Math 227 Problem Set V Solutions
If C is a curve with parametrization r(t), a ≤ t ≤ b, then we define the line integral
Z
f ds =
R
C
b
f r(t) dr
dt (t) dt
a
C
1. Evaluate the line integral
Z
f (x, y, z) ds for
(a) f (x, y, z) = x cos z, C the curve with parametrization r(t) = tı̂ı + t2̂, 0 ≤ t ≤ 1.
2 3/2
(b) f (x, y, z) = x+y
, t , 1 ≤ t ≤ 2.
y+z , C the curve with parametrization r(t) = t, 3 t
√
Solution. (a) In this case r(t) = tı̂ı + t2̂, so that v(t) = ı̂ı + 2t̂ and ds
1 + 4t2 . Hence
dt =
Z
f (x, y, z) ds =
C
Z
1
0
x(t) cos z(t) ds
dt
dt =
Z
1
0
(b) In this case r(t) = t, 32 t3/2 , t , so that v(t) = 1, t1/2 , 1 and
Z
2
ds
dt
=
=
53/2 −1
12
√
2 + t. Hence
Z
2
R
f (x, y) ds along the curve C given in polar coordinates by r = r(θ),
f (x, y, z) ds =
1
C
2. (a) Show that the integral
θ1 ≤ θ ≤ θ2 , is
2 3/2 1
1 (1+4t )
8
3/2
0
p
t(cos 0) 1 + 4t2 dt =
C
Z
θ2
θ1
x(t)+y(t) ds
y(t)+z(t) dt
dt =
Z
1
t+ 32 t3/2 √
2
2 3/2
+t
3t
+ t dt =
q
f r(θ) cos θ, r(θ) sin θ r(θ)2 +
2
(2+t)3/2 3/2 1
2
dr
dθ (θ)
8−33/2
3/2
=
dθ
(b) Compute the arc length of r = 1 + cos θ, 0 ≤ θ ≤ 2π.
Solution. (a) The curve is r(θ) = x(θ)ı̂ı + y(θ)̂ with x(θ) = r(θ) cos θ, y(θ) = r(θ) sin θ and 0 ≤ θ ≤ 2π.
On this curve
v(θ) = x′ (θ)ı̂ı + y ′ (θ)̂ = r′ (θ) cos θ − r(θ) sin θ ı̂ı + r′ (θ) sin θ + r(θ) cos θ ̂
q
2 2
ds
=⇒ dθ
r′ (θ) cos θ − r(θ) sin θ + r′ (θ) sin θ + r(θ) cos θ
=
p
= r′ (θ)2 + r(θ)2
Hence
Z
f (x, y) ds =
Z
=
Z
C
θ2
ds
dθ
f x(θ), y(θ) dθ
θ1
θ2
θ1
q
f r(θ) cos θ, r(θ) sin θ r(θ)2 +
2
dr
dθ (θ)
dθ
(b) In the case f (x, y) = 1, r(θ) = 1 + cos θ, θ1 = 0 and θ2 = 2π,
Z
Z
2π
ds =
2π
=
Z
C
0
0
0
Z
p
[1 + cos θ]2 + [− sin θ]2 dθ =
q
4 cos2
θ
2
dθ = 2
Z
0
2π
2π
p
2(1 + cos θ) dθ
cos θ dθ = 4
2
1
Z
0
π
π
cos 2θ dθ = 8 sin θ2 = 8
0
3. Find the mass of a wire formed by the intersection of the sphere x2 + y 2 + z 2 = 1 and the plane
x + y + z = 0 if the density at (x, y, z) is given by ρ(x, y, z) = x2 grams per unit length of wire.
Solution. The curve in question is the intersection of a sphere with a plane through the centre of the
sphere. So the curve is a circle (in fact, a great circle on the sphere). A simple way to parametrize such
a circle is to pick two unit vectors ı̂ı′ , ̂′ that are mutually perpendicular and perpendicular to (1, 1, 1)
(so that they both lie in the plane x + y + z = 0). For example ı̂ı′ = √12 (1, −1, 0) and ̂′ = √16 (1, 1, −2).
