But when
Φ
(
u,v
) = (
v
+ 2
,
3
u
+ 6
v,
−
2
u
−
2
v
)
,
we see that
T
u
=
∂
Φ
∂u
= (0
,
3
,
−
2)
,
while
T
v
=
∂
Φ
∂v
= (1
,
6
,
−
2)
,
In this case
T
u
×
T
v
=
vextendsingle
vextendsingle
vextendsingle
vextendsingle
vextendsingle
vextendsingle
i
j
k
0
3
−
2
1
6
−
2
vextendsingle
vextendsingle
vextendsingle
vextendsingle
vextendsingle
vextendsingle
= 6
i
−
2
j
−
3
k
,

baek (mjb3994) – HW13 – goddard – (54330)
5
and so,
bardbl
T
u
×
T
v
bardbl
=
radicalBig
(6)
2
+ (
−
2)
2
+ (
−
3)
2
.
Consequently,
dS
=
√
49
dudv
= 7
dudv
.
008
10.0 points
The surface
S
shown in
is the portion of the graph of
z
=
f
(
x,y
) =
x
2
−
y
2
lying inside the cylinder
x
2
+
y
2
= 2
Determine the surface area of
S
.
1.
Surface Area =
16
3
π
sq. units
2.
Surface Area = 5
π
sq. units
3.
Surface Area =
17
3
π
sq. units
4.
Surface Area =
14
3
π
sq. units
5.
Surface Area =
13
3
π
sq. units
correct
Explanation:
As
S
is enclosed by the cylinder
x
2
+
y
2
= 2
,
it is the graph of the function
f
(
x,y
) =
x
2
−
y
2
,
over the disk
D
=
{
(
x,y
) :
x
2
+
y
2
≤
2
}
in the
xy
-plane. Its surface area element is
dS
= (
f
2
x
+
f
2
y
+ 1)
1
/
2
dxdy
where
f
x
= 2
x,
f
y
=
−
2
y.
Thus
dS
= (4
x
2
+ 4
y
2
+ 1)
1
/
2
dxdy,
and so its surface area is given by the integral
I
=
integraldisplay integraldisplay
D
(4
x
2
+ 4
y
2
+ 1)
1
/
2
dxdy.
Because of rotational symmetry, the inte-
gral is most easily evaluated using polar coor-
dinates. For then
D
=
{
(
r,θ
) : 0
≤
r
≤
2
,
0
≤
θ
≤
2
π
}
so that
I
=
integraldisplay
2
0
integraldisplay
2
π
0
(4
r
2
+ 1)
1
/
2
rdθdr
= 2
π
integraldisplay
√
2
0
(4
r
2
+ 1)
1
/
2
rdr.
To evaluate this last integral we use the sub-
stitution
u
2
= 1 + 4
r
2
. For then
I
=
π
2
integraldisplay
3
1
u
2
du
=
π
6
bracketleftBig
u
3
bracketrightBig
3
1
=
13
3
π
.
009
10.0 points

baek (mjb3994) – HW13 – goddard – (54330)
6
If
S
is the surface
{
Φ
(
u,v
) : 0
≤
u
≤
1
,
0
≤
v
≤
3
}
,
determine the surface area of
S
when
∂
Φ
∂u
= (
−
6
,
1
,
−
2)
,
∂
Φ
∂v
= (3
,
0
,
2)
,
for all
u,v
.
1.
surface area = 25 sq units
2.
surface area = 21 sq units
correct
3.
surface area = 22 sq units
4.
surface area = 23 sq units
5.
surface area = 24 sq units
Explanation:
The surface area of any surface
{
Φ
(
u,v
) : 0
≤
u
≤
1
,
0
≤
v
≤
3
}
is given by the integral
I
=
integraldisplay
1
0
integraldisplay
3
0
dS
=
integraldisplay
1
0
parenleftBig
integraldisplay
3
0
bardbl
T
u
×
T
v
bardbl
dv
parenrightBig
du
But when
T
u
=
∂
Φ
∂u
= (
−
6
,
1
,
−
2)
,
and
T
v
=
∂
Φ
∂v
= (3
,
0
,
2)
,
we see that
T
u
×
T
v
=
vextendsingle
vextendsingle
vextendsingle
vextendsingle
vextendsingle
vextendsingle
i
j
k
−
6
1
−
2
3
0
2
vextendsingle
vextendsingle
vextendsingle
vextendsingle
vextendsingle
vextendsingle
= 2
i
+ 6
j
+ 3
k
.
Thus
bardbl
T
u
×
T
v
bardbl
=
radicalBig
(2)
2
+ (6)
2
+ (3)
2
.
Consequently,
I
=
√
49
integraldisplay
1
0
integraldisplay
3
0
dvdu
= 21
.
010
10.0 points
The surface
S
shown in
is the portion of the sphere
x
2
+
y
2
+
z
2
= 16
where
x
2
+
y
2
≥
12
.
Determine the surface area of
S
.
1.
Surface Area = 24 sq. units
2.
Surface Area = 16 sq. units
3.
Surface Area = 32 sq. units
4.
Surface Area = 16
π
sq. units
5.
Surface Area = 32
π
sq. units
correct
6.
Surface Area = 24
π
sq. units
Explanation:
The sphere
x
2
+
y
2
+
z
2
= 16
is parametrized in spherical polar coordinates
by
Φ
(
θ,φ
) = 4(cos
θ
sin
φ,
cos
θ
sin
φ,
cos
φ
)

baek (mjb3994) – HW13 – goddard – (54330)
7
with
0
≤
θ
≤
2
π,
0
≤
φ
≤
π.
We have to determine what conditions need
to be imposed on
θ,φ
so that only the points
where
x
2
+
y
2
≥
12
are included.
Since
S
doesn’t change with
rotation by
θ
from 0 to 2
π
around the
z
-axis,
there will be no restriction on
θ
.
On the
other hand, spherical caps at the north and
south poles are missing, so restrictions on
φ
are needed. Now a vertical cross-section of
S
through the origin looks like
π
3
π
3
z
4
2
√
3
and
S
is generated by rotating the arc of the
circle between the two solid arrows around
the
z
-axis. Thus
S
is parametrized by
Φ
(
θ,φ
) = 4(cos
θ
sin
φ,
cos
θ
sin
φ,
cos
φ
)
,
with
0
≤
θ
≤
2
π,
π
3
≤
φ
≤
2
π
3
,
and the surface area element is
dS
= 16 sin
φdθdφ.