Leonhard Euler was able to find the exact sum of the series in Problem 5. ... In
this problem we ask you to prove this fact by evaluating the double integral in ...
CHALLENGE PROBLEMS
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CHALLENGE PROBLEMS: CHAPTER 12 A Click here for answers.
Click here for solutions.
S
1. If 冀x冁 denotes the greatest integer in x, evaluate the integral
yy 冀x ⫹ y冁 dA R
ⱍ
where R 苷 兵共x, y兲 1 艋 x 艋 3, 2 艋 y 艋 5其. 2. Evaluate the integral 1
yy 0
1
0
2
2
e max兵x , y 其 dy dx
where max 兵x 2, y 2 其 means the larger of the numbers x 2 and y 2. 3. Find the average value of the function f 共x兲 苷
xx1 cos共t 2 兲 dt on the interval [0, 1].
4. If a, b, and c are constant vectors, r is the position vector x i ⫹ y j ⫹ z k, and E is given by the
inequalities 0 艋 a ⴢ r 艋 ␣, 0 艋 b ⴢ r 艋 , 0 艋 c ⴢ r 艋 ␥, show that 共␣␥兲2
yyy 共a ⴢ r兲共b ⴢ r兲共c ⴢ r兲 dV 苷 8 a ⴢ 共b ⫻ c兲
ⱍ
E
ⱍ
1 dx dy is an improper integral and could be defined as the 1 ⫺ xy limit of double integrals over the rectangle 关0, t兴 ⫻ 关0, t兴 as t l 1⫺. But if we expand the integrand as a geometric series, we can express the integral as the sum of an infinite series. Show that
5. The double integral y
1
0
y
1
0
1
y y 0
1
0
⬁ 1 1 dx dy 苷 兺 2 1 ⫺ xy n苷1 n
6. Leonhard Euler was able to find the exact sum of the series in Problem 5. In 1736 he proved that ⬁
兺
n苷1
1 2 2 苷 n 6
In this problem we ask you to prove this fact by evaluating the double integral in Problem 5. Start by making the change of variables u⫺v s2
x苷
y苷
u⫹v s2
This gives a rotation about the origin through the angle 兾4. You will need to sketch the corresponding region in the uv-plane. [Hint: If, in evaluating the integral, you encounter either of the expressions 共1 ⫺ sin 兲兾cos or 共cos 兲兾共1 ⫹ sin 兲, you might like to use the identity cos 苷 sin共共兾2兲 ⫺ 兲 and the corresponding identity for sin .] 7. (a) Show that 1
1
yyy 0
0
1
0
⬁ 1 1 dx dy dz 苷 兺 3 1 ⫺ xyz n n苷1
(Nobody has ever been able to find the exact value of the sum of this series.) (b) Show that Thomson Brooks-Cole copyright 2007
1
1
yyy 0
0
1
0
⬁ 1 共⫺1兲 n⫺1 dx dy dz 苷 兺 1 ⫹ xyz n3 n苷1
Use this equation to evaluate the triple integral correct to two decimal places.
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CHALLENGE PROBLEMS
8. Show that
y
⬁
0
arctan x ⫺ arctan x dx 苷 ln x 2
by first expressing the integral as an iterated integral. 9. If f is continuous, show that x
y
z
0
0
0
yyy
x
f 共t兲 dt dz dy 苷 12 y 共x ⫺ t兲2 f 共t兲 dt 0
10. (a) A lamina has constant density and takes the shape of a disk with center the origin and radius
R. Use Newton’s Law of Gravitation (see Section 10.9) to show that the magnitude of the force of attraction that the lamina exerts on a body with mass m located at the point 共0, 0, d 兲 on the positive z -axis is
冉
F 苷 2 Gm d
1 1 ⫺ d sR 2 ⫹ d 2
冊
[Hint: Divide the disk as in Figure 4 in Section 12.3 and first compute the vertical component of the force exerted by the polar subrectangle Rij.] (b) Show that the magnitude of the force of attraction of a lamina with density that occupies an entire plane on an object with mass m located at a distance d from the plane is F 苷 2 Gm
Thomson Brooks-Cole copyright 2007
Notice that this expression does not depend on d.
CHALLENGE PROBLEMS
ANSWERS
Thomson Brooks-Cole copyright 2007
S
Solutions
1. 30
3.
1 2
sin 1
7. (b) 0.90
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CHALLENGE PROBLEMS
SOLUTIONS
E
Exercises
1.
V5
Let R =
Ri , where
i=1
Ri = {(x, y) | x + y ≥ i + 2, x + y < i + 3, 1 ≤ x ≤ 3, 2 ≤ y ≤ 5}.
