MAT2007: Elementary Real Analysis II
Assignment #4
Reference Solution
。
1. Let O(0, 0, 0), A(1, 0, 0), B(1, 1, 0) and C(1, 1, 1) be four points in R3 . Evaluate the line
integral
!
y 2 dx + xydy + zxdz
Γ
for the following curves Γ
(i) Γ is the line segment OC joining from O to C;
(ii) Γ is the polygonal line consisting of OA, AB, BC.
(iii) Does the results in parts (i) and (ii) equal? Is the linear differential form a closed
one?
Solution. (i) A parametrization of Γ is (x(t), y(t), z(t)) = (t, t, t) 0 ≤ t ≤ 1. Hence, the line
integral is
! 1
I=
[t2 + t2 + t2 ]dt = 1.
0
(ii) The curve Γ can be split into three parts, Γ = Γ1 + Γ2 + Γ3 , with the parametrizations
Γ1 :
Γ2 :
Γ3 :
(x(t), y(t), z(t)) = (t, 0, 0),
(x(t), y(t), z(t)) = (1, t, 0),
(x(t), y(t), z(t)) = (1, 1, t),
0 ≤ t ≤ 1,
0 ≤ t ≤ 1,
0 ≤ t ≤ 1.
Hence, the line integral is evaluated as
!
!
!
I=
+
+
y 2 dx + xydy + zxdz
Γ
Γ2
Γ
! 11
! 1 3
! 1
=
0dt +
tdt +
tdt
0
0
0
=1.
(iii) The two line integrals are equal. If we denote the differential form by L = y 2 dx +
xydy + zxdz = Adx + Bdy + Cdz, then
∂A
= 2y,
∂y
∂B
= y.
∂x
1
The linear differential form L = y 2 dx + xydy + zxdz is not a closed one since the integrability
condition Ay = Bx does not hold.
8
2. Let F = (y, −x, x + y + z) be a force field. Find the work done by the force when an
object moves from A(a, 0, 0) to B(a, 0, c) along the different paths:
c
(i) x = a cos t, y = a sin t, z =
t, (0 ≤ t ≤ 2π);
2π
(ii) the line segment AB.
Solution. (i) The work done by the force is now given by the line integral
!
I1 =
ydx − xdy + (x + y + z)dz
Γ1
! 2π "
c #
=
(a sin t)(−a sin t) + (−a cos t)(a cos t) + (a cos t + a sin t +
t) dt
2π
0
=2πa2 + πc
(ii) The line segment has a parametrization Γ2 := (a, 0, t), 0 ≤ t ≤ c. In this case, the
work done by the force is given by
!
I2 =
ydx − xdy + (x + y + z)dz
Γ2
! c
=
[0 − 0 + (a + 0 + t)]dt
0
1
=ac + c2 .
2
3. Find the area of the domain enclosed by the lemniscate
(x2 + y 2 )2 = a2 (x2 − y 2 ),
a > 0.
Hint. 1) Area of the region enclosed by the simple closed curve C is given by
!
1
xdy − ydx.
2 C
2) Use the symmetry to reduce the problem to the first quadrant, and find a parametrization
of C using the polar coordinates.
Solution. By the symmetry, the area is 4A with A being the area in the first quadrant and
!
1
A=
xdy − ydx,
2 C
2
where C is the lemniscate in the first quadrant C1 and parts of the axis 0 ≤ x ≤ a, C2 . On
C1 , using the polar coordinates, we have
r2 = a2 (cos2 θ − sin2 θ) = a2 cos(2θ),
0≤θ≤
π
.
4
Hence C1 has the parametrization
C1 : x(θ) = a[cos(2θ)]1/2 cos θ,
y(θ) = a[cos(2θ)]1/2 sin θ,
0≤θ≤
π
.
4
The parametrization of C2 is simply
C2 : x(t) = t,
y(t) = 0,
0 ≤ t ≤ a.
Thus
!
