1052微微微甲甲甲01-04班班班期期期末末末考考考解解解答答答和和和評評評分分分標標標準準準
1. (12%) Determine whether the statement is true of false. Fill T (true) or F (false) in the blanks. If the statement is false, write down a reason, or give a correct statement, or find a counterexample.
(a) Let F(x, y) = P (x, y) i + Q(x, y) j. If ∂Q
∂x = ∂P
∂y throughout the domain of F(x, y), then the line integrals of F(x, y) is independent of path on the domain.
(b) Let f (x, y) be a smooth function. Suppose that a smooth curve C gives an orientation from initial point p to terminal point q. If −C denotes the curve consisting of the same points as C but with the opposite orientation (from initial point q to terminal point p), then
ˆ
−C
f (x, y) ds = − ˆ
C
f (x, y) ds.
(c) For a unit circle C : x2+ y2= 1, we have
˛
C
x dy = 0 by symmetry.
(d) Any smooth function f (x, y, z) satisfies div(∇f) = 0.
Solution:
(a) F (1 point)
The domain of F(x, y) should be open simply-connected (2 points) (b) F (1 point)
ˆ
−C
f (x, y)ds = ˆ
C
f (x, y)ds (2 points) (c) F (1 point)
˛
C
xdy =the area of the unit circle= π (2 points) (d) F (1 point)
A counterexample by f (x, y, z) = x2, then div(∇f) = 2 (2 points)
2. (12%) Evaluate the following integrals.
(a) ˆ 1
0
ˆ 1 y
tan(x2) dx dy.
(b) ˆ 1
1
√2
ˆ x
√1−x2
1 dy dx + ˆ
√2
1
ˆ x 0
1 dy dx + ˆ 2
√2
ˆ
√4−x2
0
1 dy dx.
Solution:
(a) ˆ
tan x2dx is not an elementary function, so we must change the order of integration.
ˆ 1 0
ˆ 1 y
tan x2dxdy
= ˆ 1
0
ˆ x 0
tan x2dydx (3pts)
= ˆ 1
0
x tan x2dx (1pt)
= 1
2ln
sec x2
1
0
= 1
2ln sec(1)
(2pts)
(b) The region is easily described using polar coordinates.
ˆ 1
1
√2
ˆ x
√1−x2
1dydx + ˆ
√2
1
ˆ x 0
1dydx + ˆ 2
√2
ˆ
√4−x2
0
1dydx
= ˆ π4
0
ˆ 2 1
rdrdθ (4pts)
= ˆ 2
1
rdr ˆ
π 4
0
dθ
= 1
2r2
2 1· θ
π 4
0
= 3
8π (2pts)
3. (12%) Evaluate the following integrals.
(a)
˚
Ey cos((y − z)2) dV , where E is the solid tetrahedron bounded by four planes x = 1, y = 1, z = 0, and x + y − z = 1, as Figure 1.
x
y z
Figure 1: The tetrahedron.
(b) ˆ √21
0
ˆ
√1−x2
x
ˆ 1+x2+y2 2√
x2+y2
1 dz dy dx.
Solution:
(a).
ˆ 1 0
ˆ y 0
ˆ 1
1−y+zy cos (y − z)2 dxdzdy (4%, including the correct order)
= ˆ 1
0
ˆ y
0 (y − z)y cos (y − z)2 dzdy
= ˆ 1
0
1
2y sin y2 dy
= 1 4−1
4cos 1 (2%) (b).
ˆ π/2 π/4
ˆ 1 0
ˆ 1+r2
2r 1 · rdzdrdθ (4%)
= π 4
ˆ 1 0
r 1 + r2− 2r dr
= π 48 (2%)
4. (10%) Evaluate the double integral
¨
R
e3x2+y2dA, where R is the region inside the ellipse 3x2+ y2= 1 and above the lines y = x and y = −√
3 x, as Figure 2.
x y
y = x y = −√
3 x 3x2+ y2= 1
R
Figure 2: The region R.
