The University of Sydney
School of Mathematics and Statistics
Solutions to Tutorial (and Practice Class) for Week 3
MATH2021: Vector Calculus and Differential Equations Semester 1, 2023
Lecturers: Eduardo G. Altmann and Fernando Viera
Material covered
Vector fields F : Rn → Rn
Sketching vector fields in R2
Integral of a scalar valued function along a curve
Integral of a vector field along a curve
Summary of essential material
Vector fields A vector field in Rn is a function F : Rn → Rn. In the case n = 2 we can visualise
F : R2 → R2 by drawing the vector F (x, y) at the point (x, y) ∈ R2 in the plane.
Line integrals, I (scalar function) Let C be a curve in Rn, and let γ : [a, b]→ Rn be a parametri-
sation of the curve. The integral of the scalar valued function f : Rn → R along C is
∫
C
f ds =
∫ b
a
f(γ(t))‖γ ′(t)‖ dt.
Line integrals, II (vector fields) Let C be an oriented curve in Rn (ie, with a choice of start and
end), and let γ : [a, b] → Rn be a parametrisation of the curve consistent with orientation (that is,
γ(a) is the start, and γ(b) is the end). The integral of the vector field F : Rn → Rn along C is
∫
C
F · ds =
∫ b
a
F (γ(t)) · γ ′(t) dt.
Another notation for this integral is as follows. If n = 3 and F = (F1, F2, F3) then we write∫
C
F1 dx+ F2 dy + F3 dz =
∫
C
F · ds.
Copyright © 2023 The University of Sydney 1
Questions to complete during the tutorial
1. Sketch the following vector fields F : R2 → R2.
(a) F (x, y) = −2i+ 3j
Solution: The vector F (x, y) at (x, y) has tail at (x, y) and head at (x− 2, y+3). Here
is the picture:
(b) F (x, y) = (x,−y)
Solution: The following are generated on the computer. For example, google Wolfram
Alpha and type in:
VectorPlot[{x,-y},{x,-1,1},{y,-1,1}]
- 1.0 - 0.5 0.0 0.5 1.0
- 1.0
- 0.5
0.0
0.5
1.0
(c) F (x, y) = (−x,−y)
Solution:
2
- 1.0 - 0.5 0.0 0.5 1.0
- 1.0
- 0.5
0.0
0.5
1.0
2. Let C be the curve in R3 with parametrisation γ(t) = (t, 2t, t), t ∈ [0, 1]. Compute the following.
(a)
∫
C
y
√
1 + xz ds
Solution: This is an integral of a scalar valued function along a curve. We have
γ ′(t) = (1, 2, 1) for t ∈ [0, 1],
and so ‖γ ′(t)‖ = √1 + 4 + 1 = √6. Therefore
∫
C
y
√
1 + xz ds =
∫ 1
0
2t
√
1 + t2
√
6 dt.
Let u = 1 + t2, so that du = 2t dt. Then
∫
C
y
√
1 + xz ds =
√
6
∫ 2
1
u1/2 du =
2
√
6
3
(2
√
2− 1).
(b)
∫
C
(−y, x,−z) · ds
Solution: This is an integral of a vector field F (x, y, z) = (−y, x,−z) along C, and so
∫
C
F · ds =
∫ 1
0
(−2t, t,−t) · (1, 2, 1) dt
=
∫ 1
0
(−2t+ 2t− t) dt
= −1.
(c)
∫
C
xy dx+ z dy − y dz
Solution: This is an integral of the vector field F (x, y, z) = (xy, z,−y) along C, and so
∫
C
xy dx+ z dy − y dz =
∫ 1
0
(2t2, t,−2t) · (1, 2, 1) dt
=
∫ 1
0
(2t2 + 2t− 2t) dt
=
2
3
.
3
3. Find a parametrisation for each curve C, and use it to find
∫
C
f ds for each function f .
(a) C the straight line segment from (0, 1, 2) to (1, 0, 4), and f(x, y, z) = x2y + 2yz
Solution: A parametrisation of the straight line segment in R3 from P = (0, 1, 2) to
Q = (1, 0, 4) is given by γ(t) = (0, 1, 2) + t
−−→
PQ with t ∈ [0, 1], and so
γ(t) = (0, 1, 2) + t(1,−1, 2) = (t, 1− t, 2 + 2t).
