Name: Date: 04/08/2020
M20550 Calculus III Tutorial
Worksheet 9
1. Calculate the line integral
Z
C
(y
2
+x) dx +4xy dy where C is the arc of x = y
2
from (1, 1)
to (4, 2).
Solution: First, we need to parametrize the curve C. Since C is a part of the curve
x = y
2
, we can let y = t; then we have x = t
2
. Moreover, since the curve C is
the part from (1, 1) to (4, 2), we get 1 ≤ y ≤ 2. So, we have 1 ≤ t ≤ 2. Thus, a
parametrization of C is as follows:
x(t) = t
2
, y(t) = t for 1 ≤ t ≤ 2.
Now,
Z
C
(y
2
+ x) dx + 4xy dy is a line integral with respect to x and y because we see
the dx and dy. Here,
dx = x
0
(t) dt = 2t dt and dy = y
0
(t) dt = 1 dt.
So, for 1 ≤ t ≤ 2,
Z
C
(y
2
+ x) dx + 4xy dy =
Z
2
1
t
2
+ t
2
2t + 4(t
2
)(t)
dt
=
Z
2
1
8t
3
dt
=
2t
4
2
1
= 2
5
− 2 = 30.
2. Evaluate the line integral
Z
C
z
2
dx +x dy + y dz where C is the line segment from (1, 0, 0)
to (4, 1, 2).
Solution: First, we parametrize C, the line segment from (1, 0, 0) to (4, 1, 2). For
0 ≤ t ≤ 1, C can be written as the vector function
r(t) = h1, 0, 0i + t
h4, 1, 2i − h1, 0, 0i
= h1, 0, 0i + t h3, 1, 2i.
So, x(t) = 1 + 3t, y(t) = t, and z(t) = 2t for 0 ≤ t ≤ 1. Then,
dx = x
0
(t) dt = 3 dt, dy = y
0
(t) dt = 1 dt, dz = z
0
(t) dt = 2 dt.
Name: Date: 04/08/2020
Hence, for 0 ≤ t ≤ 1,
Z
C
z
2
dx + x dy + y dz =
Z
1
0
(2t)
2
(3) + (1 + 3t)(1) + t(2)
dt
=
Z
1
0
12t
2
+ 5t + 1
dt
=
4t
3
+
5
2
t
2
+ t
1
0
=
15
2
.
Name: Date: 04/08/2020
3. Compute
Z
C
x
2
ds where C is the intersection of the surface x
2
+ y
2
+ z
2
= 4 and
the plane z =
√
3.
Solution: The intersection of the sphere x
2
+ y
2
+ z
2
= 4 and
the plane z =
√
3 is the circle
x
2
+ y
2
+
√
3
2
= 4, z =
√
3
or simply x
2
+ y
2
= 1, z =
√
3.
Thus, a parametrization of C could be
r(t) =
D
cos t, sin t,
√
3
E
for 0 ≤ t ≤ 2π.
Then, r
0
(t) = h−sin t, cos t, 0i =⇒ |r
0
(t)| =
p
(−sin t)
2
+ cos
2
t = 1.
So ds = |r
0
(t)|dt = 1 dt. Finally, for 0 ≤ t ≤ 2π,
Z
C
x
2
ds =
Z
2π
0
cos
2
t
dt
=
Z
2π
0
1
2
1 + cos 2t
dt
=
1
2
t +
1
2
sin(2t)
2π
0
= π.
4. Determine whether or not the following vector fields are conservative:
(a) F = (3 + 2xy) i + (x
2
− 3y
2
) j
(b) F = i + sin z j + y cos z k
Solution: (a) Since F is a vector field on R
2
, we use the criterion
∂P
∂y
?
=
∂Q
∂x
to see
if F is conservative or not. We have F = h3 + 2xy, x
2
− 3y
2
i. So, P = 3 + 2xy and
Q = x
2
− 3y
2
and
∂P
∂y
= 2x =
∂Q
∂x
.
Since
∂P
∂y
=
∂Q
∂x
, F is a conservative vector field on R
2
.
Name: Date: 04/08/2020
(b) Since F is a vector field on R
3
, we use the criterion curl F
?
= 0 to see if F is
conservative or not. We have F = h1, sin z, y cos zi. And
curl F = ∇ × F =
i j k
∂
∂x
∂
∂y
∂
∂z
1 sin z y cos z
= hcos z − cos z, 0, 0i = h0, 0, 0i = 0.