Solution:
Let x = 1
√3 · r cos θ, y = r sin θ
¨
R
e3x2+y2dxdy
= ˆ 3π/4
π/6
ˆ 1 0
er2 · 1
√3 · rdrdθ (2% for deformation, 2% for polar coordinate, 1% for Jacobian determinant, and 3% for correct intervals)
= 1
√3· 7
24π(e − 1)(2%)
5. (12%) Evaluate the surface integral
¨
S
px2+ y2dS, where S is the part of the surface z = tan−1y x
inside the circular cylinder x2+ y2= 1 and in the first octant.
Solution:
Step1.
We can write the parametric equations of S as
x = r cos θ y = r sin θ z = θ where the parameter domain is
D = {(r, θ)
0 < r ≤ 1, 0 ≤ θ ≤ π 2} and the vector equation is
r(r, θ) = r cos θi + r sin θj + θk (3pts)
Step2.
Find |rr× rθ|.
rr= cos θi + sin θj + 0k rθ= −r sin θi + r cos θj + 1k
⇒ rr× rθ= sin θi + (− cos θ)j + rk
⇒ |rr× rθ| =p
1 + r2 (3pts)
Step3.
Evaluate
¨
S
px2+ y2dS.
¨
S
px2+ y2dS =
¨
D
p(r cos θ)2+ (r sin θ)2· |rr× rθ|drdθ (3pts)
= ˆ
π 2
0
ˆ 1 0
rp
1 + r2drdθ
= ˆ 1
0
rp
1 + r2dr ˆ π2
0
dθ
= 1
3(1 + r2)32
1 0· θ
π 2
0
= π
6(232 − 1) (3pts)
6. (12%) Let F(x, y) = x + y
x2+ y2i + −x + y x2+ y2j.
(a) Is F(x, y) conservative on the half plane D = {(x, y)|x > 0}?
(b) Evaluate the line integral ˆ
C1
F•dr, where C1 is the part of the parabola y = (x − 2)2 from (2, 0) to (4, 4).
(c) Evaluate the line integral ˆ
C2
F•dr, where C2is the unit circle x2+ y2= 1 oriented counterclockwise. Is F(x, y) conservative on R2− {(0, 0)}?
Solution:
F =
x + y
x2+ y2,−x + y x2+ y2
(a)
(Method 1)
ˆ −x + y
x2+ y2 dy = 1
2ln(x2+ y2) − tan−1(y
x) + C(x) Let ∂
∂x
1
2ln(x2+ y2) − tan−1(y
x) + C(x)
= x + y x2+ y2
=⇒ x
x2+ y2−
−y x2
1 + (yx)2 + C′(x) = x + y x2+ y2
=⇒ C′(x) = 0
=⇒ C(x) = K is a constant.
We already find a function
φ(x, y) = 1
2ln(x2+ y2) − tan−1(y
x) + K (3%) such that
F = ∇φ for all (x, y) in D. Therefore, F is conservative on D. (1%)
(Method 2) Let
P (x, y) = x + y
x2+ y2 and Q(x, y) = −x + y x2+ y2
Since D is an open simply-connected region, P and Q have continuous first-order partial derivatives, and
∂P
∂y =x2− 2xy − y2 (x2+ y2)2 =∂Q
∂x (3%)
through out D, we conclude that F is conservative on D. (1%)
(b)
(Method 1) Since the path C1lies in D, we have ˆ
C1
F · dr = φ(4, 4) − φ(2, 0) (3%)
=3
2ln 2 − π
4 (1%) (Method 2) Because F is conservative on D,
ˆ
CF · dr is independent of path on D. We can choose paths α : 2 ≤ x ≤ 4, y = 0, and β : x = 4, 0 ≤ y ≤ 4 such that the path α ∪ β goes from (2, 0) to (4, 0) horizontally and then goes from (4, 0) to (4, 4) vertically. Therefore,
ˆ
C1
F · dr = ˆ
α
F · dr + ˆ
β
F · dr
= ˆ 4
x=2
1 x dx +
ˆ 4 y=0
−4 + y
16 + y2 dy (2%)
= ln |x|
4
2− tan−1(y 4)
4 0+1
2ln(16 + y2)
4
0 (1%)
= 3
2ln 2 −π
4 (1%)
(c) We can parametrize C2 : r(θ) = (cos θ, sin θ), 0 ≤ θ ≤ 2π, and the vector field becomes F(cos θ, sin θ) = (cos θ + sin θ, − cos θ + sin θ) on C2. (1%)
Then,
˛
C2
F · dr = ˆ 2π
θ=0(cos θ + sin θ, − cos θ + sin θ) · (− sin θ, cos θ)dθ
= ˆ 2π
θ=0− sin2θ − cos2θdθ = −2π (2%)
Remark: You can also observe that the normal vector (x, y) of the circle is perpendicular to (dx, dy), and then simplify
˛
C2
F · dr =
˛
C2
y
x2+ y2, −x x2+ y2
· (dx, dy) = ˆ 2π
θ=0− sin2θ − cos2θdθ = −2π.