Thus
ds = ‖γ ′(t)‖ dt =
√
12 + (−1)2 + 22 dt =
√
6 dt,
and so
∫
C
(x2y + 2yz) ds =
∫ 1
0
(t2(1− t) + 2(1− t)(2 + 2t))
√
6 dt
=
√
6
∫ 1
0
(4− 3t2 − t3) dt =
√
6(4− 1− 1
4
) =
11
√
6
4
.
(b) C the straight line segment from (2, 3, 1) to (0, 5, 5), and f(x, y, z) = ex+y+z.
Solution: A parametrisation of the straight line segment in R3 from P = (2, 3, 1) to
Q = (0, 5, 5) is given by γ(t) = (2, 3, 1) + t
−−→
PQ with t ∈ [0, 1], and so
γ(t) = (2− 2t, 3 + 2t, 1 + 4t); t ∈ [0, 1].
Thus
ds = ‖γ ′(t)‖ dt =
√
(−2)2 + 22 + 42 dt = 2
√
6 dt
and so
∫
C
ex+y+z ds =
∫ 1
0
e((2−2t)+(3+2t)+(1+4t))2
√
6 dt
= 2
√
6
∫ 1
0
e6+4t dt = 2
√
6
(
e6+4t
4
) ∣∣∣∣∣
1
0
=
√
6
2
(e10 − e6).
(c) C the part of the graph y = 13x
3 between x = 0 and x = 1, and f(x, y) =
√
x3y.
Solution: A parametrisation is given by γ(t) = (t, t3/3), t ∈ [0, 1]. Then
ds = ‖γ ′(t)‖ dt =
√
1 + t4 dt,
and so ∫
C
√
x3y ds =
∫ 1
0
√
t6/3
√
1 + t4 dt =
1√
3
∫ 1
0
t3
√
1 + t4 dt.
Make the change of variable u = 1 + t4. Then du = 4t3 dt, and so
∫
C
√
x3y ds =
1
4
√
3
∫ 2
1
√
u du =
1
6
√
3
(2
√
2− 1).
4. Let C be the triangle in the plane with vertices (0, 0), (1, 0) and (1, 2) oriented counterclockwise.
Evaluate the following integrals.
(a)
∫
C
(x2y,−x) · ds
Solution:
4
y
2
x1C1
C2
C3
We consider each edge of the triangle separately and write C = C1 ∪C2 ∪C3 as shown in
the figure.
Now γ(t) = (t, 0), t ∈ [0, 1], is a parametrisation of C1 consistent with the orientation of
C1. Hence ∫
C1
x2y dx− x dy =
∫ 1
0
t2 · 0 · 1− t · 0 dt = 0.
Next, γ(t) = (1, t), t ∈ [0, 2], is a parametrisation of C2 consistent with the orientation of
C2. Hence ∫
C2
x2y dx− x dy =
∫ 2
0
1 · t · 0− 1 · 1 dt = −2.
Finally, γ(t) = (t, 2t), t ∈ [0, 1], is a parametrisation of C3 against the orientation of C3.
Hence
∫
C3
x2y dx− x dy = −
∫ 1
0
t2(2t) · 1− t · 2 dt = −2
∫ 1
0
t3 − t dt
= −2
[ t4
4
− t
2
2
]1
0
= −1
2
+ 1 =
1
2
.
The total integral is the sum of the three integrals, so
∫
C
x2y dx− x dy = 0− 2 + 1
2
= −3
2
.
(b)
∫
C
(x− y) ds
Solution: With C1, C2 and C3 as above, we have∫
C
(x− y) ds =
∫
C1
(x− y) ds+
∫
C2
(x− y) ds+
∫
C3
(x− y) ds.
So we have
∫
C
(x− y) ds =
∫ 1
0
t dt+
∫ 2
0
(1− t) dt+
∫ 1
0
(t− 2t)
√
5 dt =
1−√5
2
.
(Note that we we do not have to put in an extra minus sign when doing the C3 integral,
even though the parametrisation is against orientation. This is because orientation is
irrelevant for integrals of scalar valued functions along curves).
5. Compute
∫
C
y dx+ z dy + x dz where C is given by (t, t2, 2t), t ∈ [0, 1].
Solution: Using the parametrisation of C given we have
∫
C
y dx+ z dy + x dz =
∫ 1
0
t2 · 1 + (2t)(2t) + t · 2 dt
=
∫ 1
0
5t2 + 2t dt =
5
3
t3 + t2
∣∣∣1
0
=
5
3
+ 1 =
8
3
.
