MATH 253 WORKSHEET 30
TRIPLE INTEGRALS AND APPLICATIONS
y=1
z=1−y
dx y=√x dy z=0
x=0
coming from changing the order of integration.
(1) Consider the iterated integral
Solution:
x=1
dzf .
Write the other
5
equivalent integrals
See WS 29.
(2) Find the volume and the center-of-mass of the solid bounded by the parabolic cylinder
xy
y + z = 1.
plane y + z = 1 intersects
y = x2 ,
the
plane, and the plane
Solution:
xy plane (where z = 0) in the line y = 1. Let R be
y = x2 and the line y = 1. The solid then consists
of the points above R and below the plane y + z = 1 [why? either draw a picture (which is enough
for this course) or compare with the solid bound by the cylinder, the xy plane, and the plane y = 1
which contains the original solid (since the plane y = 1 is always farther out than y + z = 1 if
z ≥ 0) and by construction has base R]. Considering a point (x, y) in R, the set of points (x, y, z) in
our solid lying above it is a z -line beginning at the base (xy plane, z = 0) and ending at the roof plane y + z = 1. Converting the two endpoints to statements about z , integrals over the solid will be
z=1−y
of the form
dx dy z=0 dzf (x, y, z). We can slice the region R itself horizontally or vertically.
R
• Slicing vertically, the x range is [−1, 1]. In this range every verticle y line begins at at the
2
parabolic y = x and ends at the line y = 1. The full integral is then
The
the
the region in the plane bounded by the parabola
x=1
Slicing horizontally, the
y
z=1−y
dy
dzf (x, y, z) .
y=x2
x=−1
•
y=1
dx
range is
z=0
[0, 1].
In this range every horizontal x line beings at the left
y = x2
y = x2 but
arm of the parabola (at a point where
x=
√
y ).
Both endpoints satisfy
so
√
x = − y)
and ends at the right arm (where
are not the same point, since
y
has
two
square
roots. The full integral is also
y=1
dy
y=0
Remark:
√
x= y
√
x=− y
Note that the equalities
dzf (x, y, z) .
z=0
y = x2
the domain.
Volume:
The volume
x=1
y=1
dx
x=−1
V
z=1−y
dz · 1
z=0
endpoints
of our x line, not
=
=
y=1
dy(1 − y)
dx
=
x=1
=
=
Date
only hold at the
of the region is
dy
y=x2
z=1−y
dx
x=−1
y=x2
y=1
y2
dx y −
2 y=x2
x=−1
x=1
1
x4
2
dx 1 − − x +
2
2
x=−1
3
5 x=1
1
x
x
x−
+
2
3
10 x=−1
1 1
1
2(15 − 10 + 3)
8
2
− +
=
=
.
2 3 10
30
15
: 22/11/2013.
1
x=1
inside
Or equivalently
y=1
dy
y=0
√
x= y
√
x=− y
z=1−y
dz · 1
dx
y=1
dy
=
z=0
y=0
y=1
=
√
x= y
√
x=− y
dx (1 − y)
√
dy(1 − y)2 y
y=0
y=1
2 3/2 2 5/2
2 y
− y
3
5
y=0
1 1
8
4
−
=
.
3 5
15
=
=
Center of mass:
The region is symmetric under reection in the
x2 ),
the denitions (which only involve
x-variable.
x2 ),
symmetric and any later dependence is on
so
x̄ = 0.
=
=
=
=
=
=
1
volume
1
8/15
x=1
x=−1
x=1
y=1
z=1−y
dz · y
dy
dx
y=x2
y=1
z=0
dy(1 − y)y
dx
y=x2
x=−1
y=1
y2
y3
−
dx
2
3 y=x2
x=−1
x=1
15
1 1 x4
x6
dx
− −
+
8 x=−1
2 3
2
3
x=1
15 x x5
x7
−
+
8 6
10 21 x=−1
15 1
1
1
15 35 − 21 + 10
3
2
−
+
=
=
8
6 10 21
4
210
7
15
8
x=1
and
z̄
=
=
=
=
15
8
x=1
y=x2
y=1
x=−1
x=1
=
=
z=1−y
dz · z
dy
z=0
(1 − y)2
2
x=−1
y=x2
x=1
y=1
15
(y − 1)3
dx
8 x=−1
6
y=x2
x=1
15
(x2 − 1)3
dx −
8 x=−1
6
15
8
dx
dy
=
y=1
dx
x=1
15
·2
dx 1 − 3x2 + 3x4 − x3
8·6
x=0
x=1
5
3x5
x7
x − x3 +
−
8
5
7 x=0
5
3 1
5 21 − 5
2
1−1+ −
=
= .
8
5 7
8
35
7
2
x
integral are
We will need to integrate to nd
Using the rst form of the integral, we have
ȳ
This is clear from
or from the integral (where the bounds on the
ȳ , z̄ .
In other other order of integration the integrals would look like
ȳ
=
=
15
8
=
15
4
and
=
=
=
=
=
15
8
=
=
z̄
15
8
15
8
15
8
y=1
dy
y=0
y=1
dy
y=0
y=1
y=0
y=1
√
x= y
√
x=− y
x=√y
√
x=− y
z=1−y
dz · y
dx
z=0
dxy(1 − y)
√
dx2 y · y(1 − y)
y 3/2 − y 5/2
y=0
15 2 2
15 7 − 5
3
−
=
=
4 5 7
2
35
7
y=1
dy
y=0
y=1
dy
y=0
√
x= y
√
x=− y
x=√y
√
x=− y
dz · z
z=0
dx
(1 − y)
2
2
y=1
z=1−y
dx
2
√ (1 − y)
dy2 y
2
y=0
y=1 15
y 5/2 − 2y 3/2 + y 1/2 dy
8 y=0
2 2
15 15 − 42 + 35
2
15 2
− 2· +
=
·
=
8 7
5 3
4
105
7
15
8
3