Note that C2is a closed path on R2\{(0, 0)} but
˛
C2
F · dr 6= 0, which implies that ˆ
CF · dr is not independent of path on R2\{(0, 0)}. Therefore, F is not conservative on R2\{(0, 0)}. (1%)
7. (12%) Find the value k ∈ R such that the line integral I(k) =
ˆ
Ck
(1 + y2+ y exy) dx + (2x + y + x exy) dy achieves the minimum value, where Ck is the curve y = k sin x from (0, 0) to (π, 0).
Solution:
Let F(x, y) = (1 + y2+ yexy) i + (2x + y + xexy) j and C1: r(t) = ti where t from π to 0
P (x, y) = 1 + y2+ yexy, Q(x, y) = 2x + y + xexy
⇒ I(k) + ˆ
C1
F·dr =
¨
D
(∂P
∂y −∂Q
∂x)dA (5 points) To achieve I(k) minimum value is to achieve
¨
D
(∂P
∂y −∂Q
∂x)dA minimum value
¨
D
(∂P
∂y −∂Q
∂x)dA =
¨
D(2y − 2)dA
= ˆ π
0
ˆ k sin x
0 (2y − 2)dydx
= ˆ π
0
(k2sin2x − 2k sin x)dx
= π
2k2− 4k
= π
2(k − 4 π)2−8
π (6 points)
∴k = 4
π (1 point)
8. (13%) Compute the line integral
ˆ
C
z2dx − x2dy + 2yz dz,
where C is the curve of intersection of the upper half sphere z =p4 − x2− y2and the circular cylinder x2+ y2= 2y, orientated counterclockwise viewed from the above, as Figure 3.
x
y z
C
Figure 3: The curve C.
Solution:
Method I Straightforward Computation with F = z2, −x2, 2yz Solution 1 Parameterize C by r(t) =
cos t, 1 + sin t,√
2 − 2 sin t
, t from 0 to 2π.
Then F =
2 − 2 sin t, − cos2t, 2(1 + sin t)√
2 − 2 sin t r′(t) =
− sin t, cos t,√− cos t 2 − 2 sin t
So we compute ˆ
CF · dr = ˆ 2π
0 −2 sin t + 2 sin2t − cos3t − 2 cos t − 2 sin t cos t dt
= ˆ 2π
0
2 sin2t dt = ˆ 2π
0 1 − cos 2t dt = 2π.
Solution 2 Parameterize C by r(t) =
sin t, 1 + cos t, 2 sint 2
, t from 2π to 0.
Then F =
4 sin2 t
2, − sin2t, 4(1 + cos t) sint 2
r′(t) =
cos t, − sin t, cost 2
So we compute ˆ
CF · dr = ˆ 0
2π
4 cos t sin2 t
2 + sin3t + 4(1 + cos t) sin t 2cos t
2dt
= ˆ 0
2π
4 cos t sin2 t 2dt =
ˆ 0
2π2 cos t(1 − cos t) dt
= ˆ 2π
0
2 cos2t dt = ˆ 2π
0
1 + cos 2t dt = 2π.
