1021微
微甲 07-11班
班期 末 考 解 答
√
√ 1. (15%) Let h(x) = x − x2 + sin−1 x , 0 ≤ x ≤ 1.
(a) Find the length of the curve y = h(x).
(b) Find the area of the surface generated by rotating the curve y = h(x) about the x-axis.
Solution:
h(x) =
√
√
x − x2 + sin−1 ( x)
r
1
√
1−x
1
−
2x
2(1
−
x)
2
x
⇒ h0 (x) = √
+ √
, 0 < x < 1.................(3pts)
=p
=
x
2 x − x2 2 1 − x
x(1 − x)
p
ds = 1 + [h0 (x)]2 ........................................................................................(3pts)
r
1
1
= √ ............................................................................................(2pts)
=
x
x
ˆ 1p
ˆ 1
√
1
0
2
√ dx = 2 x |10 = 2.....................................(2pts)
Length =
1 + [h (x)] dx =
x
0
0
ˆ 1
p
2πh(x) 1 + [h0 (x)]2 dx...................................................................(3pts)
Area of the surface =
0
ˆ 1 √
√
1
= 2π[ ( x − x2 + sin−1 ( x)) √ dx]
x
0
√
ˆ 1
ˆ 1
−1
√
sin ( x)
√
= 2π[
1 − xdx +
]dx
x
0
0
√
ˆ 1
3
−2
sin−1 ( x)
1
√
= 2π[
dx ]
(1 − x) 2 |0 +
3
x
0
ˆ 1
√ √
2
sin−1 ( x)d x ]
= 2π[ + 2
3
0
ˆ 1
1
√
√
√ 1
√
2
2 x
−1
= 2π[ + 2( x sin ( x) |0 −
x√
dx) ]
3
1−x
0
ˆ 1
2
1
−1
√
= 2π[ + 2 sin (1) −
dx ]
3
1−x
0
√
2
= 2π[ + π + 2 1 − x |10 ]
3
4
2
= 2π[ + π − 2 ] = 2π(π − ).................................................................(2pts)
3
3
Page 1 of 8
2. (10%) Solve xy 0 − 3y = 5x3
(a) with the initial condition y(1) = 2.
(b) with the initial condition y(−1) = 2.
Solution:
xy 0 − 3y = 5x3
3
⇒ y 0 − y = 5x2 (∗)
x
ˆ
And since
3
− dx = −3 ln |x| + K
x
Let
I = e−3 ln x = x−3 (3%)
(*) Multiply by I on both sides, then we have
(x−3 y)0 = 5x−1 ⇒ x−3 y = 5 ln |x| + C(3%)
2(a)
∵ y(1) = 2 ⇒ 2 = 0 + C ⇒ C = 2
∴ y = 5x3 ln |x| + 2x3 (2%)
2(b)
∵ y(−1) = 2 ⇒ −2 = 0 + C ⇒ C = −2
∴ y = 5x3 ln |x| − 2x3 (2%)
Page 2 of 8
π
π
3. (15%) Let y = h(x) be decreasing on [0, ) and is continuously differentiable on (0, ) with h(0) =
2
2
0. Let s(x) denote the arc length of y = h(x) from (0, 0) to (x, h(x)).
(a) Write down the formula for s(x).
ˆ
x
e−h(t) dt. Find the function h(x) explicitly.
(b) Suppose that s(x) is also given by s(x) =
0
(c) Find the function s(x) explicitly.
Solution:
(a)
ˆ
x
p
1 + (h0 (t))2 dt (3%)
s(x) =
0
(b)
ˆ
x
ˆ
p
0
2
1 + (h (t)) dt =
0
x
e−h(t) dt
0
0
−2h(x)
2
⇒ 1 + (h (x)) = e
(3%)
√
dy
y 0 = − e−2y − 1 =
(2%)
dx
ˆ
ˆ
1
√
−
dx =
dy (1%)
−2y
e
−1
Let e−y = sec θ ⇒ −e−y = sec θ tan θdθ
ˆ
tan θ
dθ
tan θ
= −θ + C
−x = −
⇒ x = θ + C = sec−1 (e−y ) + C
take into (x, y) = (0, 0), then
0 = sec−1 1 + C = C
Therefore
(1%)
e−y = sec x
⇒ −y = ln sec (−x)
⇒ h(x) = y = ln cos x (2%)
(c)
ˆ
s(x) =
x
sec t dt
0
= ln | sec x + tan x| (3%)
Page 3 of 8
4. (15%) Let Ω be the region bounded by y = cos x, y = 0, x = 0 and x =
(a) Find the volume of the solid obtained by revolving Ω about x-axis.
(b) Find the volume of the solid obtained by revolving Ω about y-axis.
(c) Find the centroid of Ω.
Solution:
ˆ π/2
ˆ π/2
2
(a) V =
πy dx = π
cos2 xdx (2%)
0
ˆ π/2 0
cos 2x + 1
dx (1%)
=π
2
0
sin 2x x π/2
= π(
+ )|0 (1%)
4
2
π2
(1%)
=
4
ˆ
ˆ
1
π/2
2
πx dy = −π
x2 (− sin x)dx (1%)
0
ˆ π/20
ˆ π/2
π/2
2
= −π(x cos x|0 − 2
x cos xdx) (1%) (or V = 2π
xydx)
0
ˆ π/20
π/2
sin xdx) (1%)
= 2π(x sin x|0 −
0
π
π/2
= 2π( + cos x|0 ) (1%)
2
= π 2 − 2π (1%)
(b) V =
ˆ
π/2
π/2
cos xdx = sin x|0 = 1(1%)
(c) A =
0
ˆ π/2
1
π
x̄ =
x cos xdx(1%) = − 1(1%)
A 0
2
ˆ π/2
1
1
π
ȳ =
cos2 xdx(1%) = (1%)
A 0
2
8
or use Pappus Theorem.
