MA22S1: SOLUTIONS TO TUTORIAL 6
1. Compute the line integrals
R
(a) C 2xy ds
R
(b) C y dx − x dy
where C is the circle with centre (0, 0) and radius 2.
Solution: To compute these integrals we need to first parametrise
the circle C. (Note that line integrals with respect to arc length are
independent of the parametrisation so for (i) any parametrisation of
C will do). Let’s use
C : r(t) = h2 cos t, 2 sin ti, 0 ≤ t ≤ 2π
We have x(t) = 2 cos t and y(t) = 2 sin t.
(a) Note that
s
dx
dt
2
+
dy
dt
2
=
p
(−2 sin t)2 + (2 cos t)2 =
√
4=2
Thus the line integral is
Z
2xy ds =
2π
Z
C
2 x(t) y(t)
0
=
dx
dt
2
+
2π
Z
2(2 cos t)(2 sin t)(2) dt
0
= 8
s
Z
2π
sin 2t dt
0
2π
cos 2t
= 8 −
2
0
= 0
1
dy
dt
2
dt
2
MA22S1: SOLUTIONS TO TUTORIAL 6
(b) Note that
dx
dy
= −2 sin t and
= 2 cos t
dt
dt
Thus the line integral is
Z
y dx − x dy =
Z
=
Z
C
2π
y(t)
0
dx
dy
dt − x(t)
dt
dt
dt
2π
2 sin t(−2 sin t) dt − 2 cos t(2 cos t) dt
0
= −4
Z
= −4
Z
2π
sin2 t + cos2 t dt
0
2π
1 dt
0
= −8π
(Note that line integrals with respect to x and y depend on
the orientation of the curve C so if we instead used the reverse
orientation we would get 8π).
2. Use the component test to determine if the following vector field is
conservative:
F(x, y, z) = h y, x, xyz i
Solution: First we write
P (x, y, z) = y, Q(x, y, z) = x, R(x, y, z) = xyz
Then compute the partial derivatives
Since
∂Q
∂z
6=
∂R
∂y
and
∂P
∂y
=1
and
∂Q
∂x
=1
∂Q
∂z
=0
and
∂R
∂y
= xz
∂R
∂x
= yz
and
∂P
∂z
=0
∂R
∂x
6=
∂P
∂z
we conclude that F is not conservative.
MA22S1: SOLUTIONS TO TUTORIAL 6
3
3. Find a potential for the vector field
F(x, y, z) = h y 2 cos z, 2xy cos z, −xy 2 sin z i
and use it to compute the line integral
Z
F · dr
C
where C is the space curve with vector equation
r(t) = ht2 , sin t, ti, 0 ≤ t ≤ π.
Solution: We need to find a function f (x, y, z) such that
F(x, y, z) = ∇f (x, y, z)
Recall that ∇f (x, y, z) is the gradient of f
∇f (x, y, z) = hfx (x, y, z), fy (x, y, z), fz (x, y, z)i
Starting with the first components we require
fx (x, y, z) = y 2 cos z
Integrating both sides with respect to x gives
Z
fx (x, y, z) dx =
Z
y 2 cos z dx =⇒ f (x, y, z) = xy 2 cos z + c(y, z)
Note that the constant of integration c(y, z) is actually a function
of y and z.
Next we need to find c(y, z). Taking the partial derivative with
respect to y on both sides gives,
fy (x, y, z) = 2xy cos z + cy (y, z)
4
MA22S1: SOLUTIONS TO TUTORIAL 6
We require that this equals the second component of F(x, y, z),
2xy cos z + cy (y, z) = 2xy cos z
Hence
cy (y, z) = 0
Integrating both sides with respect to y gives
Z
cy (y, z) dy =
Z
0 dy =⇒ c(y, z) = d(z)
where again the constant of integration d(z) is actually a function
of z. Thus we have
f (x, y, z) = xy 2 cos z + c(y, z) = xy 2 cos z + d(z)
We need to find d(z). Taking the partial derivative with respect
to z on both sides gives,
fz (x, y, z) = −xy 2 sin z + d0 (z)
We require that this equals the third component of F,
−xy 2 sin z + d0 (z) = −xy 2 sin z
Hence
d0 (z) = 0
and so d(z) = k where k is a constant. Putting this together we
have
f (x, y, z) = xy 2 cos z + d(z) = xy 2 cos z + k
MA22S1: SOLUTIONS TO TUTORIAL 6
We can check to make sure f is a potential for F,
fx (x, y, z) = y 2 cos z
fy (x, y, z) = 2xy cos z
fz (x, y, z) = −xy 2 sin z









=⇒ ∇f = F
5