CHAPTER 5

Section 5.3
1. (a) From the given end points (0, 0), (2, 2) it follows that we can represent the curve
C in the form y = x, 0 ≤ x ≤ 2. Hence, by (5.6) we find
ˆ (2,2) ˆ 2 2
2 2 x3 8
y dx = x dx = =
(0,0) 0 3 0 3
(b) Given the end points (2, 1), (1, 2) we will parameterise the curve C according to:
x = 2 − t, y = 1 + t, 0 ≤ t ≤ 1. Then by (5.4) we find
ˆ (1,2) ˆ 1 1
t2

3
y dx = − (1 + t) dt = − t + =−
(2,1) 0 2 0 2
(c) Given the end points (1, 1), (2, 1) we will parameterise the curve C according to
x = 1 + t, y = 1, 0 ≤ t ≤ 1. Then by (5.5) we find
ˆ (2,1) ˆ 1
x dy = (1 + t) (0) dt = 0
(1,1) 0
p
2. (a) Let us represent the curve C : x = 1 − y 2 in the form x = cos t, y = sin t, −π/2 ≤
t ≤ π/2. Then by (5.4) and (5.5)
ˆ (0,1) ˆ π/2
2 2
y dx + x dy = − sin3 t dt + cos3 t dt
(0,−1) −π/2
ˆπ/2
− 1 − cos2 t sin t + 1 − sin2 t cos t dt



=
−π/2
π/2
cos3 t sin3 t

4
= cos t − + sin t − =
3 3 −π/2 3

(b) Let C be the parabola y = x2 . Then by (5.6) and (5.7) we find

ˆ (2,4) ˆ 2  3 2
2 2

x 2 3
y dx + x dy = x + 2x dx = + x =8
(0,0) 0 3 3 0

(c) Let C be the curve x = cos3 t, y = sin3 t, 0 ≤ t ≤ π/2 and let us use the
substitution u = tan3 t. Then by (5.4) and (5.5) we can rewrite the integral as
ˆ (0,1) ˆ π/2 ˆ π/2
y dx − x dy sin4 t cos2 t + sin2 t cos4 t −3 sin2 t cos2 t
= −3 dt = dt
(1,0) x2 + y 2 0 cos6 t + sin6 t 0 cos6 t + sin6 t
ˆ ∞ ˆ ∞ ˆ b
cos6 t du du
=− 6 du = − = lim −
0 cos6 t + sin t 0 1 + u2 b→∞ 0 1+u
2

b π
= lim − tan−1 u 0 = lim − tan−1 b = −
b→∞ b→∞ 2

1
3. (a) Let C be the square with vertices (1, 1), (−1, 1), (−1, −1), (1, −1). Then the
integral ‰
y 2 dx + xy dy
C
can be evaluated by computing the sum of the four integrals
ˆ (−1,1) ˆ (−1,−1) ˆ (1,−1) ˆ (1,1)
2
y dx xy dy y 2 dx xy dy
(1,1) (−1,1) (−1,−1) (1,−1)
| {z } | {z } | {z } | {z }
dy=0 dx=0 dy=0 dx=0

Hence,
‰ ˆ −1 ˆ −1 ˆ 1 ˆ 1
2
y dx + xy dy = dx − y dy + dx + y dy
C 1 1 −1 −1
2 −1 2 1
y y
= x|−1
1 − + x|1−1 + =0
2 1 2 −1

(b) Let C be the circle x2 + y 2 = 1. Using the parameterization x = cos t, y = sin t

where 0 ≤ t ≤ 2π, then by (5.4) and (5.5) the integral
‰
y dx − x dy
C

may be written as
‰ ˆ 2π ˆ 2π ˆ 2π
2 2 2 2


y dx − x dy = − sin t dt − cos t dt = − sin t + cos t dt = − dt
C 0 0 0
= −2π

(c) Let C be the triangle with vertices (0, 0), (1, 0), (1, 1). Then the integral
‰
x2 y 2 dx − xy 3 dy
C

can be evaluated by computing the sum of the three integrals

ˆ (1,0) ˆ (1,1) ˆ (0,0)
2 2 3
x y dx = 0 − xy dy x2 y 2 dx − xy 3 dy
(0,0) (1,0) (1,1)
| {z } | {z }
dy=0 dx=0

Hence,
‰ ˆ 1 ˆ 1 ˆ 1
2 2 3 3 4
x y dx − xy dy = − y dy + x dx − y 4 dy
C 0 0 0
4 1 5 1 5 1
y x y 1
=− + − =−
4 0 5 0 5 0 4

2
4. (a) Let C be the circle x2 + y 2 = 4. Then using the parametrisation x = 4 cos t, y =
4 sin t, where 0 ≤ t ≤ 2π and (5.12) the integral
‰
x2 − y 2 ds