Since curl F = 0, F is a conservative vector field on R
3
.
5. Evaluate
Z
C
F·dr, where F(x, y, z) = −2xy i +4y j+k and r(t) = t i +t
2
j+k, 0 ≤ t ≤ 2.
Solution: Since x = t, y = t
2
, z = 1, we have
F(r(t)) = −2t
3
i + 4t
2
j + k = h−2t
3
, 4t
2
, 1i,
and
r
0
(t) = i + 2tj = h1, 2t, 0i
The line integral of F along C is
Z
C
F · dr =
Z
2
0
F(r(t)) · r
0
(t) dt
=
Z
2
0
h−2t
3
, 4t
2
, 1i · h1, 2t, 0i dt
=
Z
2
0
−2t
3
+ 8t
3
dt
=
Z
2
0
6t
3
dt
=
6t
4
4
2
0
=
3 · 2
4
2
− 0
= 24.
Remark: Note that F is not a conservative vector field, so we cannot apply the
Fundamental Theorem of Line Integrals in this example. To see this note that
curl F = ∇ × F =
i j k
∂
∂x
∂
∂y
∂
∂z
−2xy 4y 1
= h0, 0, 2xi 6= 0.
Name: Date: 04/08/2020
6. Evaluate
Z
C
F ·dr, where F = (y
2
cos(xy
2
) + 3x
2
) i + (2xy cos(xy
2
) + 2y) j is a conserva-
tive vector field and C is any curve from the point (−1, 0) to (1, 0).
Solution: Since we know F is a conservative vector field, F = ∇f for some scalar
function f (x, y). So,
R
C
F · dr =
R
C
∇f · dr. Then, by the fundamental theorem of
line integral (FTLI), we have
R
C
∇f ·dr = f(1, 0) −f (−1, 0). So, let’s go about and
find the potential function f(x, y) of F first.
We know F = ∇f , so hy
2
cos(xy
2
)+3x
2
, 2xy cos(xy
2
)+2yi = hf
x
, f
y
i. Thus, we have
f
x
= y
2
cos(xy
2
) + 3x
2
(1)
f
y
= 2xy cos(xy
2
) + 2y (2)
Using equation (1), we have f =
R
(y
2
cos(xy
2
)+3x
2
) dx = sin(xy
2
)+x
3
+g(y). Now,
we need to find g(y) to complete f.
With f = sin(xy
2
) + x
3
+ g(y), we compute f
y
= 2xy cos(xy
2
) + g
0
(y). Then from
equation (2) above, we must have
2xy cos(xy
2
) + g
0
(y) = 2xy cos(xy
2
) + 2y =⇒ g
0
(y) = 2y =⇒ g(y) = y
2
+ C.
We only need a potential function to apply FTLI, so we can pick C = 0. So, a
potential function f(x, y) of the vector field F is
f(x, y) = sin(xy
2
) + x
3
+ y
2
.
Finally,
Z
C
F · dr =
Z
C
∇f · dr
FTLI
= f (1, 0) − f (−1, 0)
= (sin 0 + 1
3
+ 0
2
) − (sin 0 + (−1)
3
+ 0
2
)
= 2.
7. Use Green’s Theorem to evaluate
Z
C
−
y
3
3
+ sin x
dx +
x
3
3
+ y
dy,
where C is the circle of radius 1 centered at (0, 0) oriented counterclockwise when viewed
from above.
Name: Date: 04/08/2020
Solution: Let D be the region enclosed by the unit circle C in this problem. By
Green’s Theorem, we have
Z
C
−
y
3
3
+ sin x
dx +
x
3
3
+ y
dy =
ZZ
D
x
2
− (−y
2
) dA.
Here, we have P = −
y
3
3
+ sin x and Q =
x
3
3
+ y, so
∂P
∂y
= −y
2
and
∂Q
∂x
= x
2
.
So, instead of computing the line integral
Z
C
−
y
3
3
+ sin x
dx +
x
3
3
+ y
dy, we
are going to compute the double integral
ZZ
D
x
2
+ y
2
dA, where D is the unit disk
as shown below.
Using polar coordinates,
ZZ
D
x
2
+ y
2
dA =
Z
2π
0
Z
1
0
r
3
dr dθ = 2π
1
4
=
π
2
.
Hence,
Z
C
−
y
3
3
+ sin x
dx +
x
3
3
+ y
dy =
π
2
.
