MA2E01: Problem Set 8
Solutions
Due at the end of the tutorial, 26-28 November.
You must show details of your work to get full marks.
1. Consider the line integral
I
x2 y dx + (y + xy 2 ) dy ,
C
where C is the boundary of the region enclosed by y = x2
and x = y 2
(a) Sketch the region and its boundary.
(b) Use Green’s Theorem to evaluate the integral.
Solution:
√
(a) The curves y = x2 and x = y 2 ⇒ y = x intersect at
(0, 0) and (1, 1). This is sketched in Figure 1.
1.0
0.8
y2 = x
0.6
0.4
y = x2
0.2
0.2
0.4
0.6
0.8
1.0
Figure 1: The region between y = x2 and y =
√
x.
(b) Using Green’s Theorem, we need to evaluate over the
√
region with x2 ≤ y ≤ x, 0 ≤ x ≤ 1 using a double
1
integral. Green’s Theorem give us
I
x2 y dx + (y + xy 2 ) dy
C
ZZ ∂
∂
(y + xy 2 ) − (x2 y) dA
=
∂x
∂y
R
Z 1 Z √x
=
y 2 − x2 dy dx
0
x2
√x
Z 1 3
y
=
− x2 y dx
3
0
x2
Z 1 3/2
x
x6
5/2
4
=
−x −
+ x dx
3
3
0
5/2
1
2x
2x7/2 x7 x5 =
−
−
+
15
7
21
5 0
2
2
1
1
=
− −
+
15 7 21 5
5
7
1 1
2+3 6+1
−
=
−
= − = 0.
=
15
21
15 21 3 3
2. Use Green’s Theorem to find the area of the triangular region with vertices (0, 0), (1, 0) and (1, 1).
Solution: The region is shown in Figure 2. The area enclosed by the curve by Green’s Theorem is
ZZ
I
I
I
1
A=
dA = x dy = − y dx =
−y dx + x dy .
2
R
We can use which ever one of these we prefer, but for this
problem, let’s us the last formula. Note that although the
curve is only piecewise smooth, we can use Green’s Theorem and then integrate along the three curves which we
2
y
(1,1)
1.0
0.8
0.6
0.4
0.2
0.2
0.4
0.6
0.8
1.0
x
Figure 2: The triangular region with vertices (0, 0), (1, 0) and (1, 1).
will denote C1 , C2 , C3 . To parameterise the curves we can
use the vector form of a line segment. This gives us
C1 :(0, 0)(1 − t) + (1, 0)t = (t, 0) ,
C2 :(1, 0)(1 − t) + (1, 1)t = (1, t) ,
C3 (1, 1)(1 − t) + (0, 0)t = (1 − t, 1 − t) .
For C1 we have x = t, y = 0 and so dx/dt = 1, dy/dt = 0.
For C2 we have x = 1, y = t and so dx/dt = 0, dy/dt = 1.
For C3 we have x = 1 − t, y = 1 − t and so
dx/dt = −1, dy/dt = −1.
The first line integral gives
Z
Z
1 1
1
−y dx + x dy =
(−(0)(1) + (t)(0)) dt = 0 .
2 C1
2 0
The second line integral gives
Z
Z
Z
1
1 1
1 1
1
−y dx+x dy =
(−(t)(0) + (1)(1)) dt =
dt = .
2 C2
2 0
2 0
2
3
The third line integral gives
Z
Z
1 1
1
−y dx + x dy =
(−(1 − t)(−1) + (1 − t)(−1)) dt
2 C3
2 0
Z
1 1
=
(0) dt = 0 .
2 0
Therefore, the are is given by
Z
Z
Z
1
1
1
A=
−y dx + x dy +
−y dx + x dy +
−y dx + x dy
2 C1
2 C2
2 C3
1
1
=0+ +0= .
2
2
This is exactly the area you would expect for a triangle of
height 1 with a base of length 1.