(Check the orthogonality requirements by dotting ı̂ı′ and ̂′ with each other and with (1, 1, 1).) Then
√ θ , − cos
√ θ , 0 + sin
√ θ , sin
√ θ , −2 sin
√θ
r(θ) = cos θ ı̂ı′ + sin θ ̂′ = cos
2
2
6
6
6
is a parametrization of the circle. As ı̂ı′ and ̂′ are mutually perpendicular unit vectors, r ′ (θ) = − sin θ ı̂ı′ +
ds
cos θ ̂′ is a unit vector and dθ
= 1. So
Z 2π
Z
Z 2π
cos θ sin θ 2
2 ds
√ + √
ρds =
x(θ) dθ dθ =
mass =
dθ
2
6
C
2π
=
Z
2
0
As
R 2π
0
cos2 θ dθ =
the mass is
π
2
+
π
6
R 2π
0
=
sin2 θ dθ =
2
3π
0
cos2 θ
1
2
.
R 2π
0
0
√ θ sin
√θ +
+ 2 cos
2
6
2
sin θ
6
dθ
[sin2 θ+cos2 θ] dθ = π and
R 2π
0
cos θ sin θ dθ =
1
2
R 2π
0
sin 2θ dθ = 0,
R
4. Evaluate C F · dr for
(a) F(x, y) = xyı̂ı − x2̂ along y = x2 from (0, 0) to (1, 1).
(b) F(x, y, z) = (x − z)ı̂ı + (y − z)̂ − (x + y)k̂ along the polygonal curve from (0, 0, 0) to (1, 0, 0) to
(1, 1, 0) to (1, 1, 1).
Solution. (a) We may parametrize the curve by r(t) = tı̂ı + t2̂ with 0 ≤ t ≤ 1. Then v(t) = ı̂ı + 2t̂ and
F x(t), y(t) = t3ı̂ı − t2 ̂ so
Z 1
Z 2
Z
3
− t3 dt = − 14
t ı̂ı − t2̂ · ı̂ı + 2t̂ dt =
F · dr =
0
0
C
(b) On the first segment of the curve y = z = 0 so F simplifies to xı̂ı − xk̂ and dr = ı̂ıdx (i.e. we can
~ = x dx. On the second
parametrize the first segment of the curve by r(x) = xı̂ı with 0 ≤ x ≤ 1), so F · dr
~ = y dy.
segment of the curve x = 1 and z = 0 so F simplifies to ı̂ı + y̂ − (1 + y)k̂ and dr = ̂dy, so F · dr
On the final segment of the curve x = y = 1 so F simplifies to (1 − z)ı̂ı + (1 − z)̂ − 2k̂ and dr = k̂dz, so
~ = −2 dz. So
F · dr
Z 1
Z 1
Z 1
Z
F · dr =
x dx +
y dy +
(−2) dz = 12 + 21 − 2 = −1
0
C
5. Evaluate
(0, 1).
H
C
0
0
x2 y 2 dx + x3 y dy counterclockwise around the square with vertices (0, 0), (1, 0), (1, 1) and
Solution. On the first side, y = 0 and dy = 0. That is, we may parametrize the first side by r(x) = xı̂ı
with 0 ≤ x ≤ 1. On the second side x = 1 and dx = 0. On the third side y = 1 and dy = 0. We may
parametrize the third side by r(x) = xı̂ı with x running from 1 to 0. On the final side x = 0 and dx = 0.
So
I
Z 1
Z 1
Z 0
Z 0
x2 y 2 dx + x3 y dy =
x2 × 0 dx +
13 × y dy +
x2 × 1 dx +
03 × y dy = 21 − 31 = 16
C
0
0
1
2
1
6. Let a < b and c < d. Let r : [a, b] → IRn be a parametrization for some curve C. Let u : [c, d] → [a, b]
be increasing, differentiable and obey u(c) = a and u(d) = b. Set R(t) = r(u(t)). It is another
parametrization of C.
