(9/28/10)
Math 20C. Lecture Examples (Revised)
Sections 15.1 and 15.2. Double integrals
†
A region
R
in the
xy
plane is
bounded
if it can be enclosed in a sufficiently large circle. Its boundary is
piecewise smooth
if it consists of a finite number of graphs
y
=
y
(
x
) or
x
=
x
(
y
) of functions that are
defined on finite closed intervals and have continuous first derivatives on those intervals.
A
partition
,
R
1
, R
2
, . . . , R
N
(1)
of such a region is created by slicing it with piecewise smooth curves. Figure 2 shows a partition of the
region
R
of Figure 1 into four subregions. To measure the size of a subregion we use its
diameter
, which
is the diameter of the smallest circle that contains it.
FIGURE 1
FIGURE 2
The
interior
of a region is obtained by removing any boundary points from it. The
closure
is obtained by adding any boundary points that are not in it. A function
z
=
f
(
x, y
) is
piecewise
continuous
on a region if there is a partition of the region such that in the interior of each subregion,
f
is equal to a function that is continuous on the closure of the subregion.
Definition 1 (Riemann sums)
Suppose that
z
=
f
(
x, y
)
is piecewise continuous on a bounded region
R
with a piecewisesmooth boundary and that
(1)
is a partition of
R
. A
Riemann sum
for
integraldisplay integraldisplay
R
f
(
x, y
)
dx dy
corresponding to the partition
(1)
is a sum of the form
N
summationdisplay
j
=1
f
(
x
j
, y
j
) [Area of
R
j
]
(2)
where for each
j
= 1
,
2
, . . . , N
,
(
x
j
, y
j
)
is a point in
R
j
where
f
is defined.
Double integrals
over such regions are the limits of Riemann sums:
Definition 2 (Double integrals)
If
z
=
f
(
x, y
)
is piecewise continuous on a bounded region
R
with
a piecewisesmooth boundary, then the double integral
integraldisplay integraldisplay
R
f
(
x, y
)
dx dy
is the limit of Riemann sums
(2)
as the number of subintervals in the partition tends to infinity and their diameters tend to zero.
†
Lecture notes to accompany Sections 15.1 and 15.2 of
Calculus, Early Transcendentals
by Rogawski
1
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Math 20C. Lecture Example Solutions (9/28/10).
Sections 15.1 and 15.2, p. 2
For the region
R
of Figure 1, the partition of Figure 2, and the positive function
z
=
f
(
x, y
)
of Figure 3, the Riemann sum is equal to the total volume of four prismshaped solids as in Figure 4,
whose sides are vertical, whose bases are the four subregions of the partition, and whose horizontal tops
intersect the graph of
f
at
x
=
x
j
, y
=
y
j
. Because such collections of solids approximate the solid under
the graph, we are led to the following definition.
[Volume]
=
integraldisplay integraldisplay
R
f
(
x, y
)
dx dy
[Volume]
=
A Riemann sum]
FIGURE 3
FIGURE 4
Definition 3 (Volume)
If
R
is a bounded region in an
xy
plane with a piecewisesmooth boundary
and
z
=
f
(
x, y
)
is piecewise continuous and has nonnegative values on
R
, then the volume of the solid
above
R
and below the graph
z
=
f
(
x, y
)
is
integraldisplay integraldisplay
R
f
(
x, y
)
dx dy
(Figure 3).
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 Spring '08
 Helton
 Calculus, Integrals, dx dy, Riemann

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