Solution 3 Parameterize C by r(t) = sin 2t, 2 sin2t, 2| cos t|, t from 0 to π.
Then F = 4 cos2t, − sin22t, 8 sin2t| cos t| r′(t) =
2 cos 2t, 2 sin 2t,− sin 2t
| cos t|
So we compute ˆ
C
F · dr = ˆ π
0
8 cos 2t cos2t − 2 sin32t − 8 sin2t sin 2t dt
= ˆ π
0
8 cos 2t cos2t dt = ˆ π
0
4 cos 2t(1 + cos 2t) dt
= ˆ π
0
4 cos 2t + 4 cos22t dt = ˆ π
0
4 cos22t dt
= ˆ π
0
2 + 2 cos 4t dt = 2π.
Method II Applying the Stokes Theorem with F = z2, −x2, 2yz, curlF = (2z, 2z, −2x)
Solution 1 Let S be the portion of the sphere with boundary C. By Stokes’ Theorem, we have ˆ
C
F · dr =
¨
S
curlF · dS =
¨
D
curlF · n 1
|n · k|dxdy
=
¨
D
h(2z, 2z, −2x) ·x 2,y
2,z 2
i2
zdxdy, D : x2+ (y − 1)2≤ 1
=
¨
D
2y dxdy = ˆ 2π
0
ˆ 1 0
2(1 + r sin θ)r drdθ = ˆ 2π
0
ˆ 1 0
2r drdθ = 2π.
Solution 2 Let S be the portion of the sphere with boundary C. Parameterize S by r(x, y) where r(x, y) =
x, y,p4 − x2− y2
, x2+ (y − 1)2≤ 1 rx(x, y) = 1, 0, −x
p4 − x2− y2
!
ry(x, y) = 0, 1, −y p4 − x2− y2
!
rx× ry= x
p4 − x2− y2, y
p4 − x2− y2, 1
!
, pointing upward curlF =
2p4 − x2− y2, 2p4 − x2− y2, −2x
By Stokes’ Theorem, we have ˆ
CF · dr =
¨
ScurlF · dS =
¨
ScurlF · rx× rydxdy
=
¨
x2+(y−1)2≤1
2y dxdy = ˆ 2π
0
ˆ 1 0
2(1 + r sin θ)r drdθ = ˆ 2π
0
ˆ 1 0
2r drdθ = 2π.
Solution 3 Let S be the portion of the sphere with boundary C. Parameterize S by r(u, v) where r(u, v) =
u cos v, 1 + u sin v,p
3 − u2− 2u sin v
, 0 ≤ u ≤ 1, 0 ≤ v ≤ 2π ru(u, v) =
cos v, sin v, −u − sin v
√3 − u2− 2u sin v
rv(u, v) =
−u sin v, u cos v, −u cos v
√3 − u2− 2u sin v
ru× rv =
u2cos v
√3 − u2− 2u sin v, u + u2sin v
√3 − u2− 2u sin v, u
, pointing upward curlF =
2p
3 − u2− 2u sin v, 2p
3 − u2− 2u sin v, −2u cos v By Stokes’ Theorem, we have
ˆ
CF · dr =
¨
ScurlF · dS =
¨
ScurlF · ru× rvdudv
= ˆ 2π
0
ˆ 1 0
2u + 2u2sin v dudv = ˆ 2π
0
ˆ 1 0
2u dudv = 2π.
Solution 4 Let S be the portion of the sphere with boundary C. Parameterize S by r(u, v) where r(u, v) =
u cos v, u sin v,p 4 − u2
, 0 ≤ u ≤ 2 sin v, 0 ≤ v ≤ π ru(u, v) =
cos v, sin v, − u
√4 − u2
rv(u, v) = (−u sin v, u cos v, 0) ru× rv= u2cos v
√4 − u2, u2sin v
√4 − u2, u
, pointing upward curlF =
2p
4 − u2, 2p
4 − u2, −2u cos v By Stokes’ Theorem, we have
ˆ
CF · dr =
¨
ScurlF · dS =
¨
ScurlF · ru× rvdudv
= ˆ π
0
ˆ 2 sin v 0
2u2sin v dudv = 16 3
ˆ π 0
sin4v dv
= 16 3
ˆ π 0
1
8(2 − 4 cos 2v + 1 − cos 4v) dv = 2π.
