MA22S1: SOLUTIONS TO TUTORIAL 7
1. Compute the curl and the divergence of the following vector field,
F(x, y, z) = x − y 2 , 2z + 1, x2
Solution:
curl F = ∇ × F
i
j
k
∂
∂
∂
= ∂x
∂y
∂z
x − y 2 2z + 1 x2
= h 0 − 2, 0 − 2x, 0 + 2y i
= −2 h 1, x, −y i
div F =
∂
∂
∂
(x − y 2 ) +
(2z + 1) + (x2 )
∂x
∂y
∂z
= 1
2. A potential for the vector field
F(x, y, z) = h y 2 cos z, 2xy cos z, −xy 2 sin z i
is given by
f (x, y, z) = xy 2 cos z
Use this to find the work done by F along the curve C with vector
equation
r(t) = h t2 , sin t, t i, 0 ≤ t ≤ π
1
2
MA22S1: SOLUTIONS TO TUTORIAL 7
Solution: By the fundamental theorem for line integrals we have
Z
F · dr =
Z
C
∇f · dr
C
= f (r(π)) − f (r(0))
= f (π 2 , 0, π) − f (0, 0, 0)
= 0
3. Find the work done by the force
F(x, y) = hx + y, x2 + y 2 i
along the triangle C with vertices (1, 0), (0, 1) and (−1, 0) where C
is traversed anti-clockwise.
Solution: Note that (by the component test) F is not a conservative
vector field. We break C into three line segments,
C1 : r1 (t) = h1 − t, ti , 0 ≤ t ≤ 1
C2 : r2 (t) = h−t, 1 − ti , 0 ≤ t ≤ 1
C3 : r3 (t) = h2t − 1, 0i , 0 ≤ t ≤ 1
First we find the work done along C1 ,
F · r01 (t) =
1, (1 − t)2 + t2 · h−1, 1i
= 2t2 − 2t
MA22S1: SOLUTIONS TO TUTORIAL 7
Z
F · dr =
Z
=
Z
=
C1
F · r01 (t) dt
C1
1
(2t2 − 2t) dt
0
2t3
− t2
3
= −
1
0
1
3
Next we find the work done along C2 ,
F · r02 (t) =
1 − 2t, t2 + (1 − t)2 · h−1, −1i
= 4t − 2t2 − 2
Z
F · dr =
Z
=
Z
C2
F · r02 (t) dt
C2
1
(4t − 2t2 − 2) dt
0
1
2t3
2
= 2t −
− 2t
3
0
= −
2
3
Finally we find the work done along C3 ,
F · r03 (t) =
2t − 1, (2t − 1)2 · h2, 0i
= 4t − 2
3
4
MA22S1: SOLUTIONS TO TUTORIAL 7
Z
F · dr =
Z
=
Z
C3
F · r03 (t) dt
C3
1
(4t − 2) dt
0
=
2
1
2t − 2t 0
= 0
Combining these three line integrals we get
Z
C
F · dr =
Z
F · dr +
C1
1 2
= − − +0
3 3
= −1
Z
C2
F · dr +
Z
C3
F · dr