(a) Prove that
Z b
Z d
F r(t)) · r ′ (t) dt =
F R(t)) · R′ (t) dt
a
c
(b) Set c = 0 and d = 1. Find a function u : [c, d] → [a, b] which is increasing, infinitely differentiable
and obeys u(c) = a and u(d) = b. So, it is always possible to parametrize a curve in such a way that
the parameter runs from 0 to 1.
(c) Set c = 0 and d = 1. Find a function u : [c, d] → [a, b] which is strictly increasing, infinitely
differentiable and obeys u(c) = a, u(d) = b, u′ (c) = 0 and u′ (d) = 0. So, it is always possible to
parametrize a curve in such a way that the parameter runs from 0 to 1 and the velocity is zero at
both end points.
(d) When we assign a parametrization r : [a, b] → IRn to a curve, we are implicitly designating one of
the two end points of the curve (namely r(a)) as the beginning of the curve and the other end point
(namely r(b)) as the end of the curve. This is called assigning an orientation to the curve. Given a
curve C with orientation determined by a parametrization r : [0, 1] → IRn , define −C to be the curve
with orientation determined by the parametrization w(t) = r(1 − t). The beginning of C is the end
of −C and vice versa. This called reversing the orientation. Show that
Z
Z
F · dr = − F · dr
C
−C
Solution. (a) Let f (u) = F r(u)) · r ′ (u) and g(t) = F R(t)) · R′ (t). By the chain rule R′ (t) =
r ′ (u(t)) u′ (t) so that
g(t) = F r(u(t))) · r ′ (u(t)) u′ (t) = f u(t) u′ (t)
and
Z
Z
b
a
d
c
F r(t) · r ′ (t) dt =
F R(t) · R′ (t) dt =
Z
b
Z
d
f (u) du
a
c
f u(t) u′ (t) dt
Making the change of variables u = u(t) in the first integral gives the second.
(b) u(t) = a +
b−a
d−c (t
− c)
(c) Define v(t) = t(1 − t). Observe that v(0) = 0, v(1) = 0 and v(t) > 0 for all 0 < t < 1. Define
Rt
2
3
w(t) = 0 v(t′ ) dt′ = t2 − t3 . Then w(0) = 0, w′ (0) = v(0) = 0, w(1) = 61 , w′ (1) = v(1) = 0 and w is
strictly increasing on 0 ≤ t ≤ 1. So
t−c
u(t) = a + 6(b − a)w d−c
does the trick.
(d)
Z
−C
F · dr =
Z
0
1
F w(t) · w′ (t) dt =
Z
0
1
F r(1 − t) · − r ′ (1 − t) dt
Now make the change of variables t = 1 − u.
Z
Z 0
Z 0
Z 1
′
′
F · dr =
F r(u) · − r (u) (−du) =
F r(u) · r (u) du = −
F r(u) · r ′ (u) du
−C
1
1
0
Z
= − F · dr
C
3
R
7. Find the work F · dr done by the force field F = (x + y)ı̂ı + (x − z)̂ + (z − y)k̂ in moving an object
from (1, 0, −1) to (0, −2, 3) along any smooth curve.
2
2
Solution. The point here is that F is conservative, as F = ∇φ with φ = x2 + yx − yz + z2 . Hence for
any differentiable curve C parametrized by r(t), with r(t0 ) = (1, 0, −1) and r(t1 ) = (0, −2, 3),
Z t1 h
Z t1
Z
Z t1
i
d
dt
∇φ r(t) · r ′ (t) dt =
F r(t) · r ′ (t) dt =
F · dr =
dt φ r(t)
C
t0
t0
t0
by the chain rule. So, by the Fundamental Theorem of Calculus
Z
F · dr = φ r(t1 ) − φ r(t0 ) = φ(0, −2, 3) − φ(1, 0, −1) = [0 + 0 + 6 + 29 ] − [ 12 + 0 − 0 + 21 ] = 9 21
C
R
8. Evaluate C yexy sin(y + z) dx + exy x sin(y + z) + cos(y + z) dy + exy cos(y + z) dz along the straight
line segment from (0, 0, 0) to 1, π4 , π4 .
