95下
微積分甲統一教學一組期末考題暨參考答案2007.6.20
1. (20%) Consider the transformation u(x, y) = x + y, v(x, y) = x
y, which maps R2+ = {(x, y), x > 0, y > 0} onto R2+ = {(u, v), u > 0, v > 0}.
(a) Find the inverse of the above transformation. That is, solve (x, y) in terms of (u, v).
(b) Evaluate the Jacobian ∂(x, y)
∂(u, v). (c) Evaluate the double integral I2 =
¨
R2+
x + y y
35
e−2x−2ydA.
Ans. x = , y = ,
∂(x, y)
∂(u, v) = , I2 = .
Solution:
(a) u = x + y
v = xy ⇒ x = 1+vuv y = 1+vu (b) ∂(x,y)∂(u,v) =
v 1+v
1 u 1+v
(1+v)2 −(1+v)u 2
= −(1+v)uv 3 −(1+v)u 3 = −(1+v)u 2
(c)
I2 = ˆ ∞
0
ˆ ∞ 0
(v + 1)35e−2u· u
(1 + v)2 dudv
= ˆ ∞
0
(v + 1)−75 dv · ˆ ∞
0
u · e−2udu
= (−5
2(v + 1)−25 |∞0 )(−1 2
ˆ ∞ 0
u de−2u)
= −5 2· (−1
2)(u · e−2u− (−1
2e−2u) |∞0 ) = 5 4· 1
2 = 5 8
2. (20%) Consider the vector field F =
2xyz
x2y + z + y(cos x)ey sin x , x2z
x2y + z + (sin x)ey sin x, z
x2y + z + ln(x2y + z)
, which is defined on D+= {(x, y, z), x > 0, y > 0, z > 0}.
(a) Find a potential function f (x, y, z) of F if F is conservative on D+, or show that F is not conservative on D+.
(b) Evaluate the line integral ˆ
C
F·T ds = ˆ
C
F·dr, where C is the line segment from (π, 1, 1) to π2, 8, 2.
Ans. (a) F is conservative and f (x, y, z) = ,
or F is not conservative because ,
(b) ˆ
C
F · T ds = .
Solution:
(a) Let F = (M, N, P ) F is conservative ⇔ ∂N∂x = ∂M∂y,∂P∂y = ∂N∂z,∂P∂x = ∂M∂z
∂N
∂x = 2xz(x2y + z) − x2z(2xy)
(x2y + z)2 + cosx · ey sin x+ sin x · y cos x · ey sin x= ∂M
∂y
∂P
∂y = x4y
(x2y + z)2 = ∂N
∂z
∂M
∂z = 2x3y2
(x2y + z)2 = ∂P
∂x
∴ F is conservative f =
ˆ
M dx =
ˆ 2xyz
x2y + z + y(cos xey sin x) dx
= z
ˆ 2xy
x2y + zdx + ˆ
ey sin xd(y sin x) = z · ln(x2y + z) + ey sin x+ h(y, z)
∂f
∂y = x2z
x2y + z + sin x · ey sin x+∂h
∂y ⇒ ∂h
∂y = 0, so we let h(y, z) = g(z)
∂f
∂z = ln(x2y + z) + z · 1
x2y + z ⇒ g0(z) = 0
⇒ f (x, y, z) = z ln(x2y + z) + ey sin x+ C.
(b) ´
CF · T ds = F (π2, 8, 2) − F (π, 1, 1) = 2 ln 2 + ln(π2+ 1) + e8− 1.
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3. (12%) Given a > 0, consider the iterated integral
I(a) = ˆ a
0
ˆ √a2−x2
−√ a2−x2
ˆ √a2−x2−y2 0
ze−(x2+y2+z2)dz dy dx.
By Fubini’s theorem I(a) =
˚
D(a)
ze−(x2+y2+z2)dV . Specify the region of inte- gration D(a), and evaluate I(a).
Ans. D(a) = ,
I(a) = .
Solution:
(a) D(a) = {(x, y, z) | x2+ y2+ z2 ≤ a2, x > 0, z > 0}.
