MATHEMATICS 226, FALL 2014, PROBLEM SET 6
Solutions1
Section 13.5, Question 11: 8 marks: 2 for setting up the quantity to be
minimized, 3 for the correct equations for the critical points, 3 for solving
these equations.
n
X
We want to minimize S =
(pexi + qe−xi − yi )2 . We look for critical points:
i=1
n
0=
X
∂S
=2
(pexi + qe−xi − yi )exi ,
∂p
i=1
n
X
∂S
(pexi + qe−xi − yi )e−xi ,
0=
=2
∂q
i=1
We rewrite it as the system of linear equations
n
n
X
X
2xi
e )p + nq =
exi yi
(
i=1
i=1
n
n
X
X
−2xi
e
)q =
e−xi yi
np + (
i=1
i=1
The solutions (using Cramer’s Rule) are
P
P
P
( ni=1 e−2xi )( ni=1 exi yi ) − n( ni=1 e−xi )
P
P
p=
( ni=1 e2xi )( ni=1 e−2xi ) − n2
P
P
P
( ni=1 e2xi )( ni=1 e−xi yi ) − n( ni=1 exi )
P
P
q=
( ni=1 e2xi )( ni=1 e−2xi ) − n2
Section 14.1, Question 17: - Not marked
RR
RR
Let D = {(x, y) : x2 + y 2 ≤ 1}. We have We have D 4x2 y 3 = D x = 0 by
symmetry, so that
ZZ
(4x2 y 3 − x + 5) = 5( area of D) = 5π.
D
1
c Laba. Not to be copied, used, or revised without explicit written permission from
I.
the copyright owner.
1
Section 14.1, Question 19: - Not marked
The integral represents the volume of the cone with vortex (0, 0, a) and base
D = {(x, y) : x2 + y 2 ≤ a2 } in the xy-plane. That volume is equal to πa3 /3.
Section 14.2, Question 15: 8 marks: 4 for setting up the correct integral
with reversed order of integration, 4 for evaluating it.
The domain of integration is the triangle D with vertices (0, 0), (0, 1), (1.1).
We reverse the order of integration:
Z 1
ZZ
Z 1Z x
2
−x2
−x2
e dA =
xe−x dx
e dy dx =
D
0
0
0
1 −x2 1
1
=− e = − (1 − e−1 )
2
2
x=0
Section 14.2, Question 21: - Not marked.
1
1−x
1
y 3 1−x
dx
V =
(1 − x − y )dy dx =
y−x y−
3 y=0
0
0
0
Z 1
Z 1
1
2
4 2
3
2
3
=
(1 − x − x + x − (1 − 3x + 3x − x ))dx =
− 2x2 + x3 dx
3
3
3
0
0
2
2
1
2 2 1
1
1
x − x3 + x4 = − + = .
=
3
3
3
3 3 3
3
x=0
Z
Z
2
Z
2
2
Section 14.2, Question 27: 8 marks: 4 for setting up the correct integral,
4 for evaluating it.
Let D be the solid in question and D1 the part of it that lies in the first
octant, then by symmetry
Z
a
√
Z
Vol(D) = 8Vol(D1 ) = 8
0
Z
=8
0
a
a2 −x2
√
a2 − x2 dy dx
0
x3 a 16a3
(a2 − x2 )dx = 8 a2 x −
=
3 0
3
Section 14.2, Question 29: 8 marks
2
We have
Z
x
x
Z
G(u)du =
a
Z
=
d
Z
du
f1 (u, t)dt =
a
Z
d
c
d
dt
f1 (u, t)du
c
a
Z
f (x, t) − f (a, t) dt = g(x) −
c
x
Z
d
f (a, t)dt.
c
Rx
d
The last integral does not depend on x, so differentiating in x we get dx
G(u)du =
a
g 0 (x) − 0 = g 0 (x).R On the other hand, by the Fundamental Theorem of Calx
d
culus we have dx
G(u)du = G(x). Thus G(x) = g 0 (x), as required.
a
Section 14.3, Question 5: 8 marks: 2 for the correct answer (divergent),
6 for the justification.
Z 1 Z ∞
x2 + y 2
dx.
Consider the integral
dy
(1 + x2 )(1 + y 2 )
0
1
• For 0 ≤ y ≤ 1, we have y 2 ≥ 0 and 1 + y 2 ≤ 2, so that
x2 + y 2
x2
x2
≥
≥
(1 + x2 )(1 + y 2 )
(1 + x2 )(1 + y 2 )
2(1 + x2 )
• For x ≥ 1, we have 1 ≤ x2 , so that
x2
x2
1
≥
≥ .
2
2
2(1 + x )
2(2x )
4
Hence
ZZ
Q
x2 + y 2
dA ≥
(1 + x2 )(1 + y 2 )
1
Z
Z
dy
0
1
∞
x2 + y 2
dx ≥
(1 + x2 )(1 + y 2 )
Z
1
Z
dy
0
1
∞
1
dx = ∞
4
so that the integral is divergent.
Section 14.3, Question 7: Not marked.
ZZ
−(|x|+|y|)
e
R2
Z
∞
Z
∞
dA = 4
−(|x|+|y|)
e
0
Z
dx dy = 4
0
∞
−x
e dx
0
∞ ∞ −x −y =4 −e −e = 4(1)2 = 4.
x=0
y=0
Clearly, this one is convergent.
3
Z
0
∞
e−y dy