Spring 2004 Math 253/501–503
13 Multiple Integrals
13.8 Triple Integrals [in rect coords]
c
Tue, 09/Mar
2004,
Art Belmonte
Table Notes
1. The rectangular volume differential is given. In §13.10, we’ll
see the cylindrical volume differential d V = r dz dr dθ and
the spherical volume differential d V = ρ 2 sin φ dρ dφ dθ . In
§13.11, we’ll encounter both the general volume differential
d V = J du dv dw and the Jacobian determinant J .
Definition of the triple Riemann integral
2. Other first-order moments are symmetrically defined.
ZZZ
ZZZ
Mxz =
yδ dV,
Mxy =
zδ dV
To define the triple integral of a function f (x,y,z) over a
rectangular box
E
B = {(x,y,z) : a ≤ x ≤ b, c ≤ y ≤ d, r ≤ z ≤ s} ,
3. Other second-order moments are symmetrically defined.
ZZZ ZZZ x 2 + z 2 δ dV,
Iz =
x 2 + y 2 δ dV
Iy =
first partition the box. That is, split [a, b] into l subintervals, [c, d]
into m subintervals, and [r, s] into n subintervals. (Typically, the
x-intervals are equal-length, as are the y- and z-subintervals.) The
norm kPk of the resulting partition P is the length of the longest
diagonal among the sub-boxes of the partition. We then form a
triple Riemann sum and take the limit as kPk shrinks to 0. If this
limit exists, we obtain the triple integral of f over B.
ZZZ
Z
B
f (x,y,z) dV =
= lim
kPk→0
m X
n
l X
X
sZ
r
c
f
dZ
b
E
Aids to our hand work are the TAMUCALC Muint menu and the
smi command. The former contains a variety of multiple integral
templates that facilitate typing. The latter shows the steps involved
in integration.
f (x,y,z) d x d y dz
E
Hand / MATLAB Examples
a
x i∗ , y ∗j , z k∗
E
Pay careful attention to the 2-D and 3-D graphics in the examples.
These will help you to set up the limits of integration in your triple
integrals. With practice, you’ll be able to draw rough sketches that
will suffice. In upcoming computer assignments, you will draw
some graphs yourselves with MATLAB.
1x i 1y j 1z k
i=1 j=1 k=1
Fubini’s Theorem
842/4
If f is continuous on E = [a, b] × [c, d] × [r, s], a rectangular
Z sZ dZ b
ZZZ
f (x,y,z) dV =
f (x,y,z) d x d y dz.
box, then
r
E
c
Z
1 Z 2x
Z
x+y
Evaluate the iterated integral
a
2x y dz d y d x.
0
(There are 5 other orders in which an iterated integral may be
expressed. Moreover, the region of integration E may be
nonrectangular; i.e., its boundaries may be curved surfaces.)
x
0
Solution
Z
1 Z 2x
Z
x+y
23
≈ 1.53. Verify this
15
0
x
0
on your TI-89 using Muint template #9. You may subsequently
Let E be the region of integration, δ the mass density, and σ the
show the steps involved using smi in the Calc menu. Here is a
charge density. The center of mass is also called the centroid when MATLAB diary file with code that computes the integral and
the density is constant.
shows you the steps involved.
We have
Applications of Triple Integrals
Application
3-D
differential1
dV =RRR
d x d y dz
V = RRR E 1 dV
volume
m = RRRE δ dV
Q = RRR
E σ dV
M yz =
dV
E xδ
RRR
1
[x̄, ȳ, z̄] = m
y, z] δ dV
E [x,
RRR
2 + z 2 δ dV
Ix =
y
p
√ E
√
x̄¯ = Ix /m, ȳ¯ = I y /m, z̄¯ = Iz /m
measure
total mass
electric charge
moments2
center of mass
moments of inertia3
radii of gyration
2x y dz d y d x =
%
% Stewart 842/4
%
syms x y z
f = 2*x*y;
%
exact answer = int(int(int(f, ...
z,0,x+y), y,x,2*x), x,0,1)
exact answer =
23/15
1
onto the x y-plane. Together these will help us to determine the
limits of integration involved in the triple integral.
floated = eval(exact answer)
floated =
1.5333
%
% Step-by-step!
exact answer = smi(f, [z,0,x+y; y,x,2*x; x,0,1]);
STEPWISE (MULTIPLE) INTEGRATION!
