CIVE50006 Mathematics
Solutions to Progress Test 1 – Sample Test
1. Consider the real function
(↵,) =
↵3 + 3↵2 3↵2 3
6
where (↵,) define the cartesian coordinate system.
(a) Show that (↵,) is harmonic.
(b) Determine the real function (↵,) such that f(z) = + i .
(c) Find the most general expression for the complex function f(z).
Solution:
(a) To be harmonic, the function (↵,) must satisfy the Laplace equation
@2
@↵2
+
@2
@2
= 0
Calculating the first and second partial derviatives
@
@↵
=
↵2 + 2↵ 2
2
@
@
=
↵2 2↵ 2
2
) @
2
@↵2
= ↵+ ) @
2
@2
= ↵
) LHS = ↵+ ↵ = 0
) LHS = RHS
Therefore, the Laplace equation is satisfied and (↵,) is harmonic.
(b) As (↵,) is harmonic, it must form part of an analytic function f(z) = + i , and therefore,
we can use the Cauchy-Riemann equations to determine the function (↵,).
@
@↵
=
@
@
@
@
= @
@↵
) =
Z
@
@↵
d ) =
Z
@
@
d↵
) = 1
2
Z
(↵2 + 2↵ 2) d ) = 1
2
Z
(↵2 2↵ 2) d↵
) = 3↵
2 + 3↵2 3
6
+ g(↵) ) = ↵
3 + 3↵2 + 3↵2
6
+ h()
) g(↵) = ↵
3
6
+ c ) h() =
3
6
+ c
) (↵,) = 3↵
2 + 3↵2 ↵3 3
6
+ c
1
CIVE50006 Mathematics Solutions to Progress Test 1 – Sample Test
(c) We need to determine the most general function f(z), therefore, this needs to be expressed in
terms of z.
f(z) = + i
) f(z) = ↵
3 + 3↵2 3↵2 3
6
+ i
✓
3↵2 + 3↵2 ↵3 3
6
+ c
◆
We know that z = ↵+ i
that z3 = ↵3 3↵2 + i 3↵2 3
and that iz3 = 3↵2 3 + i ↵3 + 3↵2
) f(z) = (1 i) z
3
6
+ ic
2
CIVE50006 Mathematics Solutions to Progress Test 1 – Sample Test
2. Consider a vector p, a scalar field B and a vector field M expressed as
p(t) = 3t2 i+ tanh (t) j t sin(2t)k
B(x, y, z) = x2y + cos (xy) + ex/z
M(x, y, z) = 2 sinh (x) i ln (y2) j+ z5/2 k
(a) Determine dp/dt.
(b) Calculate rB.
(c) What type of vector field is M? Show your working.
Solution:
(a)
dp
dt
=
dp1
dt
i+
dp2
dt
j+
dp3
dt
k
) dp
dt
= 6t i+ sech2 (t) j (2t cos (2t) + sin (2t))k
(b)
rB = @B
@x
i+
@B
@y
j+
@B
@z
k
) rB =
✓
2xy y sin (xy) + 1
z
ex/z
◆
i+ (x2 x sin (xy)) j x
z2
ex/z k
(c) Vector fields can be conservative (curl (M) = 0) or solenoidal (div (M) = 0). Let’s calculate
curl (M) and div (M) to determine what type of vector field we have.
curl (M) = r⇥M
) curl (M) =

i j k
@/@x @/@y @/@z
2 sinh (x) ln (y2) z5/2

) curl (M) =

@
@y
⇣
z5/2
⌘
@
@z
ln (y2) i


@
@x
⇣
z5/2
⌘
@
@z
(2 sinh (x))

j
+

@
@x
ln (y2) @
@y
(2 sinh (x))

k
) curl (M) = 0
Let’s now calculate div (M)
div (M) = r.M
) div (M) = @
@x
(2 sinh (x)) +
@
@y
ln (y2)+ @
@z
⇣
z5/2
⌘
) div (M) = 2 cosh (x) 2
y
+
5
2
z3/2
M is a conservative vector field, as curl (M) = 0
3
CIVE50006 Mathematics Solutions to Progress Test 1 – Sample Test
3. Directly evaluate the following double integrals
(a)
ZZ
P
du dv (b)
ZZ
P
uv du dv
over the region P defined as the first quadrant of a circle with radius a that is centred at the origin
and where (u, v) respresent the cartesian coordinate system.
Solution:
Let’s first sketch the region P , which is common to both parts of the question.
u
v
O a
a
P
As this is part of a circle, the easiest way to determine the double integrals is to transform the (u, v)
variables into polar coordinates (r, ✓).
u = r cos ✓ v = r sin ✓
@u
@r
= cos ✓
@v
@r
= sin ✓
@u
@✓
= r sin ✓ @v
@✓
= r cos ✓
The Jacobian is given by
J(r, ✓) =
@v@r · @u@✓ @u@r · @v@✓

) J(r, ✓) =
r sin2 ✓ r cos2 ✓
) J(r, ✓) = |r| = r
The transformed region S is given by
r
✓
O a
⇡
2
S
4
CIVE50006 Mathematics Solutions to Progress Test 1 – Sample Test
Now let’s evaluate the specific integrals for each part of the question.
(a)
RR
P
du dv
Ia =
ZZ
P
du dv
) Ia =
ZZ
S
J(r, ✓) dr d✓
) Ia =
Z ⇡/2
0
Z a
0
r dr d✓
) Ia =
Z ⇡/2
0

r2
2
a
0
d✓
) Ia =
a2
2
Z ⇡/2
0
d✓
) Ia =
a2
2
[✓]⇡/20
) Ia =
⇡a2
4
(b)
RR
P
uv du dv
Ib =
ZZ
P
uv du dv
) Ib =
ZZ
S
r cos ✓ r sin ✓ J(r, ✓) dr d✓
) Ib =
Z ⇡/2
0
Z a
0
r3 cos ✓ sin ✓ dr d✓
) Ib =
Z ⇡/2
0

r4
4
a
0
cos ✓ sin ✓ d✓
) Ib =
a4
4
Z ⇡/2
0
cos ✓ sin ✓ d✓
Now sin(2✓) ⌘ 2 cos ✓ sin ✓
) Ib = a
4
8
Z ⇡/2
0
sin(2✓) d✓
) Ib =
a4
8

cos(2✓)
2
⇡/2
0
) Ib =
a4
8
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