5
Questions for the practice class
6. Let C be the ellipse x2/a2 + y2/b2 = 1, traversed once counterclockwise. Calculate∫
C
F · ds, where F (x, y) = (x+ y, y − x).
Solution: We use the parametrization x = a cos t, y = b sin t, 0 ≤ t ≤ 2pi. Hence the integral
equals ∫ 2pi
0
(
(a cos t+ b sin t)(−a sin t) + (b sin t− a cos t)(b cos t)
)
dt
=
∫ 2pi
0
−ab+ (b2 − a2) sin t cos t dt = −2piab.
7. Let C be the lower arc of the circle x2 + y2 = 2 between (1, 1) and (−1, 1), oriented such that
(1, 1) is the starting point and (−1, 1) the endpoint. Calculate
(a)
∫
C
x3y ds
Solution:
A parametrisation is given by
γ(t) = (
√
2 cos t,
√
2 sin t), t ∈ [3pi/4, 9pi/4].
This is orientation reversing: So in part (b) we need
to correct with a minus sign (but in (a) orientation is
not important).
y
1
x1−1
Now γ ′(t) = (−√2 sin t,√2 cos t), so ‖γ ′(t)‖ =
√
2 sin2 t+ 2 cos2 t =
√
2. Hence
∫
C
x3y ds =
∫ 9pi/4
3pi/4
(
√
2 cos t)3(
√
2 sin t)
√
2 dt
= 4
√
2
∫ 9pi/4
3pi/4
cos3 t sin t dt = −4
√
2
4
cos4 t
∣∣∣9pi/4
3pi/4
=
√
2
(( 1√
2
)4
−
(
− 1√
2
)4)
= 0.
(b)
∫
C
(x− y) dx+ (x+ y) dy.
Solution: Since γ was orientation reversing, we have∫
C
(x− y) dx+ (x+ y) dy
= −
∫ 9pi/4
3pi/4
(
√
2 cos t−
√
2 sin t)(−
√
2 sin t) + (
√
2 cos t+
√
2 sin t)(
√
2 cos t) dt
= −2
∫ 9pi/4
3pi/4
1 dt = −3pi.
8. A vector field F is called conservative if∫
C
F · ds = 0 for all closed curves C
(ie, paths that start and end at the same place). A vector field F is called path independent if∫
C1
F · ds =
∫
C2
F · ds
6
whenever C1 and C2 are curves with start(C1) = start(C2) and end(C1) = end(C2). Show that
F is conservative if and only if F is path independent.
Solution: Suppose that F is conservative. Let C1 and C2 be paths both starting at the same
place and ending at the same place. Then C = C1 ∪ (−C2) is a closed curve, and∫
C1
F ds−
∫
C2
F ds =
∫
C1
F · ds+
∫
−C2
F · ds as
∫
−C2
F · ds = −
∫
C2
F · ds
=
∫
C
F · ds as C = C1 ∪ (−C2)
= 0 as F is conservative by assumption,
and so ∫
C1
F ds =
∫
C2
F ds,
showing that F is path independent.
Conversely, suppose that F is path independent. Let C be any close curve. Let A and B be
any two points on C, and divide C into two curves C1 (joining A to B) and C2 (joining A to B
along the “other” arc of C). Then C = C1 ∪ (−C2), and we have∫
C
F · ds =
∫
C1
F · ds+
∫
−C2
F · ds
=
∫
C1
F · ds−
∫
C2
F · ds
= 0 as F is path independent,
and so F is conservative.
Extra questions for further practice
9. Sketch the following vector fields F : R2 → R2.
(a) F (x, y) = xi+ yj
Solution: The vector at (x, y) has tail at (x, y) and head at (2x, 2y).
(b) F (x, y) = (−x, x)
Solution: The following are generated on the computer. For example, google Wolfram
Alpha and type in:
VectorPlot[{-x,x},{x,-1,1},{y,-1,1}]
7
- 1.0 - 0.5 0.0 0.5 1.0
- 1.0
- 0.5
0.0
0.5
1.0
(c) F (x, y) = (−y, x)
Solution:
- 1.0 - 0.5 0.0 0.5 1.0
- 1.0
- 0.5
0.0
0.5
1.0
10. For the given vector fields F and curves C, evaluate the line integral
∫
C
F · ds.
(a) F = zi+ xj + yk and C the curve with parametrisation γ(t) = ti+ t2j + 3k, t ∈ [0, 1].
Solution:
∫
C
F · ds =
∫
C
z dx+ x dy + y dz =
∫ 1
0
(3 + 2t2 + 0) dt =
11
3
.
