MATH 2004 Homework Solution
Han-Bom Moon
Homework 11 Model Solution
Section 15.3.
15.3.1 Evaluate the iterated integral
Z 4Z
0
Z 4Z
√
y
xy dx dy =
0
xy 2 dx dy.
0
2
0
0
y
√
4 2 2 y
x y
Z
2
√
4
Z
=
0
y3
2
dy =
0
0
dy =
4
Z
4
y4
8
√
( y)2 y 2 02 y 2
−
dy
2
2
= 32
0
15.3.8 Evaluate the double integral
ZZ
y
dA, D = {(x, y) | 0 ≤ x ≤ 1, 0 ≤ y ≤ x2 }.
5+1
x
D
ZZ
D
x2
y2
dx
2(x5 + 1) 0
0 0
0
Z 1
Z 1
(x2 )2
02
x4
=
−
dx
=
dx
5
5
2(x5 + 1)
0 2(x + 1)
0 2(x + 1)
Z 2
1
=
du (using substitution u = x5 + 1)
10u
1
2
1
1
1
ln 2
=
ln u =
ln 2 −
ln 1 =
10
10
10
10
1
Z 1Z
y
dA =
x5 + 1
x2
y
dy dx =
x5 + 1
1
Z
15.3.18 Evaluate the double integral
ZZ
(x2 + 2y) dA, D is bounded by y = x, y = x3 , x ≥ 0.
D
x = x3 , x ≥ 0 ⇔ x = 0, 1
Also if 0 ≤ x ≤ 1, x ≥ x3 .
ZZ
Z 1Z x
Z 1
2
x
2
2
(x + 2y) dA =
x + 2y dy dx =
x y + y 2 x3 dx
D
0 x3
0
Z 1
=
x2 · x + x2 − x2 · x3 + (x3 )2 dx
0
Z 1
=
x3 + x2 − x6 − x5 dx
0
=
x4 x3 x7 x6
+
−
−
4
3
7
6
1
1
=
0
1 1 1 1
23
+ − − =
4 3 7 6
84
MATH 2004 Homework Solution
Han-Bom Moon
15.3.19 Evaluate the double integral
ZZ
y 2 dA, D is the triangular region with vertices (0, 1), (1, 2), (4, 1).
D
(x, y) ∈ D ⇔ 1 ≤ y ≤ 2, y − 1 ≤ x ≤ −3y + 7
ZZ
−3y+7
Z 2Z
2
Z
2
y dx dy =
y dA =
1
D
Z
2
xy 2
1
y−1
2
−3y+7
y−1
Z
dy
2
−4y 3 + 8y 2 dy
(−3y + 7)y − (y − 1)y dy =
1
1
2 8
11
8
8
4
3
4
3
= −y + y
=
= −2 + · 2 − −1 +
3
3
3
3
1
2
=
2
15.3.24 Find the volume of the given solid under the surface z = 1 + x2 y 2 and above
the region enclosed by x = y 2 and x = 4.
(x, y) is in the region ⇔ −2 ≤ y ≤ 2, y 2 ≤ x ≤ 4
4
x3 2
dy
=
1 + x y dx dy =
x+ y
3
−2 y 2
−2
y2
Z 2
Z 2
64 2
(y 2 )3 2
y 8 61
2
=
4+ y − y +
y
dy =
− + y 2 + 4 dy
3
3
3
3
−2
−2
9
2
y
61
2336
= − + y 3 + 4y
=
27
9
27
−2
Z
volume
2Z 4
Z
2 2
2
15.3.27 Find the volume of the given solid bounded by the coordinate planes and the
plane 3x + 2y + z = 6.
The solid is a tetrahedron over a triangle T bounded by x = 0, y = 0, and 3x+2y =
6 under z = 6 − 3x − 2y.
3
(x, y) ∈ T ⇔ 0 ≤ x ≤ 2, 0 ≤ y ≤ − x + 3
2
ZZ
volume
6 − 3x − 2y dA
=
T
Z 2Z
− 32 x+3
2
− 3 x+3
6 − 3x − 2y dy dx =
6y − 3xy − y 2 0 2
dx
0
0 0
2
Z 2 3
3
3
=
6 − x + 3 − 3x − x + 3 − − x + 3
dx
2
2
2
0
2
Z 2
9 2
3 3 9 2
=
x − 9x + 9 dx =
x − x + 9x = 6
4
2
0 4
0
=
Z
2
MATH 2004 Homework Solution
Han-Bom Moon
15.3.36 Find the volume of the solid by subtracting two volumes, where the solid is
enclosed by the parabolic cylinder y = x2 and the planes z = 3y, z = 2 + y.
