Solved exercises line integral

Lebanese University - Faculty of Sciences
Section º
Chapter 3: Line integral
Solved Problems
Dr. Kamel ATTAR
attar.kamel@gmail.com
F 2020-2021 F

2Ú19
Exercises
Solutions
1 Exercises
2 Solutions
Dr. Kamel ATTAR | Chapter 4: Line integral | Solved Problems

3Ú19
Exercises
Solutions
. Exercise 1. Evaluate the following line integrals.
a)
Z
C
ydx + dy where C is the curve of equation x = t3
− t and y = t2
from the point (0, 0) to the
point (6, 4).
b)
Z
C
(x + 2y)dx + (x − y)dy where C is the part of the ellipse of equation x = 2 cos(t) and
y = 4 sin(t) for 0 ≤ t ≤
π
4
.
c)
Z
C
xydy where C is the portion of the ellipse of equation 4x2
+ 9y2
= 36 lying in the first
quadrant, oriented clockwise.
d)
Z
C
ydx − xdy where C is the portion of the curve y =
1
x
from the point (1, 1) to the point (2,
1
2
).
e)
Z
C
(ey
+ yex
)dx + (xey
+ ex
)dy where C is given parametrically x = sin

πt
2

and
y = ln(t) for 1 ≤ t ≤ 2.
Go to Solution

4Ú19
Exercises
Solutions
. Exercise 2. For each part below compute
Z
C
−
→
F · d
−
→
M
a) C is the line segment from (2; 3) to (0; 3) and
−
→
F = x~
i − y~
j
b) C is the line segment from (5; 0; 2) to (5; 3; 4) and
−
→
F = z~
i − y~
j + x~
k.
c) C is the curve from y = ex
from (2; e2
) to (0; 1) and
−
→
F = x2~
i − y~
j.
d) C is the part of the circle of radius 3 in the first quadrant from (3; 0) to (0; 3) and
−
→
F =~
i − y~
j
e) C is the part of the curve x = cos(y) from (1; 2π) to (1; 0) and
−
→
F = y~
i + 2x~
j.
Go to Solution
. Exercise 3. We consider the vector field
−
→
F (x, y) = (x2
− y)
~
i + (y2
+ x)
~
j . Evaluate
Z
AB
−
→
F · d
−
→
M along
a) a straight line from (0, 1) to (1, 2),
b) a straight line from (0, 1) to (1, 1) and then from (1, 1) to (1, 2),
c) the parabola y = x2
+ 1,
Here A = (0, 1) and B = (1, 2).
Go to Solution

5Ú19
Exercises
Solutions
. Exercise 4. We consider the vector field
−
→
F (x, y, z) = (3x2
− 6yz)
~
i + (2y + 3xz)
~
j + (1 − 4xyz2
)~
k . Evaluate
Z
C
−
→
F · d
−
→
M from (0, 0, 0)
to (1, 1, 1) along the following paths C:
a) x = t, y = t2
and z = t3
b) The straight lines from (0, 0, 0) to (0, 0, 1), then to (0, 1, 1), and then to (1, 1, 1),
c) The straight line joining (0, 0, 0) and (1, 1, 1),
Go to Solution
. Exercise 5. Find the work done in moving a particle once around an ellipse C in the xy plane, if the
ellipse has its center at the origin with equation
x2
16
+
y2
9
= 1 and if the force field is given by
−
→
F (x, y, z) = (3x − 4y + 2z)
~
i + (4x + 2y − 3z2
)
~
j + (2xz − 4y2
+ z3
)~
k .
Go to Solution

6Ú19
Exercises
Solutions
. Exercise 6. Evaluate the following line integrals using Green’s Theorem. Unless otherwise stated,
assume that all curves are oriented counterclockwise.
a)
I
C
2xydx + y2
dy where C is the closed curve formed by y =
x
2
and y =
√
x.
b)
I
C
xydx + (x + y)dy where C is the triangle with vertices (0, 0), (2, 0) and (0, 1).
c)
I
C
(e3
x + 2y)dx + (x3
+ sin y)dy where C is the rectangle with vertices (2, 1), (6, 1), (6, 4)
and (2, 4).
d)
I
C
ln(1 + y)dx −
xy
1 + y
dy where C is the triangle with vertices (0, 0), (2, 0) and (0, 4).
Go to Solution