Solution 5 Let S be the portion of the sphere with boundary C. Parameterize S by r(u, v) where r(u, v) = (2 sin u cos v, 2 sin u sin v, 2 cos u) , 0 ≤ u ≤ π
2, u ≤ v ≤ π − u.
ru(u, v) = (2 cos u cos v, 2 cos u sin v, −2 sin u) rv(u, v) = (−2 sin u sin v, 2 sin u cos v, 0)
ru× rv= 4 sin2u cos v, 4 sin2u sin v, 4 sin u cos u , pointing upward curlF = (4 cos u, 4 cos u, −4 sin u cos v)
By Stokes’ Theorem, we have ˆ
CF · dr =
¨
ScurlF · dS =
¨
ScurlF · ru× rvdudv
= ˆ
π 2
0
ˆ π−u
u
16 sin2u cos u sin v dvdu = ˆ
π 2
0
16 sin2u cos u [− cos v]|π−uu du
= ˆ π2
0
16 sin2u cos u [2 cos u] du = 8 ˆ π2
0
sin22u du = 4 ˆ π2
0 1 − cos 4u du = 2π.
評分標準：
1. 若直接計算線積分，寫出參數 r(t) 、計算F, r′(t) 並代入 ˆ
CF · dr 正確得7分，後續計算正確再得6分。 但若
r(t), F, r′(t) 中有錯誤、計算流程正確本題最多得7分。
2. 若使用 Stokes 定理則，寫出參數 r(u, v)、 計算curlF, ru, rv, ru× rv ( 或 n, dS ) 並代入 ˆ
S curlF · dS 正確 得8分，後續計算正確再得5分。 但若curlF, ru, rv, ru× rv 中有任何錯誤、計算流程正確本題最多得8分。
9. (15%) Consider the vector field F(x) = x
|x|3, that is, F(x, y, z) = x
(x2+ y2+ z2)32 i + y
(x2+ y2+ z2)32 j + z
(x2+ y2+ z2)32 k.
(a) Evaluate
¨
S1
F•dS, where S1 is the part of the sphere x2+ y2+ z2 = 1 inside the cone z =
rx2+ y2 3 with upward orientation.
(b) Evaluate
¨
S2
F•dS, where S2 is the part of the cone z =
rx2+ y2
3 between planes z = 1
2 and z = 2
√3 with outward orientation.
(c) Use the Divergence Theorem to evaluate
¨
S3
F•dS, where S3 is the part of the paraboloid z = 6 − x√2− y2 3 inside the cone z =
rx2+ y2
3 with upward orientation.
S1
S2
S3
Figure 4: Parts of sphere, cone, and paraboloid.