3. Consider the vector field
F(x, y, z) = xi + yj + 2zk ,
and the surface σ, which is the portion of z = 4 − x2 − y 2
above the xy-plane, oriented by upward normal vectors.
(a) Sketch the projection of the surface onto the xy-plane.
(b) Calculate the upward normal vector.
(c) Calculate the flux of F across σ.
Solution:
(a) The projection is in fact at it’s largest when z = 0,
which gives x2 + y 2 = 4, which is a circle of radius 2,
sketched in Figure 3.
4
2
1
0
-1
-2
-2
-1
0
1
2
Figure 3: The projection of z = 4 − x2 − y 2 .
(b) The upward normal vector is
n=−
∂z ∂z
i− j+k = −(−2x)i−(−2y)k+k = 2xi+2yj+k .
∂x ∂y
(c) The flux is
ZZ
ZZ
Φ=
F · n dS =
(xi + yj + 2zk) · (2xi + 2yj + k) dA
σ
ZZ
=
R
(2x2 + 2y 2 + 2z) dA .
R
However, the surface is z = 4 − x2 − y 2 , so we use this
to eliminate z,
2x2 +2y 2 +2z = 2x2 +2y 2 +2(4−x2 −y 2 ) = 2x2 +2y 2 +8−2x2 −2y 2 = 8 .
Since we are integrating over the projection in the xyplane, which is a disk, it makes sense to use polar co5
ordinates, with 0 ≤ ρ ≤ 2, giving
ZZ
Z 2π Z 2
Φ=
8 dA =
ρ dρ dθ
0
R
Z
0
2π
2
Z 2π
ρ2 4(24 ) dθ
8 dθ =
2 ρ=0
0
Z
2π
=
0
dθ = 16(2π) = 32π .
= 16
0
This is the flux of F through σ.
4. Consider the vector field
F(x, y, z) = yj + k ,
and the surface σ, which is the portion of z = x2 + y 2 below
z = 9, oriented by downward normal vectors.
(a) Sketch the projection of the surface onto the xy-plane.
(b) Calculate the downward normal vector.
(c) Calculate the flux of F across σ.
Solution:
(a) At z = 0 we get x2 + y 2 = 0 which is a point. However,
at z = 9 we get x2 + y 2 = 9 which is a circle of radius 3.
Therefore the projection is the disk of radius 3, shown
in Figure 4.
(b) The downward normal vector is
n=
∂z
∂z
i + j − k = (2x)i + (2y)k + k = 2xi + 2yj − k .
∂x ∂y
6
3
z = x2 + y2 = 9
2
1
z = x2 + y2 = 0
0
-1
-2
-3
-3
-2
-1
0
1
2
3
Figure 4: The projection of z = x2 + y 2 below z = 9, with z = 0 also
indicated.
(c) The flux is given by
ZZ
ZZ
Φ=
F · n dS =
(yj + k) · (2xi + 2yj − k) dA
σ
ZZ
=
R
(2y 2 − 1) dA .
R
We are integrating over the projection, which is a disk,
so again polar coordinates are convenient, with 0 ≤
7
ρ ≤ 3. Then
ZZ
Φ=
(2(ρ sin t)2 − 1)ρ dρ dθ
R
Z
2π
Z
3
(2ρ3 sin2 θ − ρ) dρ dθ
0
0
Z 2π 4
2 3
2ρ
ρ
=
sin2 θ −
dθ
4
2 ρ=0
0
Z 2π 9
81 2
sin θ −
dθ .
=
2
2
0
=
However,
Z 2π
Z 2π
1
1
sin2 θ dθ =
(1 − cos 2θ) dθ = (2π − 0) = π ,
2 0
2
0
since cos 2t integrated from 0 to 2π gives 0. Hence
Φ=
81
9
(81 − 18)π
63π
(π) − (2π) =
=
.
2
2
2
2
This is the flux of F through σ.
8