C

may be written as
‰ ˆ 2π ˆ 2π
2 2

2

2 2π
x − y ds = 64 cos t − sin t dt = 64 cos 2t dt = 32 sin 2t 0
=0
C 0 0

(b) Let C be the line y = x with endpoints (0, 0), (1, 1). Then by (5.14) the integral
ˆ (1,1)
x ds
(0,0)

may be written as
ˆ (1,1) √ ˆ 1
√ 1
2 2 1
x ds = 2 x dx = x =√
(0,0) 0 2 0 2

(c) Let C be the parabola y = x2 with endpoints (0, 0), (1, 1). Then by (5.14) and
using the substitution x = (1/2) tan u, such that dx = (1/2) sec2 u du the integral
ˆ (1,1)
ds
(0,0)

may be written as
ˆ (1,1) ˆ 1 √ ˆ tan−1 2
1
ds = 1+ 4x2 dx = sec3 u du
(0,0) 0 2 0

In order to solve the integral on the right hand side, let us solve the indefinite
integral
ˆ ˆ ˆ
sec x dx = sec x sec x dx = sec x tan x − sec x tan2 x dx + C
3 2
0 ˆ
= sec x tan x − sec x sec2 x − 1 dx + C

ˆ ˆ
3
= sec x tan x − sec x dx + sec x dx + C
´
Adding the term sec3 x dx to both sides and dividing by two then gives
ˆ ˆ
3 1 1
sec x dx = sec x tan x + sec x dx + C
2 2
1 1
= sec x tan x + ln|sec x + tan x| + C
2 2

3
 Substituting in the original equation then gives
ˆ (1,1) ˆ 1 √ ˆ −1
1 tan 2 3
ds = 1+ 4x2 dx = sec u du
(0,0) 0 2 0
1 tan−1 2 1 tan−1 2
= sec u tan u 0 + ln|sec u + tan u| 0
4
√ √
4
5 ln 5 + 2
= +
2 4

5. Let a path x = φ(t), y = ψ(t), h ≤ t ≤ k, where x and y are continuous and have
continuous derivatives for h ≤ t ≤ k like (5.1) be given. Next, let us make a change of
parameter by the equation t = g(τ ), α ≤ τ ≤ β, where g 0 (τ ) is continuous ´and positive
in the interval and g(α) = h, g(β) = k. Then by (5.4) the line integral C f (x, y) dx
on the path x = φ(g(τ )), y = ψ(g(τ )), such that dx = (d/dτ )φ(g(τ )) dτ , is given by
ˆ ˆ β
d
f (x, y) dx = f [φ (g (τ )) , ψ (g (τ ))] φ (g (τ )) dτ
α dτ
C
ˆ β
dφ d
= f [φ (g (τ )) , ψ (g (τ ))]
g (τ ) dτ
α dt dτ
ˆ k ˆ k
dφ dt
= f [φ (t) , ψ (t)] dτ = f [φ (t) , ψ (t)] φ0 (t) dt
h dt dτ h

´
6. (a) Using (a), the integral P dx + Q dy along the path C → ABF G may be approx-
imated as
ˆ    
1 1 1 1
P dx + Q dy ∼ (0 + 3) · 1 + (1 + 2) · 0 + (3 + 0) · 0 + (2 + 4) · 1
2 2 2 2
C
 
1 1
+ (0 + 5) · 1 + (4 + 6) · 0 = 7
2 2
´
(b) Using (a), the integral P dx + Q dy along the path C → AF GKH may be
approximated as
ˆ    
1 1 1 1
P dx + Q dy ∼ (0 + 0) · 1 + (1 + 4) · 1 + (0 + 5) · 1 + (4 + 6) · 0
2 2 2 2
C
   
1 1 1 1
+ (5 + 0) · 0 + (6 + 9) · 1 + (0 + 2) · 1 + (9 + 8) · −1
2 2 2 2
=5

4
 ´
(c) Using (a), the integral P dx + Q dy along the path C → ABCDHLSON M IEA
may be approximated as
ˆ    
1 1 1 1
P dx + Q dy ∼ (0 + 3) · 1 + (1 + 2) · 0 + (3 + 8) · 1 + (2 + 3) · 0
2 2 2 2
C
   
1 1 1 1
+ (8 + 5) · 1 + (3 + 4) · 0 + (5 + 2) · 0 + (4 + 8) · 1
2 2 2 2
   
1 1 1 1
+ (2 + 1) · 0 + (8 + 2) · 1 + (1 + 4) · 0 + (2 + 6) · 1
2 2 2 2
   
1 1 1 1
+ (4 + 3) · −1 + (6 + 2) · 0 + (3 + 7) · −1 + (2 + 8) · 0
2 2 2 2
   
1 1 1 1
+ (7 + 2) · −1 + (8 + 4) · 0 + (2 + 8) · 0 + (4 + 3) · −1
2 2 2 2
   
1 1 1 1
+ (8 + 3) · 0 + (3 + 2) · −1 + (3 + 0) · 0 + (2 + 1) · −1
2 2 2 2
=8
´
(d) Using (a), the integral P dx + Q dy along the path C → AF JN M IJF A may
be approximated as
ˆ    
1 1 1 1
P dx + Q dy ∼ (0 + 0) · 1 + (4 + 1) · 1 + (0 + 5) · 0 + (4 + 6) · 1
2 2 2 2
C
   
