1106
Chapter Sixteen /SOLUTIONS
(b) The estimates asked for are just the upper and lower sums. We partition
R
into subrectangles
R
(
a,b
)
of width 2 and
height 2, where
(
a, b
)
is the lower-left corner of
R
(
a,b
)
. The subrectangles are then
R
(0
,
0)
,
R
(2
,
0)
,
R
(0
,
2)
, and
R
(2
,
2)
.
(0
,
0)
(0
,
4)
(4
,
0)
(4
,
4)
Figure 16.2
Then we Fnd the lower sum
Lower sum
=
X
(
a,b
)
A
R
(
a,b
)
·
min
R
(
a,b
)
f
=
X
(
a,b
)
4
·
(
Min of
f
on
R
(
a,b
)
)
= 4
X
(
a,b
)
(
Min of
f
on
R
(
a,b
)
)
= 4(
f
(0
,
0) +
f
(2
,
0) +
f
(0
,
2) +
f
(2
,
2))
= 4(
√
0
·
0 +
√
2
·
0 +
√
0
·
2 +
√
2
·
2)
= 8
.
Similarly, the upper sum is
Upper sum
= 4
X
(
a,b
)
(
Max of
f
on
R
(
a,b
)
)
= 4(
f
(2
,
2) +
f
(4
,
2) +
f
(2
,
4) +
f
(4
,
4))
= 4(
√
2
·
2 +
√
4
·
2 +
√
2
·
4 +
√
4
·
4)
= 24 + 16
√
2
≈
46
.
63
.
The upper sum is an overestimate and the lower sum is an underestimate, so we can get a better estimate by averaging
them to get
16 + 8
√
2
≈
27
.
3
.
4.
(a) We Frst Fnd an over- and underestimate of the integral, using four subrectangles. On the Frst subrectangle (
0
≤
x
≤
3
,
0
≤
y
≤
4)
, the function
f
(
x, y
)
appears to have a maximum of 100 and a minimum of 79. Continuing in this
way, and using the fact that
Δ
x
= 3
and
Δ
y
= 4
, we have
Overestimate
= (100 + 90 + 85 + 79)(3)(4) = 4248
,
and
Underestimate
= (79 + 68 + 61 + 55)(3)(4) = 3156
.