Solution:
(a) Method 1:
Let D1= {(u , v ) | u2+ v2≤ (
√3
2 )2}. Then S1can be deseribed as r(u, v) = (u , v ,p
1 − u2− v2 ) on D1. (2 pts) So
¨
S1
F · dS =
¨
D1
F(r(u, v)) · (ru× rv) dA. Note that
F(r(u, v)) = r(u, v) ru× rv = ( u
√1 − u2− v2 , v
√1 − u2− v2 , 1 ) ←(2 pts)
Thus
¨
S1
F · dS =
¨
D1
√ 1
1 − u2− v2 dA = ˆ 2π
0
ˆ
√3/2
0
√ r
1 − r2 drdθ = π. (2 pts) Method 2:
Observe that S1 can be desribed as
r(u, v) = (sin u cos v , sin u sin v , cos u) (2 pts) for 0 ≤ u ≤ π/3, 0 ≤ v ≤ 2π. Since
F(r(u, v)) = r(u, v) ru× rv = sin u · r(u, v), ← (2 pts) we obtain
¨
S1
F · dS = ˆ 2π
0
ˆ π/3
0 r(u, v) · sin u · r(u, v) dudv = π (2 pts) Method 3:
Observe that div(F) = 0. So by divergence theorem,
¨
S1
F · dS =
¨
R1
F · dS, where R1is the surface
r(u, v) = (u cos v , u sin v ,1
2 ) (2 pts) for 0 ≤ u ≤
√3
2 , 0 ≤ v ≤ 2π. Note that
F(r(u, v)) = (u2+ 1/4)−3/2· r(u, v) ru× rv= (0 , 0 , u ) ←(2 pts) Thus
¨
S1
F · dS = 1 2
ˆ 2π 0
ˆ
√3/2
0
u
(u2+ 1/4)3/2 dudv = π. (2 pts) (b) Method 1:
Let D2= {(u , v ) | (
√3
2 )2≤ u2+ v2≤ (2)2}. Then S2 can be described as
r(u, v) = (u , v ,
ru2+ v2
3 ) on D2. (2 pts) Note that
F(r(u, v)) = (4
3(u2+ v2))−3/2· r(u, v)
ru× rv = (3(u2+ v2))−1/2· (−u , −v ,p3(u2+ v2) ) ←(2 pts)
So ¨
S2
F · dS =
¨
D2
0 dA = 0. (1 pt) Method 2:
Observe that S2 can be described as
r(u, v) = (u cos v , u sin v , u
√3 ) (2 pts)
for
√3
2 ≤ u ≤ 2, 0 ≤ v ≤ 2π. Note that
F(r(u, v)) = (4
3u2)−3/2· r(u, v) rv× ru= ( 1
√3u cos v , 1
√3u sin v , −u) ← (2 pts)
So ¨
S2
F · dS =
¨
F (r(u, v)) · (rv× ru) dA = 0 (1 pt) Method 3:
Observe that div(F) = 0. So by divergence theorem,
¨
S2
F · dS =
¨
R1
F · dS −
¨
R2
F · dS, where R1is as in method 3, (a), and R2is the surface
r(u, v) = (u cos v , u sin v , 2
√3) (2 pts)
for 0 ≤ u ≤ 2, 0 ≤ v ≤ 2π. Note that
F(r(u, v)) = (u2+ 4/3)−3/2· r(u, v) ru× rv= (0 , 0 , u ) ←(2 pts) So
¨
R2
F · dS = 2
√3 ˆ 2π
0
ˆ 2 0
u
(u2+ 4/3)3/2 dudv = π, (1 pt) i.e.
¨
S2
F · dS = π − π = 0
(you will lose 1 pt if your answer is not 0 in this step) (c) Method 1:
Observe that, by the divergence theorem,
¨
S3
F · dS +
¨
S2
F · dS −
¨
S1
F · dS =
˚
S1+S2+S3
div(F ) dV.
↑(1 pt) ↑(1 pt)
Since div(F) = 0 (2 pts) by a simple calculation, we obtain
¨
S3
F · dS =
¨
S1
F · dS −
¨
S2
F · dS = π − 0 = π.
Method 2:
Since div(F) = 0 (2 pts), we have
¨
S3
F · dS =
¨
R2
F · dS, where R2is as in method 3, (b). Now again by method 3, (b), we obtain
¨
S3
F · dS =
¨
R2
F · dS = π. (2 pts)
Method 3:
Observe that S3 can be described as
r(u, v) = (u cos v , u sin v ,6 − u2
√3 ). (2 pts) for 0 ≤ u ≤ 2, 0 ≤ v ≤ 2π. Note that
F(r(u, v)) = (u4− 9u2+ 36
3 )−3/2· r(u, v) ru× rv= (2u2cos v
√3 ,2u2cos v
√3 , u).
Thus
¨
S3
F · dS = 6π ˆ 2
0
u3+ 6u
(u4− 9u2+ 36)3/2 du = π (2 pts)