Two planes meet over 3y = 2 + y ⇔ y = 1.
D is the planar region that −1 ≤ x ≤ 1, x2 ≤ y ≤ 1.
On this region, 2 + y ≥ 3y.
ZZ
ZZ
ZZ
volume =
2 + y dA −
3y dA =
2 − 2y dA
D
ZZ
D
Z
1
Z
1
2 − 2y dA =
=
Z
1
2 − 2y dy dx =
x2
−1
Z 1
D
D
−1
2y − y 2
1
x2
dx
1
2 3 x5
16
1 − 2x + x dx = x − x +
=
3
5 −1 15
−1
2
4
15.3.46 Sketch the region of integration and change the order of integration.
Z Z √
4−y 2
2
f (x, y) dx dy
−2 0
p
4 − y 2 ⇒ x2 = 4 − y 2 ⇒ x2 + y 2 = 4
p
p
p
−2 ≤ y ≤ 2, 0 ≤ x ≤ 4 − y 2 ⇔ 0 ≤ x ≤ 2, − 4 − x2 ≤ y ≤ 4 − x2
Z 2 Z √4−y2
Z 2 Z √4−x2
f (x, y) dx dy =
f (x, y) dy dx
√
x=
−2 0
0
− 4−x2
15.3.47 Sketch the region of integration and change the order of integration.
Z 2Z ln x
f (x, y) dy dx
1
0
3
MATH 2004 Homework Solution
Han-Bom Moon
1 ≤ x ≤ 2, 0 ≤ y ≤ ln x ⇔ 0 ≤ y ≤ ln 2, ey ≤ x ≤ 2
Z 2Z ln x
Z ln 2Z 2
f (x, y) dy dx =
f (x, y) dx dy
1
0
ey
0
15.3.49 Evaluate
3
Z 1Z
2
ex dx dy
3y
0
by reversing the order of integration.
0 ≤ y ≤ 1, 3y ≤ x ≤ 3 ⇔ 0 ≤ x ≤ 3, 0 ≤ y ≤
Z 1Z
3
x2
e
0
x
3
Z 3Z
dx dy =
3y
0
Z
3h
Z
dy dx =
0
yex
2
0
ix
3
dx
0
1 x2
1 x2 3 1 9 1
xe dx =
e
= e −
3
6
6
6
0
3
=
0
15.3.52 Evaluate
e
x2
x
3
Z 1Z
1
x
e y dy dx
0
x
by reversing the order of integration.
0 ≤ x ≤ 1, x ≤ y ≤ 1 ⇔ 0 ≤ y ≤ 1, 0 ≤ x ≤ y
Z 1Z
1
Z 1Z
x
y
e dy dx =
0
Z
x
y
e dx dy =
x
0
=
y
0
1h
0
1
y2
e−1
(e − 1) =
2
2
0
4
ye
x
y
iy
0
Z
1
ye − y dy
dy =
0
MATH 2004 Homework Solution
Han-Bom Moon
15.3.60 Find the average value of f (x, y) = x sin y over the region D where D is enclosed by the curves y = 0, y = x2 , and x = 1.
(x, y) ∈ D ⇔ 0 ≤ x ≤ 1, 0 ≤ y ≤ x2
ZZ
Z 1Z
Z
1
2
[−x cos y]x0 dx
0
0 0
Z 1
Z 1
=
−x cos(x2 ) − (−x cos 0) dx =
x − x cos(x2 ) dx
x sin y dy dx =
x sin y dA =
D
x2
0
=
0
1
x2 1
1 1
2
− sin(x ) = − sin 1
2
2
2 2
0
1
x3
1
x dx =
area(D) =
=
3
3
0
0
ZZ
3 3
1
x sin y dA = − sin 1
average =
area(D)
2 2
D
Z
1
2
5