Self-Help Work Sheets C11: Triple Integration Problems for ...
[Pages:10]SHWS C11: TRIPLE INTEGRATION
29
Self-Help Work Sheets C11: Triple Integration
These problems are intended to give you more practice on some of the skills the chapter on Triple Integration has sought to develop. They do not cover everything so a careful review of the Chapter and your class notes is also in order.
Problems for Fun and Practice
1. For each of the following solids give a description in rectangular coordinates in the order speci?ed:
(a)
S in
is bounded above terms of range on
by z,
x2 y2 th9en+y,3t6he+n
z2 x1.6
=
1 and below by z
=
2.
Description:
(b)
S
is
bounded
above
by
x2 16
+
y2 36
+
z2 16
= 1 and below by
yz 6+4
= 1. Description:
in terms of range on z, then x, then y.
2. Each of the following iterated integrals cannot be easily done in the order given. Convince yourself that this is true and then convert each one to an equivalent iterated integral that can be done and evaluate it.
(a) !2!1!1 sinh "z2# dzdyd x
00y
!2 !4 !2
(b)
yzex3dxdydz
00z
3. Convert each of the following to an equivalent triple integ4al in cylindrical coordinates and evaluate.
!5 (a)
!25-x2!6 d zd yd x $
0 0 0 x2 + y2
(b)
!2
!4-x 2
x2+y2
!2
zdzdydx
$
00
0
x2 + y2
30
4. Convert each of the following to an equivalent triple integral in spherical coordinates and evaluate.
(a)
!1
!1-x
2 1-!x
2-
y2
dzdydx
1 + x2 + y2 + z2
00
0
!3
!9-x
2 9-!x
2-
y2
(b)
xzdzdydx
00
0
5. Convert to cylindrical coordinates and evaluate the integral
!!!$
(a)
x2 + y2d V where S is the solid in the ?rst octant
S
bounded by the coordinate plane, the plane z = 4, and the cylinder x2 + y2 =
(b)
!
!
!
"x2
+
y
2
#
3 2
dV
25. where S is the solid bounded above by
S
the low
paraboloid z = by the x y-plane,
21an"dx
2 + y2#, laterally
beby
the cylinder x2 + y2 = 4.
6. Convert to spherical coordinates and evaluate the integral.
(a)
!
!
!
"x2
+
y2
+
z
2
#
3 2
dV
where S is the solid in the ?rst oc-
S
tant bounded by the sphere x2 +
$y$2x+2 +z
2
y
= 25, 2, and
the the
cone cone
z z
= =
2 x2 + y2.
!!! $
(b)
sin x2 + y2 + z2d V where S is the solid bounded
S
above by the sphere x2 + y2 +
zz2==$4x92
and below + y2.
by
the
cone
7. Find the coordinates of the center of gravity of the solid S with indicated mass density = (x, y, z). (Choose whichever system of coordinates would be best.)
(a) S: 0 x 1, 0 y 1, 0 z 1; = 3 - x - y - z.
SHWS C11: TRIPLE INTEGRATION
31
(b)
S:
x2 + y2
a2,
b a
$ x
2
+
y2
z
b
for constants b
>
0,
a
>
0 and
= x2 + y2 + z2.
8. Find the centroid for each of the following solids S:
(a) S: x2 + y2 1, x 0, y 0, 0 z x y (b) S: 9 x 2 + y2 + z2, z 0
9. Find the designated moment for each solid S with density = (x, y, z).
(a) Mxy ? the moment relative to the x y-plane where S: 0 x 1, 0 y 1 - x, 0 z 1 = 3xy
(b) Mxz ? the moment relative to the x z-plane where S: 0 x 1, 0 y 1, 0 z x y = 2 (x + y)
(c) Myz ? the moment relative to the yz-plane where S: bounded in the ?rst octant by x + z = 1, x = y, and the coordinate planes = 5y
10. Find the moment of inertia about the z-axis for the solid S bounded by 0 x 1, 0 y 1 - x, 0 z 5 with density = (x, y, z) = 3.
11. Find the moment of inertia about the y-axis of the solid S in the ?rst octant bounded by x2 + z2 = 1, y = x, y = 0, z = 0 with density = 2z.
12. Find the moment of inertia about the central axes of a homogeneous right circular cylindrical shell with total mass m, inner radius a, outer radius b and height h.
Proposed Solutions/Answers
1. (a)
(x/3)2+(y/6)2+(z/4)2=1 the plane z=2
To ?nd the range of x and y we
substitute the value of z into the
equation will give
x2 th9e
y2 z2 l+arg3e6st+ran1g6e
= for
1 x
that and
y. From inspecting the ?gure we
see that this value is z = 2.
32
For
z
=
2,
we
have
x2 9
+
y2 36
+
(2)2 16
=
1
or
x2 9
+
y2 36
=
3 4
.
From
this
we
obtain:
% 2z 4 1- --63%34x-x932
x2 y2
9 -%36
y6
3 4
-
x2 9
(b)
the plane (y/6)+(z/4)=1
To ?nd the scope of x and y
we notice from the ?gure that
(x/4)2+(y/36)2+(z/4)2=1 the largest range of x and y
iyn thze 6+4
solid = 1.
occur on Thus we
the plane solve this
for z and substitute this into
the equation Substitution
x2 y2 z2 16 + 36 + 16 this value of z
= 1. into
Solving
y 6
+
x2 y2 z2
16 + 36 + 16
z 4
=
1 for z
we get
z
=
1,
we
get
x2 16
+
y2 36
= +
146&&11--6y '26y
16
' .
