Solution
1.
z
=
x
y
,
so
d
s
=
√
1 +
z
x
2
+
z
y
2
d
x
d
y
=
√
1 +
x
2
+
y
2
d
x
d
y
,
S
=
⌠
⌡
⌠
⌡
x
2
+
y
2
≤
a
2
√
1 +
x
2
+
y
2
d
x
d
y
=
⌠
⌡
2 π
0
⌠
⌡
a
0
√
1 +
r
2
r
d
r
d
θ
=
2 π
⌠
⌡
a
0
√
1 +
r
2
r
d
r
=
2 π
3
((1+
a
2
)
3/2
−1).
(1)
2.
Method 1
(
a
×(
b
×
c
))·(
c
×
b
)
=
(
c
×
b
) ·(
a
×(
b
×
c
))
=
a
·( (
b
×
c
) ×(
c
×
b
))
=
a
·
0
=
0,
(2)
where the following diagram was used.
Method 2
The vector, (
a
×(
b
×
c
)), is perpendicular to both
a
and
b
×
c
. Hence, it is perpendicular to
c
×
b
and their dot product is 0.
3.
Note that
n
=
⎛
⎜
⎜
⎜
⎝
x
2
y
3
z
⎞
⎟
⎟
⎟
⎠
√
x
2
+ 4
y
2
+ 9
z
2
√
x
2
+ 4
y
2
+ 9
z
2
=
⎛
⎜
⎜
⎜
⎝
x
2
y
3
z
⎞
⎟
⎟
⎟
⎠
√
x
2
+ 4
y
2
+ 9
z
2
·
⎛
⎜
⎜
⎜
⎝
x
2
y
3
z
⎞
⎟
⎟
⎟
⎠
=
n
·
u
,
where
u
=
⎛
⎜
⎜
⎜
⎝
x
2
y
3
z
⎞
⎟
⎟
⎟
⎠
.
Therefore,
⌠
(⎜)
⌡
n
·
u
dS
=
⌠
⌡
⌠
⌡
⌠
⌡
∇·
u
dV
= 6 ×
volume
= 6 ×
4 π
3
×1 ×
1
√2
×
1
√3
=
8 π
√6
=
4√6
3
π.
File translated from T
E
X by
T
T
H
, version 4.03.
On 08 Oct 2025, 22:22.