7Ú19
Exercises
Solutions
. Exercise 7.
¬ Let C be the curve joining the points A and B with parametrization
~
M(t) =

1 + (t − 1)et

~
i +

2t + cos

πt
2

~
j + sin

πt
2

~
k , 0 ≤ t ≤ 1
We consider the vector field
−
→
F (x, y, z) = (2xy3
z4
+ 2 cos x)
~
i + (sin y + 3x2
y2
z4
)
~
j + (ez
+ 4x2
y3
z3
)~
k .
a) Find the coordinates of A and B.
b) Show that
−
→
F is conservative (gradient field), then find a scalar potential f of
−
→
F .
c) Deduce the circulation of
−
→
F along the curve C.
Let Γ1 be the part of the parabola y = x2
− 3x from the point (0, 0) to the point (4, 4) and Γ2 be
the line segment y = x from the point (4, 4) to the point (0, 0). We consider the vector field
−
→
V =

ex15
− 2y + cos(x3
)

~
i +

sin(y2018
) + x2
− ln(y2
+ 1)

~
j .
a) Let Γ = Γ1 ∪ Γ2, draw the curve Γ.
b) Using the formula of Green-Riemann, evaluate
I
Γ+
−
→
V · d
−
→
M
Go to Solution

8Ú19
Exercises
Solutions
Solution 1.
a)
308
15
b) −
9
2
− π
c) 4
d) ln(4)
e) −1 + ln 2
Go Back

9Ú19
Exercises
Solutions
Solution 2.
a) Parameterization
n
x = −2t + 2
y = 3 , 0 ≤ t ≤ 1 then
n
dx = −2dt
dy = 0
Integration
Z
C
−
→
F · d
−
→
M =
Z
(x~
i − y~
j) · (dx~
i + dy~
j) =
Z 1
0
(4t − 4) dt = −2
b) Parameterization
(
x = 5
y = 3t
z = 2t + 2
, 0 ≤ t ≤ 1 then
(
dx = 0
dy = 3dt
dz = 2dt
Integration
Z
C
−
→
F · d
−
→
M =
Z
(z~
i − y~
j + x~
k) · (dx~
i + dy~
j + dz~
k) =
Z 1
0
(−9t + 10) dt =
11
2
c) Parameterization

x = t
y = et , 0 ≤ t ≤ 2 then

dx = dt
dy = et
dt
Integration
Z
C
−
→
F · d
−
→
M =
Z
(x2~
i − y~
j) · (dx~
i + dy~
j) = −
Z 2
0
(t2
− e2t
) dt = −
19
6
+
1
2
e4
d) Parameterization

x = 3 cos(t)
y = 3 sin(t) , 0 ≤ t ≤
π
2
then

dx = −3 sin(t)dt
dy = 3 cos(t)dt
Z
C
−
→
F · d
−
→
M =
Z
(
~
i − y~
j) · (dx~
i + dy~
j) =
Z π
2
0
(−3 sin(t) − 9 sin(t) cos(t)) dt = −
15
2
e) Parameterization
n
x = cos(t)
y = t , 0 ≤ t ≤ 2π then
n
dx = − sin(t)dt
dy = dt
Z
C
−
→
F · d
−
→
M =
Z
(y~
i + 2x~
j) · (dx~
i + dy~
j) = −
Z 2π
0
(−t sin(t) + 2 cos(t)) dt = −2π.
Go Back

10Ú19
Exercises
Solutions
Solution 3.
a) An equation for the line joining (0, 1) and (1, 2) in the xy plane is y = x + 1, then the parametric
equation is given by x = t and y = t + 1 and so dx = dt, dy = dt and the line integral
Z
−
→
F · d
−
→
M =
Z 1
t=0
(2t2
+ 2t)dt =
5
3
b) Along the straight line from (0, 1) to (1, 1) x = t, y = 1, then dx = dt, dy = 0 and the line
integral
Z
−
→
F · d
−
→
M =
Z 1
t=0
(t2
− 1)dt = −
2
3
Along the straight line from (1, 1) to (1, 2) x = 1, y = t, then dx = 0, dy = dt and the line
integral
Z
−
→
F · d
−
→
M =
Z 2
t=1
(t2
+ 1)dt =
10
3
Then the required value is 8/3
c) The parametric equation of the parabola is x = t and y = t2
+ 1, with 0 ≤ t ≤ 1 the line integral
Z
−
→
F · d
−
→
M =
Z 1
t=0
(2t5
+ 4t2
+ 2t2
+ 2t − 1)dt = 2
Go Back

