MATH4008 Calculus 3 for Economists (Class 11)
Homework 5 (Suggeseted Solutions)
NTU 112-2
Exercise 1(A) :
(a) (i)
Z
1
y=3
−1
dx
(1 + x2 + y 2 )1/2 y=2
0
Z 1
1
1
√
−√
dx
=
5 + x2
10 + x2
0
√ √ 1+ 6
1 + 11
√
√
= ln
− ln
.
5
10
3
Z
y
dydx =
(1 + x2 + y 2 )3/2
2
0
Z
√
where the last equality follows from
a2
1
Z
p
1
dx = ln x + x2 + a2 + C.
2
+x
(ii)
ZZ
1
Z
sin (x2 ) dA =
R
Z
y=x
0
y=0
Z
y=x2
1
Z
sin (x2 ) dydx =
x sin(x2 )dx =
0
cos(x2 )
−2
x=1
=
x=0
1 − cos(1)
2
(iii)
1
ZZ p
Z
3
x + 1 dA =
R
p
x3
Z
+ 1 dydx =
y=0
0
1
x
2
p
2(x3 + 1)3/2
+ 1dx =
9
x3
0
1
(b) The region of integration is a triangle as follows.
ZZ
Z
a
Z
y=a−x
xy dA =
Volume =
R
Z
xy dydx =
0
y=0
0
a
x(a − x)2
a4
dx =
2
24
=
0
2 3/2
(2 − 1)
9
Exercise 2(A) :
(a)
1
Z
3
Z
e
x2
dx dy
Fubini
3
Z
y=x/3
e
=
3y
0
Z
Z
x2
dy dx =
y=0
0
3
0
" 2 #3
2
xex
e9 − 1
ex
dx =
=
.
3
6
6
0
(b)
Z
1
π
4
Z
√
sec x 2 + sec x dx dy
0
Fubini
π
4
Z
tan x
Z
=
arctan y
0
Z
=
π
4
√
sec x 2 + sec x dy dx
0
√
tan x sec x 2 + sec x dx
0
3
2
=
(2 + sec x) 2
3
π4
=
0
√ 3
3
2
(2 + 2) 2 − 3 2 .
3
(c)
Note that the two regions of integration can be combined into a single, large region. So we can simply
integrate over this combined region.
Z
1
2
Z
x
√
x
sin y
dy dx +
y
Z
2
4
Z
2
√
x
Z 2 Z x=y2
sin y
sin y
Fubini
dy dx =
dx dy
y
y
1
x=y
Z 2
=
(y − 1) sin y dy
1
y=2
= [−(y − 1) cos y + sin y]y=1
= − cos 2 + sin 2 − sin 1.
Exercise 3(B) :
We first divide the given region into two subregions R1 (on which y ≥ x2 ) and R2 (on which y ≤ x2 )
as follows :
ZZ
2
ZZ
ZZ
2
Therefore,
min{y, x } dA =
x dA +
R
R1
ZZ
Z 2 Z y=2x
8
—
x2 dy dx = .
x2 dA =
5
y=x2
R1
0
ZZ
Z 2 Z y=x2
16
y dA =
.
—
y dy dx =
5
RZ
0
y=0
2 Z
8 16
24
Hence,
min{y, x2 } dA = +
=
.
5
5
5
R
y dA. For each integral,
R2
Exercise 4(C) :
(a) The volume of the solid below the surface z = u2 (x, y) and above the region D is given by
ZZ
V2 =
u2 (x, y) dA.
D
Likewise,
Z Z the volume of the solid below the surface z = u1 (x, y) and above the region D is given by
V1 =
u1 (x, y) dA. Since the volume of E is the difference of these, we have that
D
ZZ
u2 (x, y) − u1 (x, y) dA.
V =
D
(b) (i) The solid is below z = 3 and above z = y. Moreover, its projection on xy-plane is the region D
enclosed by y = x2 and y = 3. Therefore, its volume equals
ZZ
Z √3 Z y=3
3 − y dA = √
(3 − y) dy dx
− 3
√
3
D
Z
=
√
− 3
√
24 3
=
5
y=x2
9
x4
− 3x2 +
dx
2
2
(ii) The solid is below z = 0 and above z = x. Moreover, its projection on xy-plane is the region D
enclosed by x + y = 2, y = x, x = 0. Therefore, its volume equals
ZZ
Z 1 Z y=x
0 − x dA =
−x dy dx
D
y=2−x
0
Z
=
1
−2x2 + 2x dx
0
=
1
3
End of Homework 5.