1 1 1 1
+ (5 + 7) · 0 + (6 + 8) · 1 + (7 + 2) · −1 + (8 + 4) · 0
2 2 2 2
   
1 1 1 1
+ (2 + 8) · 0 + (4 + 3) · −1 + (8 + 5) · 1 + (3 + 6) · 0
2 2 2 2
   
1 1 1 1
+ (5 + 0) · 0 + (6 + 4) · −1 + (0 + 0) · −1 + (4 + 1) · −1
2 2 2 2
11
=
2
´
(e) Using (a), the integral P dx + Q dy along the path C → ABF EAEF BA may

5
 be approximated as
ˆ    
1 1 1 1
P dx + Q dy ∼ (0 + 3) · 1 + (1 + 2) · 0 + (3 + 0) · 0 + (2 + 4) · 1
2 2 2 2
C
   
1 1 1 1
+ (0 + 3) · −1 + (4 + 2) · 0 + (3 + 0) · 0 + (2 + 1) · −1
2 2 2 2
   
1 1 1 1
+ (0 + 3) · 0 + (1 + 2) · 1 + (3 + 0) · 1 + (2 + 4) · 0
2 2 2 2
   
1 1 1 1
+ (0 + 3) · 0 + (4 + 2) · −1 + (3 + 0) · −1 + (2 + 1) · 0
2 2 2 2
=0
7. Let C be a smooth curve in the xy-plane and let f (x, y) > 0 be a continuous function
defined over a region of the xy-plane containing the curve C. The equation z = f (x, y)
then is the equation of a surface that lies above the region of the xy-plane containing
the curve C. Next, we imagine moving a straight line along C perpendicular to the
xy-plane, effectively tracing out a ”wall” standing on C, orthogonal to the xy-plane.
This ”wall” cuts the surface z = f (x, y), forming a curve on it that lies above the curve
C. In fact, the curve C may be interpreted as the projection of the surface curve onto
the xy-plane. Using (5.11), the line integral
ˆ Xn
f (x, y) ds = n→∞ lim f (x∗i , yi∗ ) ∆i s
max ∆i s→0 i=1
C
then may be interpreted as an infinite sum of the length of each straight line directed
from C to the surface curve lying above it in the limit where the distance ∆s between
each subsequent line becomes infinitely small, effectively tracing out a ”wall” with
height at each point (x, y) given by f (x, y). This may be interpreted the as the area
of the cylindrical surface 0 ≤ z ≤ f (x, y), (x, y) on C.

Section 5.5
1. Let the vector v = (x2 + y 2 )i + 2xyj be given. Then by (5.25) and (5.29)
´
(a) The integral C vT ds along the path C → y = x from (0, 0) to (1, 1) may be
evaluated as
ˆ ˆ ˆ 1 ˆ 1
2 2 (5.6) (5.9) 2 4
2y 2 dy =


vT ds = x + y dx + 2xy dy = 2x dx +
0 0 3
C C
´
(b) The integral C vT ds along the path C → y = x2 from (0, 0) to (1, 1) may be
evaluated as
ˆ ˆ ˆ 1
2 2 (5.6) (5.7) 4
x2 + 5x4 dx =



vT ds = x + y dx + 2xy dy =
0 3
C C

6
 ´
(c) The integral C vT ds along the broken line from (0, 0) to (1, 1) with corner at
(1, 0) may be evaluated as
ˆ ˆ
x2 + y 2 dx + 2xy dy


vT ds =
C C
ˆ (1,0) ˆ (1,1)
2 2
x2 + y 2 dx + 2xy dy



= x +y dx + 2xy dy +
(0,0) (1,0)
ˆ 1 ˆ 1
4
= x2 dx + 2y dy =
0 0 3

2. Let v = P (x, y)i + Q(x, y)j be the same vector as given in Problem 1, and let n be
the unit normal vector 90◦ behind the tangent vector T as given by (5.37). Then the
normal component of v is given by vn = v · n = (P i + Qj) · (ys i − xs j) = −Qxs + P ys .
Then by (5.25) and (5.29)
´
(a) The integral C vn ds along the path C → y = x from (0, 0) to (1, 1) may be
evaluated as
ˆ ˆ ˆ 1 ˆ 1
2 2 (5.6) (5.9) 2
2y 2 dy = 0


vn ds = −2xy dx + x + y dy = −2x dx +
0 0
C C

´
(b) The integral C
vn ds along the path C → y = x2 from (0, 0) to (1, 1) may be
evaluated as
ˆ ˆ ˆ 1
2 2 (5.6) (5.7) 1
2x5 dx =


vn ds = −2xy dx + x + y dy =
0 3
C C

´
(c) The integral C vn ds along the broken line from (0, 0) to (1, 1) with corner at
(1, 0) may be evaluated as
ˆ ˆ
vn ds = −2xy dx + x2 + y 2 dy

C C
ˆ (1,0) ˆ (1,1)
2 2
−2xy dx + x2 + y 2 dy



= −2xy dx + x + y dy +
(0,0) (1,0)
ˆ 1
2

4
= 1+y dy =
0 3

3.