1
xdy − ydx +
xdy − ydx
2 C2
C1
! π
1 4$ 2
=
a [cos(2θ)]1/2 cos θ([cos(2θ)]−1/2 (− sin(2θ)) sin θ + [cos(2θ)]1/2 cos θ)
2 0
%
− a2 [cos(2θ)]1/2 sin θ([cos(2θ)]−1/2 (− sin(2θ)) cos θ − [cos(2θ)]1/2 sin θ) dθ
!
1 a
+
(t · 0 − 0 · 1)dt
2 0
! π
a2 4
=
cos(2θ)dθ
2 0
a2
=
4
1
A=
2
!
And thus the area is
4A = a2 .
o
4. Evaluate the line integral
!
ydx + zdy + xdz,
C
where C is the curve from (a, 0, 0) to (0, 0, c) implicitly given by
x2 y 2 z 2
+ 2 + 2 = 1,
a2
b
c
x z
+ = 1,
a c
x ≥ 0, y ≥ 0, z ≥ 0,
where a, b, c > 0 are constants.
Solution. On the curve,
z
c
=1−
x
a
and thus
(x − a2 )2
y2
+
= 1,
( a2 )2
( √b2 )2
3
so C has the parametrization
C:
x=
a a
+ cos θ,
2 2
b
y = √ sin θ,
2
z=
c c
− cos θ,
2 2
(0 ≤ θ ≤ π).
Now,
!
ydx + zdy + xdz
&'
()
( )
* )c c
*' b
*)c
*+
b
a
a a
√ sin θ
√ cos θ +
=
− sin θ +
− cos θ
+ cos θ
sin θ dθ
2
2 2
2 2
2
2
2
0
! π
! π
! π
2
2
ab
bc
ac 2
2
2
=− √
sin θ dθ − √
cos θ dθ +
sin θ dθ
2 0
2 0
2 0
ac
πb
= − √ (a + c).
2
4 2
!Cπ
。
5. Consider the linear differential form
L=
ydx − xdy
x2 + y 2
(i) Verify that L satisfies the integrability condition for D = R2 \ {(0, 0)}. Why this does
not necessarily imply that L is exact in D?
(ii) Let C be a simple smooth curve from (1, 0) to (0, 2) that lies entirely
! in the first
quadrant (except for the initial and final points). Evaluate the line integral
L.
C
Solution. (i)
Write L = Adx + Bdy with
A=
y
,
x2 + y 2
B=
−x
.
x2 + y 2
Now,
Ay =
(x2 + y 2 ) − y ∗ 2y
x2 − y 2
=
,
(x2 + y 2 )2
(x2 + y 2 )2
Bx =
−(x2 + y 2 ) − (−x) ∗ 2x
x2 − y 2
=
,
(x2 + y 2 )2
(x2 + y 2 )2
so Ay = Bx for (x, y) &= (0, 0) and L satisfies the integrability condition (and thus closed) in
D.
L is not necessary exact since the domain D is not simply-connected.
(ii) On the simply-connected domain D# = R2 \ (−∞, 0] the differential form L is exact,
Thus as along as the curve is in the first quadrant, the line integral of L depends only on
the initial and final point, and is independent of the particular course. We may choose the
curve to be
C = C1 + C2 ,
4
with
C1 : x = cos θ,
y = sin θ,
and
C2 : x = 0,
L=
C
!
L+
C1
!
L=
C2
!
π
2
0
π
2
1 ≤ t ≤ 2.
y = t,
Therefore,
!
0≤θ≤
2
2
[− sin θ − cos θ] dθ +
!
2
1
t·0+0·1
π
dt
=
−
.
02 + t2
2
6. Consider the linear differential form
。
L=
ydx − xdy
.
− 2xy + 3y 2
3x2
(i) Verify that L satisfies the integrability condition in the upper-half plane D = {(x, y) ∈
R2 ; y > 0}.
(ii) Find f (x, y) such that
! L = df in D.
Hint. Use the fact that
L is path-independent for any curve Γ in D.
Γ
Solution. (i)
Write L = Adx + Bdy with
A=
Now,
y
,
3x2 − 2xy + 3y 2
B=
−x
.