=
1
or
x2 16
+
y2 18
+
y 3
=
1
or
x2 16
+
(y
- 3)2 18
=
1 2
.
From
this
we
see
that
y
will
be
maximized when x = 0. Thus we obtain the desired description.
4
& 1
y -(6
'
z
% 41
-
x2 16
-
y2 36(
-22
1
-
(y
- 3)2 9
x
22
1
-
(y
- 9
3)2
0 y 6.
SHWS C11: TRIPLE INTEGRATION
33
2. (a)
!2 !1 !1
&'
sinh z2 d zd yd x
00y
!2 !1 !z
&'
=
sinh z2 dydzd x
000
!2 !1
&'
= z sinh z2 dzd x
00
=
!2
1 2
(cosh
(1)
-
cosh
(0))
d
x
0
= cosh (1) - cosh (0)
=
e1
+ e-1 2
-1
The region indicated by the integral is
bounded by z = y, y = 0, z = 1, x = 0,
and x = 2 which is indicated by the ?gure
above. The dif?culty with integrating the
ogrriagteinsailnthri"pzl2e#i,nwteegrnaleeisd
that to a zdz
easily rather
intethan
just d z. Note that if we switch the dz and
d y, we might get a z where we need it.
(b)
!2 !4 !2 yzex3dxdydz
00z
!2 !4 !x
=
yzex3dzdydx
000
=
!2 !4
y
x2 2
ex3dyd
x
00
=
!2
0
) y2 *4 20
x2 2
ex
3
d
x
!2 = 4 x2ex3dx
0
=
4 3
ex
3
*2
0
=
4 3
"e8
-
1#
The region described by the integral is bounded by y = 0, y = 4, z = 0, z = x, and x = 2. A picture of the region is indicated above. In the original integral, if we try to integrate ex3d x we have a problems. We can easily integrate x2ex3, so this suggests
switching d x and dz.
34
3. (a)
!5 !25-x2!6 d zd yd x
$ 0 0 0 x2 + y2
=
!/2!5!6 r dzdr d r
=
15
0 00
The region described by the integral the three inequalities 0 z 6, 0
is y
thes2e5t
of -
all x2,
point and 0
(x, y, z) satisfying
x
5.
It
is
1 4
of
a cylinder as shown in the ?gure above.
(b)
!2
!4-x
2
x2+y2
!2
zd
z
d
y
d
x
00
0
$ x
2
+
y2
!/2!2 !r2/2zr d zdr d
2
=
r =5
00 0
The region or solid is in the ?rst octant between the paraboloid and the cylinder
as indicated in the picture above. It is the set of all points (x, y, z) satisfying
the
three
inequalities
0
z
x2
+ 2
y2 ,
0
y
4
-
x2,
and
0
x
2.
SHWS C11: TRIPLE INTEGRATION
4. (a)
!1
!1-x
2
1-! x
2-
y2
dzdydx
00
1 + x2 + y2 + z2
0
!/2!/2!1 2 sin ddd
=
1 + 2
0 00
=
!/2!/2!1
sin
+ 1
-
1
1 +
2
*
d
d
d
0 00
=
& 1
-
4
'
!/2 (cos
]0 /2
d
0
=
& 1
-
4
'
2
=
2 2- 8
35
The with
sroaldidiusis181ceonftetrheed
sphere at the
origin as indicated in the ?g-
ure.
(b)
!3
!9-x
2
9-! x
2-
y2
xzdzdydx
00
0
!/2!/2!3
=
( sin cos ) ( cos ) 2 sin ddd
0 00
!/2!/2!3
=
4 cos sin2 cos ddd
0 00
=
35 5
!/2 cos
, sin3 -=/2 3
d
=
34 5
0
=0
The solid is sphere with
r18adoifusth3e
centered at the origin
as depicted in the ?g-
ure in part a.
x = sin cos
z = cos
36
5. (a)
!!!. x2 + y2dV
S
!2!5 !4
=
r ? rdzdrd
4
000
=
4 3
(5)3
2
x2+y2=25
(b)
!!! &
'3
x2 + y2 2 dV
S
!2!2 !r2/2
=
r3 ? rdzdrd
00 0
=
!2!2
r6 2
dr
d
00
=
!2
0
r7 14
*2
0
d
=
)
26 7
/
(2
)
z=(x2+y2)/2 z goes upto 2
x2+y2=4
................
................
In order to avoid copyright disputes, this page is only a partial summary.
To fulfill the demand for quickly locating and searching documents.
It is intelligent file search solution for home and business.
Related download
- fun true or false questions
- conflict resolution activities for middle school skill
- resident and family engagement in nursing home quality
- self help work sheets c11 triple integration problems for
- © 1 fun ways to stay fite fun
- aidet overview why what how paetc
- myths vs facts about counseling
- confidence activities polk
- resume writing workshop activities
- personality test guide jobtestprep
Related searches
- free self help worksheets printable
- mental health self help workbooks
- free printable self help workbooks
- free pdf self help workbooks
- self help worksheets mental health
- self help journals
- calculus integration problems and solutions
- integration problems and answers
- integration problems and solutions pdf
- self help workbooks free download
- self help printable workbooks pdf
- clark county family court self help center