UU
R
S5
[[x + y]] dA =
i =1
UU
[[x + y]] dA =
Ri
S5
i=1 [[x
+ y]]
[[x + y]] = constant = i + 2 for (x, y) ∈ Ri . Therefore UU
R
S5
[[x + y]] dA =
i=1
UU
Ri
dA, since
(i + 2) [A(Ri )]
= 3A(R1 ) + 4A(R2 ) + 5A(R3 ) + 6A(R4 ) + 7A(R5 ) = 3 12 + 4 32 + 5(2) + 6 32 + 7 12 = 30
3. fave = = = =
1 b−a
U 1U 1 0
x
U 1U t 0
U1 0
0
]
b
f (x) dx = a
1 1−0
]
]
1
0
1
x
cos(t2 ) dt dx
cos(t2 ) dt dx cos(t2 ) dx dt
t cos(t2 ) dt =
1 2
[changing the order of integration] 1 sin t2 0 =
1 2
sin 1
5. Since |xy| < 1, except at (1, 1), the formula for the sum of a geometric series gives ] 1] 0
0
1
1 dx dy = 1 − xy
] 1] 0
0
∞ S
=
n=0 ∞ S
=
n=0
1 S ∞
(xy)n dx dy =
n=0
]
n=0
] xn dx
1
0
∞ S
1
0
] 1] 0
1
∞ S 1 = (xy)n , so 1 − xy n=0
(xy)n dx dy
0
∞ S y n dy =
n=0
1 1 · n+1 n+1
∞ 1 S 1 1 1 1 = 2 + 2 + 2 + ··· = 2 2 (n + 1) 1 2 3 n=0 n
7. (a) Since |xyz| < 1 except at (1, 1, 1), the formula for the sum of a geometric series gives so ] 1] 1] 0
0
0
1
1 dx dy dz = 1 − xyz =
] 1] 1] 0
Thomson Brooks-Cole copyright 2007
=
0
∞ ] [
n=0
=
0
0
∞ 1[
(xyz)n dx dy dz =
n=0
] 1 xn dx
1
y n dy
0
]
0
1
∞ ] 1] 1] [
n=0
0
z n dz
0
0
∞ [
1 1 1 · · n + 1 n + 1 n + 1 n=0 ∞ [
∞ [ 1 1 1 1 1 = + + + · · · = 3 3 3 3 3 (n + 1) 1 2 3 n n=0 n=1
1
∞ [ 1 = (xyz)n , 1 − xyz n=0
(xyz)n dx dy dz
CHALLENGE PROBLEMS
(b) Since |−xyz| < 1, except at (1, 1, 1), the formula for the sum of a geometric series gives ∞ [ 1 = (−xyz)n , so 1 + xyz n=0
] 1] 1] 0
0
0
1
1 dx dy dz = 1 + xyz
=
] 1] 1] 0
0
∞ [
0
=
∞ [
0
(−1)n
]
(−1)n
n=0
=
1
0
n=0
(−xyz)n dx dy dz
n=0
∞ ] 1] 1] [
n=0
=
0
∞ 1[
(−xyz)n dx dy dz
1
0
] xn dx
y n dy
0
]
1
z n dz
0
1 1 1 · · n+1 n+1 n+1
∞ ∞ [ [ (−1)n (−1)n−1 1 1 1 = − + − · · · = 3 3 3 3 (n + 1) 1 2 3 n3 n=0 n=1
To evaluate this sum, we first write out a few terms: s = 1 − Notice that a7 =
1
1 1 1 1 1 + 3 − 3 + 3 − 3 ≈ 0.8998. 23 3 4 5 6
1 < 0.003. By the Alternating Series Estimation Theorem from Section 8.4, we have 73
|s − s6 | ≤ a7 < 0.003. This error of 0.003 will not affect the second decimal place, so we have s ≈ 0.90. 9.
U xU y U z 0
0
0
f (t) dt dz dy =
UUU
E
f (t) dV , where
E = {(t, z, y) | 0 ≤ t ≤ z, 0 ≤ z ≤ y, 0 ≤ y ≤ x}. If we let D be the projection of E on the yt-plane then D = {(y, t) | 0 ≤ t ≤ x, t ≤ y ≤ x}. And we see from the diagram that E = {(t, z, y) | t ≤ z ≤ y, t ≤ y ≤ x, 0 ≤ t ≤ x}. So U xU y U z 0
0
0
f (t) dt dz dy = = =
Thomson Brooks-Cole copyright 2007
=
U xU xU y 0
t
t
U x 1 2
0
U x 1 2
0
1 2
Ux 0
f (t) dz dy dt =
U x U x 0
t
(y − t) f(t) dy dt
y = x Ux y 2 − ty f (t) y = t dt = 0 12 x2 − tx − 12 t2 + t2 f (t) dt
Ux x2 − tx + 12 t2 f (t) dt = 0 12 x2 − 2tx + t2 f (t) dt
(x − t)2 f(t) dt
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