2.
Let
V
be the set in the first octant in
3
R
bounded by the surface
8
4
2
3
3
3
=
+
+
z
y
x
and the
coordinate planes and let the map
be defined by
.
3
3
:
R
R
→
G
)
,
,
(
)
,
,
(
3
3
3
w
v
u
w
v
u
=
G
(a)
[7 marks]
Apply the change of variables
)
,
,
(
)
,
,
(
w
v
u
z
y
x
G
=
to the integral
.
Write
the iterated integral for
I
in the new variables in
∫∫∫
+
+
=
V
dV
z
y
x
I
)
1
ln(
such an order of integration that there are NO FRACTIONS in the limits of integration.
DO NOT EVALUATE THE INTEGRAL
We have to describe
such that
. It is easily seen that
is the tetrahedron with
vertices at (4, 0, 0), (0, 8, 0), (0, 0, 2). To avoid fractions we have to project
on
uw
-plane and
let
. On
uw
-plane we have the region bounded by coordinate axes and the line
. Hence we can describe it as
∗
V
)
(
∗
=
V
V
G
∗
V
∗
V
w
u
v
4
2
8
0
−
−
≤
≤
4
2
=
+
w
u
w
u
2
4
0
−
≤
≤
,
2
0
≤
≤
w
. So describe
as
∗
V
follows:
}
2
0
,
2
4
0
,
4
2
8
0
:
)
,
,
{(
≤
≤
−
≤
≤
−
−
≤
≤
=
∗
w
w
u
w
u
v
w
v
u
V
. Given function
s
continuous, hence integrable
and we get
R
R
V
f
→
⊂
3
:
i
dudvdw
w
v
u
w
v
u
dV
z
y
x
I
V
V
)
,
,
(
det
)]
1
ln(
[
)
1
ln(
3
3
3
G
+
+
=
+
+
=
∫∫∫
∫∫∫
∗
.
Now
, and in the required order of integration, the iterated integral can be
written as
2
2
2
27
)
(
det
w
v
u
u
=
G
dw
du
dv
w
v
w
v
u
I
ww
u
)
1
ln(
27
3
2
0
2
4
0
4
2
8
0
3
2
2
5
+
+
=
∫∫ ∫
−−
−
(b)
[2 marks]
Explain
briefly
, why the given change of variables could be applied for
I
although
0
)
,
,
(
det
=
w
v
u
D
G
on some subset of
the inverse image of
V
under
G
.
The applicability of given change of variables follows directly from “The Change of Variables
Theorem”. The map
G
is of class
on
1
C
3
R
and on any open box
)
,
0
(
)
,
0
(
)
,
0
(
0
a
a
a
U
×
×
=
,
with
a
> 8,
G
is 1-1 on
and
0
U
0
)
,
,
(
det
≠
w
v
u
D
G
on
.
The open set
0
U
}
2
0
,
2
4
0
,
4
2
8
0
:
)
,
,
{(
<
<
−
<
<
−
−
<
<
=
w
w
u
w
u
v
w
v
u
T