3x2 − 2xy + 3y 2
(3x2 − 2xy + 3y 2 ) − y(−2x + 6y)
3(x2 − y 2 )
Ay =
=
(3x2 − 2xy + 3y 2 )2
(3x2 − 2xy + 3y 2 )2
−(3x2 − 2xy + 3y 2 ) − (−x)(6x − 2y)
3(x2 − y 2 )
Bx =
=
,
(3x2 − 2xy + 3y 2 )2
(3x2 − 2xy + 3y 2 )2
We have Ay = Bx for y > 0 and thus L satisfies the integrability condition there.
(ii) Let (x, y) be a point with,y > 0. Since L is closed in the simply-connected D, L is
exact there, and the line integral C L only depends on the initial and final points as long as
C is contained in D. Now, choose C be a curve in D joining (0, 1) to (x, y) and define
!
f (x, y) =
L.
C
The curve C := C1 + C2 , with r = x2 + y 2 and
C1 : x(t) = 0, y(t) = t,
and
C2 :, x(t) = r cos θ,
y = r sin θ,
5
1≤t≤r
π
y
≤ θ ≤ arctan .
2
x
Thus
f (x, y) =
=
!
!
L=
C
! r
1
L+
C1
!
L
C2
t·0−0·1
dt +
0 + 0 + 3t2
! arctan y
!
arctan
π
2
y
x
− sin2 θ − cos2 θ
dθ
3 − 2 sin θ cos θ
1
dθ
π
3 − 2 sin θ cos θ
2
&
+
π
1
3y − x
= − √ − √ arctan √
4 2 2 2
2 2x
=0 −
x
。
7. Assume f (t) is continuous. Show that
!
f (x2 + y 2 )(xdx + ydy) = 0,
C
where C is a smooth closed curve.
Proof. Let x = x(t), y = y(t), a ≤ b be a parametrization of C, thus x(a) = x(b) and
y(a) = y(b). Now,
!
! b
2
2
f (x + y )(xdx + ydy) =
f [x(t)2 + y 2 (t)][x(t)x# (t) + y(t)y # (t)]dt
C
a
!
.
/
1 b
=
f [x(t)2 + y 2 (t)]d x(t)2 + y 2 (t)
2 a
! 2
2
1 x (b)+y (b)
=
f (u)du = 0
2 x2 (a)+y2 (a)
for x2 (a) + y 2 (a) = x2 (b) + y 2 (b).
Method II. Define
G(s) =
!
s
and
f (u)du
2
2
F (x, y) = G(x + y ) =
0
!
x2 +y 2
f (t)dt.
0
Since f is continuous, by the Fundamental Theorem of Calculus, G(s) is differentiable, thus
F (x, y) = 12 G(x2 + y 2 ) is also differentiable. The chain rule gives
1
1
dF = G# (s)ds = f (s)(2xdx + 2ydy) = f (x2 + y 2 )[xdy + ydy].
2
2
That is f (x2 + y 2 )[xdy + ydy] is an exact differential, and thus along any smooth closed
curve,
!
f (x2 + y 2 )(xdx + ydy) = 0.
C
6
。
8. Consider the linear differential form
L=
xdx − αydy
.
x2 + y 2
(i) Find the value of α such that L is closed in the domain D = R2 \ {(0, 0)}.
(ii) Prove that L is exact in D.
Solution. (i) Write L = Adx + Bdy with
A=
Now,
Ay =
x2
x
,
+ y2
B=
−2xy
,
(x2 + y 2 )2
−αy
.
+ y2
x2
Bx =
2αxy
.
(x2 + y 2 )
If L is closed, then Ay = Bx and thus α = −1 and
L=
(ii) Note that
L=
xdy + ydy
.
x2 + y 2
xdx + ydy
= df
x2 + y 2
with f = 12 ln(x2 + y 2 ) in D, and L is exact.
Method II. Show that along the unit circle C with counterclockwise orientation,
— End —
7
,
C
L = 0.
Assynment 4
11 20040066
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