Stewart 843/12: Projection onto xy−plane
Stewart 843/12: Tetrahedron
D
Top plane z = y − x
Antiderivative w.r.t. z:
1
1
0.8
0.8
When z = x+y:
0.6
y
2 x y z
0.6
z
0.4
0.4
2 x y (x + y)
C
B
When z = 0:
0.2
0
0
0
Difference:
1
A
0.5
0.2
0.5
0
y
−0.2
1 0
x
2 x y (x + y)
0
0.5
x
1
Antiderivative w.r.t. y:
3
You can find an equation (z = y − x) of the top plane of the
tetrahedron with your TI-89 by using the plane3pt in the CLP
menu followed by the solve command in the Alg menu. This
together with the pictures enables you to determine the ranges for
z, y, and x; i.e., innermost to outermost.
2
+ 1/2 x y )
2 x (1/3 y
When y = 2*x:
4
28/3 x
When y = x:
4
• For fixed x and y, we see that z varies from the plane z = 0
(the x y-plane) to the top plane z = y − x.
5/3 x
Difference:
4
• Then, for fixed x, we see y varies from the slanted line y = x
to the horizontal line y = 1.
23/3 x
Antiderivative w.r.t. x:
23 5
-- x
15
• Finally, the numerical span of x is from x = 0 to x = 1.
When x = 1:
Therefore, the integral is
23
-15
Z
When x = 0:
0
0
1 Z 1 Z y−x
x
x z dz d y d x =
0
1
≈ 0.0083.
120
Difference:
Verify this on your TI-89.
23
-15
Answer (above) and approximaton (below)
1.5333
%
%
% Stewart 843/12
%
syms x y z
f = x*z;
%
exact answer = int(int(int(f, ...
z,0,y-x), y,x,1), x,0,1)
echo off; diary off
843/12
ZZZ
exact answer =
x z d V , where E is the
Evaluate the triple integral
1/120
E
tetrahedron with vertices (corners)
A(0, 0, 0),
B(0, 1, 0),
C(1, 1, 0),
floated = eval(exact answer)
floated =
0.0083
%
D(0, 1, 1).
[A tetrahedron is a solid with 4 flat sides; i.e., a pyramid with 3
triangular sides and a triangular base.]
echo off; diary off
843/13
Solution
ZZZ
z d V , where E is the solid
We’ll draw two plots, one showing the edges of the tetrahedron in Evaluate the triple integral
E
space and the other showing the projection of the solid tetrahedron bounded by the planes x = 0, y = 0, z = 0, y + z = 1, x + z = 1.
2
Solution
Stewart 843/16: Projection of solid onto xy−plane
3.5
Stewart 843/16: Edges of solid
Cylinder y2 + z2 = 9
Here are 3-D and 2-D plots. You’re instinct is to project the solid
onto the x y-plane. That would work, but it would involve the
computation of two triple integrals which you would then add
together to give the final result. (Why?) Instead, we project onto
the back yz-plane. (We could alternatively project onto the left
x z-plane.) This involves the computation of one triple integral.
2
1
1
0.5
0
0
0
0
0.5
1
2
3 1
−0.5
−0.5
x
y
Z
Stewart 843/13: Edges of solid
1
√
1Z 3Z
9−y 2
The integral is
0
0.8
0.8
1.5
z
y
2
Stewart 843/13: Projection onto the yz−plane
1
3
2.5
3
3x
z dz d y d x =
0
0
0.5
x
1
1.5
27
= 3.375.
8
z
0.6
0.6
%
% Stewart 843/16
%
syms x y z
f = z;
%
exact answer = int(int(int(f, ...
z,0,sqrt(9-yˆ2)), y,3*x,3), x,0,1)
z
0.4
0.4
0.2
0.2
0
0
0
0
0.5
0.5
1
1
−0.2
x
y
1 Z 1−y
Z
Z
1−z
The integral is
0
0
0
z d x dz d y =
0
0.5
y
1
exact answer =
1
≈ 0.083.
