14-75
SECTION 14.7
.
.
The Divergence Theorem
1189
56.
S
yz dS
, where
S
is the portion of
x
2
+
y
2
=
1 with
x
≥
0
and
z
between
z
=
1 and
z
=
4
−
y
57.
S
(
y
2
+
z
2
)
dS
, where
S
is the portion of the paraboloid
x
=
9
−
y
2
−
z
2
in front of the
yz
-plane
58.
S
(
y
2
+
z
2
)
dS
, where
S
is the hemisphere
x
=
4
−
y
2
−
z
2
59.
S
x
2
dS
, where
S
is the portion of the paraboloid
y
=
x
2
+
z
2
to the left of the plane
y
=
1
60.
S
(
x
2
+
z
2
)
dS
, where
S
is the hemisphere
y
=
√
4
−
x
2
−
z
2
61.
S
4
dS
, where
S
is the portion of
y
=
1
−
x
2
with
y
≥
0 and
between
z
=
0 and
z
=
2
62.
S
(
x
2
+
z
2
)
dS
, where
S
is the portion of
y
=
√
4
−
x
2
be-
tween
z
=
1 and
z
=
4
63.
Explain the following result geometrically. The flux integral
of
F
(
x
,
y
,
z
)
=
x
,
y
,
z
across the cone
z
=
x
2
+
y
2
is 0.
64.
In
geometric
terms,
determine
whether
the
flux
in-
tegral
of
F
(
x
,
y
,
z
)
=
x
,
y
,
z
across
the
hemisphere
z
=
1
−
x
2
−
y
2
is 0.
65.
For the cone
z
=
c
x
2
+
y
2
(where
c
>
0), show that in spher-
ical coordinates tan
φ
=
1
c
. Then show that parametric equa-
tions are
x
=
u
cos
v
√
c
2
+
1
,
y
=
u
sin
v
√
c
2
+
1
and
z
=
cu
√
c
2
+
1
.
66.
Find the surface area of the portion of
z
=
c
x
2
+
y
2
below
z
=
1,
using
the
parametric
equations
in
exercise 65.
67.
Find the flux of
x
,
y
,
z
across the portion of
z
=
c
x
2
+
y
2
below
z
=
1. Explain in physical terms why this answer makes
sense.
68.
Find
the
flux
of
x
,
y
,
z
across
the
entire
cone
z
2
=
c
2
(
x
2
+
y
2
).
69.
Find the flux of
x
,
y
,
0 across the portion of
z
=
c
x
2
+
y
2
below
z
=
1. Explain in physical terms why this answer makes
sense.
70.
Find the limit as
c
approaches 0 of the flux in exercise 69.
Explain in physical terms why this answer makes sense.
EXPLORATORY EXERCISES
1.
If
x
=
3 sin
u
cos
v,
y
=
3 cos
u
and
z
=
3 sin
u
sin
v
, show
that
x
2
+
y
2
+
z
2
=
9. Explain why this equation doesn’t
guarantee that the parametric surface defined is the entire
sphere, but it does guarantee that all points on the surface
are also on the sphere. In this case, the parametric surface
is the entire sphere. To verify this in graphical terms, sketch
a picture showing geometric interpretations of the “spher-
ical coordinates”
u
and
v
. To see what problems can oc-
cur, sketch the surface defined by
x
=
3 sin
u
2
u
2
+
1
cos
v,
y
=
3 cos
u
2
u
2
+
1
and
z
=
3 sin
u
2
u
2
+
1
sin
v
. Explain why you
do not get the entire sphere. To see a more subtle example
of the same problem, sketch the surface
x
=
cos
u
cosh
v,
y
=
sinh
v,
z
=
sin
u
cosh
v
.
Use
identities
to
show
that
x
2
−
y
2
+
z
2
=
1 and identify the surface. Then sketch the
surface
x
=
cos
u
cosh
v,
y
=
cos
u
sinh
v,
z
=
sin
u
and use
identities to show that
x
2
−
y
2
+
z
2
=
1. Explain why the sec-
ond surface is not the entire hyperboloid. Explain in words and
pictures exactly what the second surface is.
14.7
THE DIVERGENCE THEOREM
Recall that at the end of section 14.5, we had rewritten Green’s Theorem in terms of the
divergence of a two-dimensional vector field. We had found there (see equation 5.3) that
C
F
·
n
ds
=
R
∇ ·
F
(
x
,
y
)
dA
.