Solutions to the 1st Midterm
1.
Consider the following integral
Z
1
0
Z
1

2
x

1
Z
1

y
2
0
f
(
x, y, z
)
dz dy dx.
(a)
Rewrite this integral so that the innermost integral is with respect to
z
, the
middle integral is with respect to
x
, and the outer integral is with respect to
y
.
As given in the integral, the region of integration can be viewed as the region under the
graph of
z
= 1

y
2
over a domain
D
in the
xy
plane. Further,
D
is the domain bounded
by the lines
y
=

1,
x
= 0, and
y
= 1

2
x
. If we want to switch the order of
x
and
y
integration, this means that
y
goes from

1 to 1 while
x
goes from 0 to
1
2
(1

y
). Thus
the integral can be rewritten as
Z
1

1
Z
1
2
(1

y
)
0
Z
1

y
2
0
f
(
x, y, z
)
dz dx dy.
(b)
Rewrite this integral so that the innermost integral is with respect to
x
, the
middle integral is with respect to
z
, and the outer integral is with respect to
y
.
In this case, we think of our region as the region under a graph of
x
as a function of
y
and
z
over some domain in the
yz
plane. The domain in the
yz
plane is bounded by
z
= 0
and
z
= 1

y
2
, so
y
goes from

1 to 1 while
z
goes from 0 to 1

y
2
. In the
x
direction,
the region is bounded form below
x
= 0 and from above by
x
=
1
2
(1

y
).
Thus the integral can be rewritten as
Z
1

1
Z
1

y
2
0
Z
1
2
(1

y
)
0
f
(
x, y, z
)
dz dx dy.
2.
Consider the surface given by the part of the plane
2
x
+
y
+ 2
z
= 8
that lies in the
ﬁrst octant. Find its surface area. Be sure to show the integral you set up as well
as your ﬁnal answer.
The part of the plane which lies in the ﬁrst octant is a triangle with vertices at (4
,
0
,
0),
(0
,
8
,
0), and (0
,
0
,
4); these are just the intersections of the plane with the coordinate axes.
We can view this as the graph of
z
= 4

x

1
2
y
over the triangle in the
xy
plane bounded
by
x
= 0,
y
= 0 and 2
x
+
y
= 8. To ﬁnd the area of the graph of a function, we integrate
p
(
∂f/∂x
)
2
+ (
∂f/∂y
)
2
+ 1 over the domain. In the present case, we have that