2πAx̄ = V (revolve about y-axis)
2πAȳ = V (revolve about x-axis)
Page 4 of 8
π
.
2
5. (10%) Find the area of the region that lies inside the curve r = 2 + cos 2θ but outside the curve
r = 2 + sin θ.
y
r  2  sin
x
r  2  cos2
Solution:
find intersection
2 + sin θ = 2 + cos 2θ
→ 2 sin2 θ + sin θ − 1 = 0
1
π π 5π
sin θ = −1or , θ = − , or
2
2 6 6
(2 pts)
ˆ π
1 6
Area =
(2 + cos θ)2 − (2 + sin θ)2 (3 pts)
2 ( 56 −2)π
ˆ π
6
(2 + cos θ)2 − (2 + sin θ)2 dθ)
(=
π
−
ˆ π2
6
=
4 cos 2θ + cos2 2θ − 4 sin θ − sin2 θdθ
−π
ˆ π2
6
1 + cos 4θ
1 − cos 2θ
4 cos 2θ +
=
− 4 sin θ −
dθ (3 pts)
2
2
− π2
π
9
1
6
= sin 2θ + sin 4θ + 4 cos θθ=−
π
2
4
8
√
51
=
3 (2 pts)
16
Page 5 of 8
6. (10%) Find the arc length of the curve x = t sin 2t, y = t cos 2t, 0 ≤ t ≤ 1.
Solution:
Since (+3 points)
0
x = sin 2t + 2t cos 2t,
y 0 = cos 2t − 2t sin 2t.
We have
0
x02 + y 2 = (sin 2t + 2t cos 2t)2 + (cos 2t − 2t sin 2t)2
= sin2 2t + cos2 2t + 4t2 sin2 2t + cos2 2t
= 1 + 4t2 .
The arc length is (+2 points)
ˆ
L :=
1
ˆ
p
02
02
x + y dt =
0
Let t =
Note
1
√
1 + 4t2 dt.
0
1
1
tan θ, then we have dt = sec2 θdθ and
2
2
ˆ −1
1 tan 2
L =
|sec θ| sec2 θdt
2 0
ˆ −1
1 tan 2
=
|sec θ|3 dt (+2 points)
2 0
1
−1 2
=
(tan θ sec θ + ln |sec θ + tan θ|) |tan
0
4
√
1 √
=
2 5 + ln 5 + 2 .(+3 points)
4
ˆ
ˆ
3
tan2 θ sec θdθ
sec θdθ = tan θ sec θ −
ˆ
sec2 θ − 1 sec θdθ
ˆ
ˆ
3
= tan θ sec θ − sec θ + sec θdθ.(+2 points)
= tan θ sec θ −
Hence
ˆ
ˆ
1
tan θ sec θ + sec θdθ
sec θdθ =
2
1
=
(tan θ sec θ + ln |sec θ + tan θ|) + C.
2
3
ˆ
Ps. If you didn’t write down the calculation of
sec3 but a wrong formula, you won’t get any
points.
Page 6 of 8
ˆ
3
2
3
2
7. (15%) Let P (x) = x + x + x + 1 and Q(x) = x − x + x + 1. Evaluate
Q(x)
dx. Note that
P (x)
P (−1) = 0.
Solution:
Note that P (x) = (x2 + 1)(x + 1). (2pts)
A
Bx + C
Q
=1+
+ 2
(5 pts)
P
x+1
x +1
⇒ A = B = −1, C = 1. (3 pts)
Hence
ˆ
ˆ
ˆ
ˆ
Q
−1
1−x
dx = 1dx +
dx +
dx
P
x+1
x2 + 1
1
= x − ln |x + 1| + tan−1 x − ln(1 + x2 ) + c(constant)(5 pts)
2
Page 7 of 8
ˆ
∞
8. (10%) Determine the values of α > 0 such that
1
ln x
dx is convergent.
xα
Solution:
1◦ For α 6= 1
ˆ
ln x
dx =
xα
ˆ
x1−α
1
x1−α
)=
ln x −
ln xd(
1−α
1−α
1−α
ˆ
x−α dx =
C
ˆ
∴
1
∞
ln x
dx = lim
t→∞
xα
ˆ
1
t
ln x
dx = lim
t→∞
xα
1
x1−α
ln x −
x1−α +
1−α
(1 − α)2
(5%)
1
t1−α
1
ln t −
t1−α +
2
1−α
(1 − α)
(1 − α)2
(case1) If 0 < α < 1
(1 − α) ln t − 1
lim
=∞
t→∞ (1 − α)2 tα−1
ˆ
∴
1
∞
ln x
dx is divergent.
xα
(1%)
(case2) If α > 1
1
(1 − α) ln t − 1 h ∞ i
t
=
lim
= 0 by L’Hospital Rule.
t→∞ (1 − α)2 (α − 1) tα−2
t→∞ (1 − α)2 tα−1
∞
ˆ ∞
ln x
∴
dx is convergent.
xα
1
lim
(1%)
2◦ For α = 1
ˆ
ˆ
1
ln x
dx = ln xd(ln x) = (ln x)2 + C
x
2
ˆ ∞
ˆ ∞
ln x
1
ln x
2
∴
dx = lim (ln t) = ∞ ⇒
is divergent.
t→∞
x
2
x
1
1
ˆ ∞
ln x
Thus,
dx is convergent for α > 1.
xα
1
Page 8 of 8
(2%)
(1%)