Line Integral Practice
Scalar Function Line IntegralsZwith Respect to
Z Arc Length
For each example below compute,
f (x, y) ds or
f (x, y, z) ds as appropriate.
C
Problems:
C
Z
1. C is the line segment from (1, 3) to (5, −2), compute
x − y ds
C
Z
z + y 2 ds.
2. C is the line segment from (3, 4, 0) to (1, 4, 2), compute
C
3. C is the curve from y = x2 from (0, 0) to (3, 9), compute
Z
3x ds.
C
Z
4. C is the upper half of the circle of radius 2 from (2, 0) to (−2, 0), compute
y ds.
C
Solutions:
1. (a) Parameterization: x = 1 + 4t, y = 3 − 5t, 0 ≤ t ≤ 1
Z
Z 1
√ Z
√
x − y ds =
[(1 + 4t) − (3 − 5t)] 42 + 52 dt = 41
(b) Integration:
C
0
0
−2 + 9t dt =
0
2. (a) Parameterization: x = 3 − 2t, y = 4, z = 2t, 0 ≤ t ≤ 1
Z
Z 1
√ Z
√
2
2
2
2
(b) Integration:
z + y ds =
[2t + 16] 2 + 0 + 2 dt = 8
C
1
1
√
2t + 16 dt = 17 8.
0
3. (a) Parameterization: x = t, y = t2 , 0 ≤ t ≤ 3
Z
Z
Z 3 p
3 37 √
1
2
2
(b) Integration:
3x ds =
3t 1 + (2t) dt =
u du = (373/2 − 1).
8 1
4
C
0
4. (a) Parameterization: x = 2 cos(t), y = 2 sin(t), 0 ≤ t ≤ π
Z
Z π
Z
p
2
2
(b) Integration:
y ds =
2 sin(t) (−2 sin(t)) + (2 cos(t))) dt = 4
C
0
0
π
sin(t) dt = 8.
5√
41.
2
Vector Function Line Integrals
Z
F · dr.
For each example below compute
C
Problems:
Z
F · dr
1. C is the line segment from (2, 3) to (0, 3) and F = hx, −yi, compute
C
Z
2. C is the line segment from (5, 0, 2) to (5, 3, 4) and F = hz, −y, xi, compute
F · dr
C
x
2
Z
2
3. C is the curve from y = e from (2, e ) to (0, 1) and F = hx , −yi, compute
F · dr
C
4. C is the part
Z of the circle of radius 3 in the first quadrant from (3, 0) to (0, 3) and F = h1, −yi,
F · dr
compute
C
Z
5. C is the part of the curve x = cos(y) from (1, 2π) to (1, 0) and F = hy, 2xi, compute
F · dr
C
Solutions:
1. (a) Parameterization: x = 2 − 2t, y = 3, 0 ≤ t ≤ 1
Z
Z 1
Z
h2 − 2t, −3i · h−2, 0i dt =
hx, −yi · dr =
(b) Integration:
C
1
−4 + 4t dt = −2.
0
0
2. (a) Parameterization: x = 5, y = 3t, z = 2 + 2t, 0 ≤ t ≤ 1
Z
Z 1
Z
(b) Integration:
hz, −y, xi · dr =
h2 + 2t, −3t, 5i · h0, 3, 2i dt =
C
0
1
−9t + 10 dt =
0
11
.
2
t
3. (a) Parameterization: x = t, y = e , 0 ≤ t ≤ 2. NOTE: The parameterization I’ve given has the
wrong orientation. So we can either change the parameterization (change all t’s to −t’s), or
just note that this is the parameterization of −C and change the sign of what we get. That
is what I will do below.
Z
Z
Z 2
Z 2
2
2
2
t
t
hx , −yi·dr = −
hx , −yi·dr = −
ht , −e i·h1, e i dt = −
t2 −e2t dt =
(b) Integration:
C −C
0
0
8 1 4 1
19 1 4
−
− e +
=− + e .
3 2
2
6
2
4. (a) Parameterization: x = 3 cos(t), y = 3 sin(t), 0 ≤ t ≤ π2 . NOTE: This parameterization has
the correct orientation.
Z
Z π/2
Z π/2
(b) Integration:
h1, −yi · dr =
h1, −3 sin(t)i · h−3 sin(t), 3 cos(t)i dt =
−3 sin(t) −
C
15
9 sin(t) cos(t) dt = · · · = − .
2
0
0
5. (a) Parameterization: x = cos(t), y = t, 0 ≤ t ≤ 2π. NOTE: The parameterization I’ve given
has the wrong orientation. So we can either change the parameterization (change all t’s to
−t’s), or just note that this is the parameterization of −C and change the sign of what we
get. That is what I will do below.
Z
Z 2π
Z 2π
(b) Integration:
hy, 2xi·dr = −
ht, 2 cos(t)i·h− sin(t), 1i dt = −
−t sin(t)+2 cos(t) dt =
· · · = −2π.
C
0
0