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Math 23
B. Dodson
Week 8 Homework:
15.2 iterated integrals [reprint of week 7 example]
15.3 general regions
15.4 polar coords
Week 9 Homework:
15.5 Applications of Double Integrals
15.6 Triple integrals
15.7 Cylindrical Coordinates; 15.8 Spherical Coordinates
[End of Exam 2 syllabus, Thurs April 8]
Problem 15.2.3:
Evaluate the iterated integral
Z
3
1
Z
1
0
(1 + 4
xy
)
dxdy.
Solution:
We start with the inside integral:
Z
1
0
(1 + 4
xy
)
dx
and ﬁnd an antiderivative (under
∂
∂x
)
for 1 + 4
xy.
Holding
y
constant, we ﬁnd
∂
∂x
(
x
+ 2
x
2
y
)
= 1 + 4
xy,
so
Z
1
0
(1 + 4
xy
)
dx
=
£
x
+ 2
x
2
y
/
1
0
= (1 + 2
y
)

(0 + 0) = 1 + 2
y.
We replace the inside integral with this value,
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giving the outside integral:
Z
3
1
(1 + 2
y
)
dy
=
£
y
+
y
2
/
3
1
= (3 + 9)

(1 + 1) = 10
.
We recall that this calculates the value
of the double integral
ZZ
R
(1 + 4
xy
)
dA,
over the rectangle
R
= [0
,
1]
×
[1
,
3]
,
deﬁned
as the limit of approximating sums that can
be regarded as sums of volumes of rectangular solids
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 Spring '06
 YUKICH
 Calculus, Integrals

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