11Ú19
Exercises
Solutions
Solution 4. Z
C
−
→
F · d
−
→
M =
Z
C
(3x
2
− 6yz)dx + (2y + 3xz)dy + (1 − 4xyz
2
)dz
a) If x = t, y = t
2
and z = t
3
, then dx = dt, dy = 2tdt and dz = 3t
2
dt and points (0, 0, 0) and (1, 1, 1)
correspond to t = 0 and t = 1 respectively. Then
Z
C
−
→
F · d
−
→
M =
Z 1
0
(3t
2
− 6t
5
)dt + (4t
3
+ 6t
5
)dt + (3t
2
− 12t
11
)dt = 2
b)
• Along the straight line from (0, 0, 0) to (0, 0, 1) we have x = 0, y = 0 and z = t then dx = 0, dy = 0 and
dz = dt with 0 ≤ t ≤ 1. Then the integral over this part of the path is
Z 1
0
dt = 1.
• Along the straight line from (0, 0, 0) to (0, 1, 1) we have x = 0, y = t and z = 1 then dx = 0, dy = dt and
dz = 0 with 0 ≤ t ≤ 1. Then the integral over this part of the path is
Z 1
0
2tdt = 1.
• Along the straight line from (0, 1, 1) to (1, 1, 1) we have x = t, y = 1 and z = 1 then dx = dt, dy = 0 and
dz = 0 with 0 ≤ t ≤ 1. Then the integral over this part of the path is
Z 1
0
(3t
2
− 6)dt = −5.
• Adding,
Z
C
−
→
F · d
−
→
M = 1 + 1 − 5 = −3
c) The straight line joining (0, 0, 0) and (1, 1, 1) is given in parametric form x = t, y = t, z = t with 0 ≤ t ≤ 1.
Then Z 1
0
(3t
2
− 6t
2
)dt + (2t + 3t
2
)dt + (1 − 4t
4
)dt = 6/5
Go Back

12Ú19
Exercises
Solutions
Solution 5. In the plane z = 0,
−
→
F (x, y, z) = (3x − 4y)
~
i + (4x + 2y)
~
j − 4y2~
k . and
d
−
→
M = dx~
i + dy~
j. The parametric equation is of form x = 4 cos θ, y = 3 sin θ and z = 0, then
dx = −4 sin θdθ, dy = 3 cos θdθ and dz = 0 with 0 ≤ θ ≤ 2π. Then
Z
C
−
→
F · d
−
→
M =
Z 2π
0
(12 cos θ − 12 sin θ)(−4 sin θdθ) + (16 cos θ + 6 sin θ)(3 cos θdθ)
Go Back

13Ú19
Exercises
Solutions
Solution 6.
a)
I
C
2xydx + y2
dy =
ZZ
D

∂Q
∂x
−
∂P
∂y

dxdy
=
Z 4
0
dx
Z √
x
x/2
(−2x)dy
= −
64
15
b)
I
C
xydx + (x + y)dy =
ZZ
D

∂Q
∂x
−
∂P
∂y

dxdy
=
Z 2
0
dx
Z − x
2
+1
0
(1 − x)dy
=
1
3
Go Back

14Ú19
Exercises
Solutions
c)
I
C
(e3
x + 2y)dx + (x3
+ sin y)dy =
ZZ
D

∂Q
∂x
−
∂P
∂y

dxdy
=
Z 6
2
dx
Z 4
1
(3x2
− 2)dy
= 600
d)
I
C
ln(1 + y)dx −
xy
1 + y
dy =
ZZ
D

∂Q
∂x
−
∂P
∂y

dxdy
=
Z 2
0
dx
Z −2x+4
0

−
y
1 + y
−
1
1 + y

dy
= −4
Go Back

15Ú19
Exercises
Solutions
Solution 7.
¬ a) A = ~
r(0) = (0, 1, 0) and B = ~
r(1) = (1, 2, 1).
b)
−
→
∇ ×
−
→
F =

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