1
Mathematics 1c: Solutions, Homework Set 5
Due: Monday, May 10th at 10am.
1. (10 Points) Section 5.1, Exercise 4 Using Cavalieri’s principle, compute the
volume of the structure shown in Figure 5.1.11 of the textbook; each section is
a rectangle of length 5 and width 3.
Solution. By Cavalieri’s principle the volume of the solid in Figure 5.1.11
is the same as that of a rectangular parallelepiped of dimensions 3 × 5 × 7 or
(3)(5)(7) = 105.
2. (20 Points) Section 5.2, Exercise 8 Let f be continuous on R = [a, b]×[c, d].
For a < x < b, and c < y < d, define
Z xZ y
F (x, y) =
f (u, v)dv du.
a
c
Show that
∂2F
∂2F
=
= f (x, y).
∂x∂y
∂y∂x
Use this example to discuss the relationship between Fubini’s Theorem and the
equality of mixed partial derivatives.
Solution.
By the Fundamental Theorem of Calculus
Z y
∂F
(x, y) =
f (x, v)dv
∂x
c
and applying it once again, we have
∂2F
(x, y) = f (x, y).
∂y∂x
In the reverse order, we first apply Fubini’s Theorem and then the Fundamental Theorem of Calculus twice. Thus
Z yZ x
F (x, y) =
f (u, v)du dv
c
and
∂F
(x, y) =
∂y
a
Z
x
f (u, y)du
a
and then
∂2F
(x, y) = f (x, y).
∂x∂y
Fubini’s Theorem is, in a sense, the integral version of the theorem on the
equality of mixed partials. If
∂F ∂F ∂ 2 F
,
,
∂x ∂y ∂x∂y
and
∂2F
∂y∂x
2
are all continuous, then Fubini’s Theorem, and the Fundamental Theorem of
Calculus imply that
Z xZ y 2
Z yZ x 2
∂ F
∂ F
(u, v)dv du =
(u, v)du dv
∂x∂y
a
c ∂x∂y
Zc y a
∂F
∂F
=
(x, v) −
(a, v) dv
∂y
∂y
c
= F (x, y) − F (x, c) − F (a, y) + F (a, c).
A similar calculation of the iterated integral
Z xZ y 2
∂ F
(u, v)dv du
a
c ∂y∂x
gives the same answer and thus
Z xZ y 2
Z xZ y 2
∂ F
∂ F
(u, v)dv du =
(u, v)dv du.
∂y∂x
∂x∂y
a
c
a
c
This in turn implies that
ZZ
R
∂2F
dA =
∂x∂y
ZZ
R
∂2F
dA
∂y∂x
for all rectangles R. Since R is arbitrary, this implies that
∂2F
∂2F
=
.
∂x∂y
∂y∂x
This shows how one may deduce equality of mixed partials assuming Fubini’s
Theorem has been proved.
3. (10 Points)Section 5.3, Exercise 2(a). Evaluate and sketch the region of
integration
Z 2 Z y2
(x2 + y)dx dy.
−3
Solution.
0
The region of integration is shown in the Figure.
y
x = y2
y=2
x=4
y=–3
x
3
This region is x-simple but not y-simple, since it is bounded on the left and
right by graphs, but not top and bottom (unless we broke it into two pieces,
one above the x-axis and one below. The integral is
Z
2
−3
x=y2 !
Z 2 6
7
4 2
x3
y
y
y
3
dy =
+ xy + y dy =
+
3
3
21
4 −3
−3
x=0
=
27
37
24 34
27 + 37 65
+
+
−
=
− .
21 21
4
4
21
4
4. (10 Points)Section 5.4, Exercise 2(a). Find
Z 1Z 1
(x + y)2 dx dy.
−1
|y|
Solution. Changing the order of integration (see the Figure) we see that
this iterated integral is equal to
Z 1Z x
Z
x
1 1
(x + y)2 dy dx =
(x + y)3 −x dx
3 0
0
−x
Z
2
8 1 3
x dx = .
=
3 0
3
y


x
-
5. (10 Points)Section 5.4, Exercise 8. Compute the double integral
ZZ
f (x, y)dA
D
where
√
f (x, y) = y 2 x
and D is the set of (x, y) where x > 0, y > x2 , and y < 10 − x2 .
4
Solution.
√
Z
This integral is equal to the iterated integral
√
5 Z 10−x2
y
0
2√
x2
1
x dy dx =
3
Z
1
=
3
Z
0
√
5
h
i10−x2
dx
y 3 x1/2 2
x
5
0
1
(10 − x2 )3 x1/2 dx −
3
√
Z
5
x13/2 dx.
0
The second integral equals
1 2 √
− ( )( 5)15/2 .
3 15
To slightly simplify the first integral set u2 = x so that 2udu = dx. Changing
variables, the first integral equals
2
3
Z
51/4
(10 − u4 )3 u2 du
=
2
3
Z
51/4
[1000u2 − 100u6 + 10u10 − u14 ]du
51/4
2
u3
u 7 10 11 u15
=
1000 − 100 + u −
3
3
7
11
15 0
"
#
2 1000 3/4 100 7/4 10 11/4 515/4
5 −
5 + 5
−
.
=
3
3
7
11
15
6. (10 Points)Section 5.5, Exercise 15. Evaluate
ZZZ
(x2 + y 2 + z 2 )dx dy dz,
W
where W is the region bounded by x + y + z = a (where a > 0 is a given
constant), x = 0, y = 0, and z = 0.
Solution.
The set W is defined by the inequalities
x ≥ 0,
y ≥ 0,
and is shown in the Figure.
z ≥ 0,
and
0≤x+y+z ≤a
5
z
x+y+z=a
y
x+y=a
x
Thus,
ZZZ
(x2 + y 2 + z 2 )dx dy dz
Z a Z a−x Z a−x−y
=
dx
dy
dz(x2 + y 2 + z 2 )
0
0
0
Z a Z a−x 1
3
2
2
=
dx
(x + y )(a − x − y) + (a − x − y) dy
3
0
0
Z a
1
1 2
1
2
2
4
2
4
=
x (a − x) + (a − x) − x (a − x) + (a − x) dx
12
2
12
0
a
a
Z a
1
1
1
2 2
3
4
5
5
=
(a x − 2ax + x )dx +
(x − a) +
(x − a) 2 0
60
60
0
0
1 5 1 5
1 5
1 5
1 5
1 5
= a − a + a + a + a = a . ♦
6
4
10
60
60
20
W
7. (10 Points)Section 5.5, Exercise 16. Evaluate
ZZZ
z dx dy dz
W
where W is the region bounded by the planes x = 0, y = 0, z = 0, z = 1, and
the cylinder x2 + y 2 = 1, with x ≥ 0, y ≥ 0.
Solution.
Since W is defined by the inequalities
x2 + y 2 ≤ 1,
we have
ZZZ
x ≥ 0,
y ≥ 0,
ZZ
z dx dy dz =
W
Z
dx dy
x2 +y 2 ≤1
x≥0,y≥0
0
1
and
0 ≤ z ≤ 1,
π 1 2 1 π
zdz = · z = .
4 2 0
8
♦