Solution. Trick question. This F is conservative with F = ∇ exy sin(y + z) . So
Z
(1, π4 , π4
xy
xy
xy
xy
ye sin(y + z) dx + e x sin(y + z) + cos(y + z) dy + e cos(y + z) dz = e sin(y + z)
= eπ/4
(0,0,0)
C
9. Let F be a continuous vector field on IRd . Prove that the following two statements are equivalent.
R
R
(i) If C1 and C2 are any two curves in IRd with the same initial and final points then C1 F·dr = C2 F·dr.
R
(ii) If C is any closed curve in IRd (i.e. its initial and final points are the same) then C F · dr = 0.
Solution. Assume that property (i) holds. Let C be any closed curve. Let ~a be the initial and final
points of C. Let C ′ be the curve with parametrization R(t) = ~a, 0 ≤ t ≤ 1. That is, C ′ is just a single
R1
R
point, viewed as a (degenerate) curve. Then R′ (t) = ~0, so C ′ F · dr = 0 F R(t) · R′ (t) dt = 0. The
two curves C and C ′ have the same initial and final points, so by property (i),
Z
Z
F · dr = 0
F · dr =
C′
C
and property (ii) holds as well.
Now, assume that property (ii) holds. Let C1 and C2 be any two curves with the same initial and final
points. Define C to the closed curve formed by first following C1 from its initial to its final points and
then following C2 backwards from its final point to its initial point. This C is usually denote C1 − C2 .
We can construct a parametrization for C as follows. Let r1 : [0, 1] → IRd and r2 : [0, 1] → IRd be
parametrizations for C1 and C2 respectively. Then R : [0, 2] → IRd , defined by
r1 (t)
if 0 ≤ t ≤ 1
R(t) =
r2 (2 − t) if 1 ≤ t ≤ 2
is a parametrization of C. (If you’re wrried about a discontinuity in the velocity vector R′ (t) at t = 1,
just choose r1 and r2 so that r ′1 (1) = r′2 (1) = ~0. This is always possible by part c of the last problem.)
Then
Z 2
Z
Z 2
Z 1
′
′
F R(t) · R (t) dt =
F R(t) · R (t) dt +
F R(t) · R′ (t) dt
F · dr =
0
1
C
Z
Z
Z
Z0
F · dr
F · dr −
F · dr =
F · dr +
=
C1
But, as C is a closed curve,
R
−C2
C
F · dr = 0. So
R
C1
C1
F · dr =
4
R
C2
C2
F · dr and property (i) holds.
10. Let F be a continous vector field on IRd . Fix any two points Q1 and Q2 in IRd and assume that
R
R
F · dr = D′ F · dr for all curves D and D′ that start at Q1 and end at Q2 . Let P1 and P2 be any two
D
R
R
points in IRd . Prove that C F · dr = C ′ F · dr for all curves C and C ′ that start at P1 and end at P2 .
Solution. Let P1 and P2 be any two points in IRd and C and C ′ be any two curves that start at P1 and
end at P2 . Fix any curve E1 from Q1 to P1 and any curve E2 from P2 to Q2 . Then D = E1 + C + E2 and
D′ = E1 + C ′ + E2 are curves from Q1 to Q2 . So
Z
Z
F · dr
F · dr =
D′
D
Z
Z
Z
Z
Z
Z
F · dr
F · dr +
F · dr +
F · dr =
F · dr + F · dr +
=⇒
E2
C′
E1
E2
ZC
ZE1
F · dr
F · dr =
=⇒
C′
C
11. Is F = yı̂ı + x̂ + y k̂ a conservative field? Find infinitely many closed curves C for which
Solution.
∂
∂y F3
=
∂
∂y y
=1
and
∂
∂z F2
=
∂
∂z x
R
C
F · dr = 0.
=0
are different. So the field cannot be conservative. On the other hand, for any curve C that lies in the
R
R
plane z = b, for any fixed value of the constant b, dz
dt = 0 so C F · dr = C y dx + x dy = 0 because the
~ = yı̂ı + x̂ = ∇(xy) is conservative.
vector field G
5