(b) F = x3i and C the part of the hyperbola x2− y2 = 1 in xy-plane starting at P = (1, 0, 0)
and ending at Q = (2,
√
3, 0).
Solution: The curve C has parametrisation γ(t) = (t,
√
t2 − 1), t ∈ [1, 2] (as it is part
of the graph y =
√
x2 − 1). Then γ ′(t) = (1, t√
t2−1), and so
∫
C
F · ds =
∫ 2
1
(t3, 0) · (1, t√
t2 − 1) dt =
∫ 2
1
t3dt = 15/4.
8
(c) F = eyi+ exj + ezk and C the curve traced out by γ(t) = (0, t, t2), t ∈ [0, ln 2].
Solution: Since γ ′(t) = (0, 1, 2t), t ∈ [0, ln 2], the line integral of F along C is
∫
C
F · ds =
∫ ln 2
0
(et, 1, et
2
) · (0, 1, 2t) dt
=
∫ ln 2
0
(1 + 2tet
2
) dt
=
[
t+ et
2
]ln 2
0
= ln 2 + e(ln 2)
2 − 1.
11. In each case calculate the work W done by the force field F along the given path C.
(a) F (x, y) = (2,−3xy) with C the straight line segment from A = (0, 0) to B = (2, 4).
Solution: The work done by F along a path C is
∫
C
F · ds =
∫
C
2 dx− 3xy dy.
Since the straight line segment running from A to B has the parametrisation γ(t) =
t(2i+ 4j), t ∈ [0, 1], one has that
W =
∫
C
F · ds =
∫ 1
0
(4 dt− 96t2 dt) = −28.
(b) F = (2,−3xy) with C be the path consisting of the straight line from A = (0, 0) to
D = (2, 0) followed by the straight line from D to B = (2, 4).
Solution: Let C1 be the straight line from A to D, and C2 the straight line from D to
B. Then ∫
C
F · ds =
∫
C1
F · ds+
∫
C2
F · ds.
The curve C1 has parametrisation γ1(t) = (2t, 0), t ∈ [0, 1], and so
W1 =
∫
C1
F · ds =
∫ 1
0
4 dt = 4.
The straight line from D to B has parametrisation γ(t) = (2, 4t), t ∈ [0, 1], and so
W2 =
∫
C2
F · ds =
∫ 1
0
4(−24)t dt = −48.
Therefore
W = W1 +W2 = 4− 48 = −44.
(c) F = (2,−3xy) with C be the piece of the parabola y = x2 from A = (0, 0) to B = (2, 4).
Solution: Along the curve C, y = x2, and thus a parametrisation of C is given by
γ(t) = (t, t2), t ∈ [0, 2].
Hence,
W =
∫
C
F · ds =
∫ 2
0
(2 − 6t4) dt = −172/5.
9
(d) F = (y,−x, 1) with C be the curve traced out by γ(t) = (cos t,− sin t, t2pi ), t ∈ [0, 2pi].
Solution: Since γ ′(t) = (− sin t,− cos t, 1/2pi), one has that
W =
∫ 2pi
0
F (γ(t)) · γ ′(t) dt
=
∫ 2pi
0
(
sin2 t+ cos2 t+
1
2pi
)
dt
= 2pi + 1.
12. Let C be the curve in R3 with parametrisation γ(t) = (t, 2t, t), t ∈ [0, 1]. Compute the following.
(a)
∫
C
xyz2 ds
Solution: This is an integral of a scalar valued function along a curve. We have
γ ′(t) = (1, 2, 1) for t ∈ [0, 1],
and so ‖γ ′(t)‖ = √1 + 4 + 1 = √6. So the line integral of f(x, y, z) = xyz2 along C is
∫
C
f ds =
∫ 1
0
2t4
√
6 dt = 2
√
6
[
t5
5
]1
0
=
2
√
6
5
.
(b)
∫
C
(x, z, y2) · ds
Solution: This is an integral of a vector valued function along C. From the definitions,
we have
∫
C
f · ds =
∫ b
a
f(γ(t)) · γ ′(t) dt
=
∫ 1
0
(t, t, 4t2) · (1, 2, 1) dt
=
∫ 1
0
(t+ 2t+ 4t2) dt
=
∫ 1
0
(3t+ 4t2) dt
=
[
3t2/2 + 4t3/3
]1
0
= 3/2 + 4/3
=
17
6
10