(b)
I(a) = ˆ π
2
0
ˆ π
2
−π
2
ˆ a 0
ρ cos φ · e−ρ2 · ρ2sin φ dρ dθ dφ
= ˆ a
0
ρ3e−ρ2dρ · ˆ π
2
−π2
dθ · ˆ π
2
0
cos φ sin φ dφ
= 1
2(−ρ2e−ρ2 − e−ρ2) |a0 ·π · (sin2φ 2 ) |
π 2
0
= 1
2(−a2e−a2 − e−a2 + 1) · π · 1 2
= π
4(1 − a2e−a2 − e−a2)
4. (15%) Apply Green’s theorem to find the area of the region bounded by the curve parametrized by r(t) = (cos t, sin3t), 0 ≤ t ≤ 2π.
Ans. The area is .
Solution:
By Green’s theorem, Let M = −y, N = x.
¨
D
dA = 1 2
ˆ
C
−y dx + x dy
= 1 2
ˆ 2π 0
− sin3t d cos t + cos t d sin3t
= 1 2
ˆ 2π
0
sin4t + 3 sin2t cos2t dt
= 1 2
ˆ 2π 0
3 sin2t − 2 sin4t dt
= 1 2
ˆ 2π
0
(3 · 1 − cos 2t
2 − 1
2+ cos 2t − 1 + cos 4t 4 ) dt
= 1 8
ˆ 2π 0
3 − 2 cos 2t − cos 4t dt
= 1
8 · 6π = 3π 4
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5. (15%) Evaluate I5 =
˛
Γ
(y + sin x) dx + (z2 + cos y) dy + x3dz, where Γ is the closed curve r(t) = (sin t, cos t, sin 2t), 0 ≤ t ≤ 2π, oriented by increasing t.
Hint. You may either evaluate the line integral directly, or apply Stokes’ theorem.
For the latter approach, use the relation sin 2t = 2 sin t cos t to find a surface on which the curve Γ lies.
Ans. I5 = .
Solution:
Let F = (y + sin x, x2+ cos y, x3) ∇ × F = −2zi − 3x2j − k I5 =
ˆ F dr
= ˆ 2π
0
(cos t + sin(sin t), sin2(2t) + cos(cos t), sin3t) · (cos t, − sin t, 2 cos 2t) dt
= ˆ 2π
0
cos2+ cos t · sin(sin t) − sin t sin22t − sin t · cos(cos t) + 2 cos 2t · sin3t dt
= ˆ 2π
0
cos2+ cos t · sin(sin t) − sin t · cos(cos t)
− sin t · (4 sin2t cos2t) + 2(2 cos2t − 1) · sin3t dt
= ˆ 2π
0
cos2+ cos t · sin(sin t) − sin t · cos(cos t) − 2 sin3t dt
= 1 + sin 2t
2 − cos(sin t) + sin(cos t) + 2 cos t − 2
3cos3t |2π0
= π
6. (18%) Let a > 0, b, c ∈ R, b < c, and S be the part of the cylinder S = {(x, y, z), x2 + y2 = a2, b ≤ z ≤ c}, which is oriented with the unit normal pointing toward the z-axis. Evaluate the surface integral I6 =
¨
S
x3dydz + x2y dzdx + x2z dxdy.
Ans. I6 = .
Solution:
Let F = x3i + x2yj + x2zk, divF = 3x2+ x2+ x2 = 5x2. By divergence theorem
˚
I
5x2dxdydz = −I6+
¨
x2+y2≤a,z=c
x2z dxdy −
¨
x2+y2≤a2,z=b
x2z dxdy.
I6 = − ˆ c
b
ˆ 2π 0
ˆ a 0
5r2cos2θ · r drdθdz + ˆ 2π
0
ˆ a 0
cr2cos2θ · r drdθ
− ˆ 2π
0
ˆ a
0
br2cos2θ · r drdθ
= 5a4π
4 (b − c) + a4π
4 c − a4π 4 b
= πa4(b − c)
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