by integrating f (x, y ) dA
over D:
¨
Volume =
f (x, y ) dA.
D
? Double Integrals and Mass Chop up the region D into infinitesimal
rectangles, each having area dA. If σ(x, y ) ≥ 0 represents the
surface density (density per unit area) at each point of the planar
region D, then σ(x, y ) dA = density × (infinitesimal) area = the
mass of an infinitesimal rectangle with area dA. To find the total
mass mD of D, add up the masses of all the rectangles by integrating
σ(x, y ) dA over D:
¨
mD =
σ(x, y ) dA.
D
(This application
can be extended to probability. If σ(x, y ) ≥ 0 over
˜
D and D σ(x, y ) dA = 1, then σ is a probability density function,
and can be used to measure bivariate probabilities.)
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Technical Details: Double Integrals as Limits of Riemann
Sums
The definition of a double integral is analogous to that of the definite integral
of a function of a single variable from Calc I.
Recall: Calc I In the single-variable case, the definite integral
defined as follows:
´b
a
f (x) dx was
◦ The region of integration is an interval [a, b] on the real line R. This
interval is is partitioned (“chopped up”) into n sub-intervals of length
∆x.
◦ In each sub-interval, the integrand f (x) is evaluated at a point xi , and
multiplied by ∆x (the length of the sub-interval). These products f (x1 ) ∆x
are added up (a Riemann sum) to approximate the integral:
n
X
ˆ
b
f (xi ) ∆x ≈
f (x) dx
a
i=1
◦ The smaller the subintervals, the better the approximation. If the integral
exists, then the exact value of the definite integral is the limit of the
Riemann sums as n → ∞) (which forces the length ∆x → 0):
lim
n→∞
n
X
ˆ
f (xi ) ∆x =
f (x) dx
a
i=1
Double Integrals The double integral
b
˜
D
f (x, y ) dA is defined as follows:
◦ The region of integration is a two-dimensional region D in the plane R2 .
This region is partitioned (“chopped up”) into small rectangles of area
∆A = ∆x ∆y = ∆y ∆x.
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◦ In each sub-rectangle, the integrand f (x, y ) is evaluated at a point (xi , yj ),
and multiplied by ∆A (the area of the small rectangle):
f (xi , yj ) ∆A
◦ These products f (xi , yj ) ∆A are added up (a Riemann sum) to approximate the double integral:
M X
N
X
¨
f (xi , yj ) ∆A ≈
f (x, y ) dA
D
j=1 i=1
◦ If the integral exists, then the smaller the rectangles in the partition of D,
the better the approximation, and the exact value of the double integral
is the limit of the Riemann sums as M, N → ∞ (in such a way that
∆x, ∆y → 0):
lim
M,N→∞
N
M X
X
¨
f (xi , yj ) ∆A =
f (x, y ) dA
D
j=1 i=1
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