12
27/8
floated = eval(exact answer)
floated =
3.3750
%
%
% Stewart 843/13
%
syms x y z
f = z;
%
exact answer = int(int(int(f, ...
x,0,1-z), z,0,1-y), y,0,1)
echo off; diary off
843/18
Use a triple integral to find the volume of the solid E bounded by
the elliptic cylinder 4x 2 + z 2 = 4 and the planes y = 0 and
y = z + 2.
exact answer =
1/12
floated = eval(exact answer)
floated =
0.0833
%
%
Solution
echo off; diary off
Here are two 3-D views of the solid’s faces, followed by its
projection onto the x z-plane.
Faces of solid
Projection onto xz−plane
Backside view
2
2
843/16
1
2
0
1
z
1
Evaluate the triple integral
z
z
ZZZ
0
0
−1
−1
−1
z d V , where E is the solid
4
−2
0
2
E
y2 + z2 =
4
bounded by the cylinder
9 and the planes x = 0,
y = 3x, and z = 0 in the first octant.
0
x
1
y
−2
−2
−1
−1
2
x
0
1
0
−1
y
0
x
1
The volume is
Z
ZZZ
1 dV =
V =
E
Solution
1
Z
√
4−4x 2
√
−1 − 4−4x 2 0
%
% Stewart 843/18
%
syms x y z
f = 1; s = sqrt(4 - 4*xˆ2);
Here are 3-D and 2-D plots.pSince we are in the first octant
y 2 + z 2 = 9 implies z = + 9 − y 2 .
3
Z
z+2
1 d y dz d x = 4π ≈ 12.57 cm3 .
%
volume = int(int(int(f, ...
y,0,z+2), z,-s,s), x,-1,1)
Stewart 843/16: Projection of solid onto xy−plane
3.5
Stewart 843/16: Edges of solid
Cylinder y2 + z2 = 9
volume =
3
4*pi
2
3
2.5
z
y
2
1.5
floated = eval(volume)
floated =
12.5664
%
%
1
1
0
0
0
1
0
0
4−y 2
0
3 1
x
1
=
20
≈ 6.67 m3 .
1 d x dz d y =
3
0
=
CM [x̄, ȳ, z̄]
Solution
=
=
The projection of the solid onto the back plane is a triangle whose
slanted boundary is z = 2 − y. For fixed y and z, x varies from
the back plane x = 0 to the parabolic cylinder x = 4 − y 2 . Here is
a 3-D picture.
=
Iz
=
Stewart 843/26
=
0
0.5
x
1
1.5
=
1.5
9−y 2
3
3x
x 2 + y 2 dz d y d x
0
56
= 11.2
5 ZZZ
1
[x, y, z] δ dV
m
E
Z 1 Z 3 Z √9−y 2
1
[x, y, z] x 2 + y 2 dz d y d x
56/5 0 3x 0
h
i
855 π, 45 π, 15 ≈ [0.3747, 2.2089, 0.9375]
7168
64
16
ZZZ x 2 + y 2 δ dV
E
Z 1 Z 3 Z √9−y 2 2
x 2 + y 2 dz d y d x
0
2
z
−0.5
−0.5
The limits of integration are exactly the same as in 843/16.
We simply change the integrands. Do these on your TI-89!
ZZZ
mass m =
δ dV
E
Z Z Z √
Sketch the solid whose volume is given by the iterated integral
Z
2
y
843/26
2 Z 2−y
0
0.5
echo off; diary off
Z
0.5
3x
0
10464
= 59.7943
175
1
843/48
0.5
0
0
ZZZ
1
Find the average value f avg =
f d V of the function
V
E
f (x,y,z) = x + y + z over the solid tetrahedron with vertices
(0, 0, 0), (1, 0, 0), (0, 1, 0), and (0, 0, 1).
0
2
1
2 4
x
y
Solution
844/40+44
Draw a rough sketch of theRRR
tetrahedron, then set up your limits of
1
integration. The volume is
E 1 d V = 6 . The average value is
RRR
1
3
f avg = 1/6
E x + y + z d V = 4 . Make sure you can compute
this on your TI-89. Ask if you need help with the sketch or limits
of integration!
Assume that the variable density of the solid in Exercise 843/16 is
δ = x 2 + y 2 . Compute the solid’s (a) mass, (b) center of mass
(CM), and (c) moment of inertia Iz with respect to the z-axis.
Solution
We recall the pix from 843/16. “Our creed is firepower!”
4