Math 234 Worksheet 3: Line integrals I
1. Let C be the helix with theRparametrization ~x(t) = (cos t, sin t, t) and 0 ≤ t ≤ 2π.
Compute the line integral C xyzds. (Hint: use the double angle formula for sine
and integration by parts.)
2. Let F~ (x, y) = (ex , xy) be a vector field. Let C be the curve enclosing the region
bounded by y = x and y = x2 with the counter clockwise orientation.
(i) We may decompose the curve C by C = C1 − C2 (i.e. C = C1 + (−C2 )),
where both C1 and C2 are curves from (0, 0) to (1, 1). Find parametrizations of C1 and C2 .
R
R
(ii) Compute C1 F~ · d~x and C2 F~ · d~x by using the parametrizations.
H
(iii) Compute
F~ · d~x.
C
(iv) Is F~ a conservative vector field?
3. Continued from Question 2.
(i) Repeat the above questions (ii) and (iii) with the vector field F~ (x, y) =
(cos x + 2xy, x2 ).
(ii) Verify that F~ is a conservative vector field, and then find a function f such
~ .
that F~ = ∇f
R
(iii) Compute again C1 F~ · d~x by using the fundamental theorem of calculus
for line integrals.
Review 1 (Line integrals) Let C be a curve with a parametrization ~x(t), a ≤ t ≤ b.
The line integral of a function f over the curve C is given by
Z b
Z
f (~x(t))k~x 0 (t)kdt.
f ds =
a
C
The line integral of a vector field F~ over the curve C is given by
Z b
Z
~
F~ (~x(t)) · ~x 0 (t)dt.
F · d~x =
a
C
Review 2 (Fundamental theorem of calculus) Let C be a curve with the initial
point A = ~x(a) and the terminal point B = ~x(b). Then,
Z
~ · d~x = f (B) − f (A).
∇f
C
This shows that the line integral of a gradient vector field is independent of the path
taken (this is not necessarily the case for a non-gradient vector field).
Review 3 (Conservative vector field) A vector field F~ is called conservative if
I
F~ · d~x = 0
C
for any closed curve C. It was shown that F~ is conservative if and only if F~ is a
~ for some function f .
gradient vector field, i.e. F~ = ∇f
Answers
√ Z
1. 2
√
2π
t cos t sin tdt = −
0
2.
2
π.
2
(i) C1 is the path from (0, 0) to (1, 1) along the curve y = x2 . We may
parametrize C1 by ~x(t) = (t, t2 ) for 0 ≤ t ≤ 1.
C2 is the path from (0, 0) to (1, 1) along the curve y = x. We may
parametrize C2 by ~x(t) = (t, t) for 0 ≤ t ≤ 1.
(ii)
Z 1 t Z
3
1
e
~
dt = e − .
F · d~x =
2 ·
2t
t·t
5
0
C
Z 1 t Z 1
2
e
1
·
dt = e − .
F~ · d~x =
t·t
1
3
0
C2
I
Z
Z
3
2
1
~
~
F · d~x =
F · d~x −
F~ · d~x = (e − ) − (e − ) = .
(iii)
5
3
15
C
C1
C2
H
(iv) No, since
F~ · d~x 6= 0. We could also argue by showing that Py 6= Qx .
C
3.
(i)
1
cos t + 2t · t2
1
·
dt = sin 1 + 1.
2
t
2t
0
C1
Z 1
Z
cos
t
+
2t
·
t
1
F~ · d~x =
·
dt = sin 1 + 1.
2
t
1
0
C2
Z
F~ · d~x =
I
Thus
Z
F~ · d~x =
C
Z
C1
F~ · d~x −
Z
F~ · d~x = 0.
C2
(ii) It is conservative since Py = 2x = Qx . f (x, y) = sin x + x2 y.
Z
Z
~ · d~x = f (1, 1) − f (0, 0) = sin 1 + 1.
~
∇f
F · d~x =
(iii)
C1
C1
2