THE DOUBLE INTEGRAL AS THE LIMIT OF RIEMANN SUMS; POLAR COORDINATES
1
THE DOUBLE INTEGRAL AS THE LIMIT OF
RIEMANN SUMS; POLAR COORDINATES
The Double Integral as the Limit of Riemann Sums
In the one-variable case we can write the integral as the limit of Riemann sums:
Z b
a
f (x) dx =
lim
max ∆xi →0
n
X
f (xi∗ )∆xi .
i=1
The same approach works with double integrals. To explain it we need to explain
what we mean by the diameter of a set. Suppose that S is a bounded closed set
(on the line, in the plane, or in three-space). For any two points P and Q of S we
can measure their separation, d(P, Q). The greatest separation between points of S is
called the diameter of S:
diam S = max d(P, Q).
P,Q∈S
For a circle, a circular disc, a sphere, or a ball, this sense of diameter agrees with
the usual one.
Now let’s start with a basic region Ω and decompose it into a finite number of basic
subregions Ω1 , · · · , ΩN .
If f is continuous on Ω, then f is continuous on each Ωi . Now from each Ωi , we
pick an arbitrary point (x∗i , yi∗ ) and form the Riemann sum
N
X
f (x∗i , yi∗ )(area of Ωi )
i=1
The double integral over Ω can be obtained as the limit of such sums; namely, given
any  > 0, there exists δ > 0 such that, if the diameters of the are all Ωi less than δ
then
Œ
Œ
N
ŒX
Œ
f (x∗i , yi∗ )(area
Œ
Œ
i=1
of Ωi ) −
ZZ
Ω
Œ
f (x, y) dxdy ŒŒŒ < 
no matter how the (x∗i , yi∗ ) are chosen within the Ω. We express this by writing
ZZ
Ω
f (x, y) dxdy =
lim
diamΩi →0
N
X
f (x∗i , yi∗ )(area of Ωi ).
i=1
Evaluating Double Integrals Using Polar Coordinates
Here we explain how to calculate double integrals
ZZ
f (x, y) dxdy
Ω
using polar coordinates (r, θ). Throughout we take r ≥ 0 and restrict θ to [0, 2π].
March 17, 2001
THE DOUBLE INTEGRAL AS THE LIMIT OF RIEMANN SUMS; POLAR COORDINATES
2
We will work with the type of region consisting of all points (x, y) that have polar
coordinates (r, θ) in the set
Γ : α ≤ θ ≤ β, ρ(θ) ≤ r ≤ P (θ).
According to the formula in one-variable case the area of Ω is given by
area of Ω =
Z β
α
1 2
[P (θ) − ρ2 (θ)] dθ.
2
We can write this as a double integral over Γ:
ZZ
area of Ω =
r drdθ.
Γ
Proof. Simply note that
1 2
[P (θ) − ρ2 (θ)] =
2
and therefore
area of Ω =
Z β Z P (θ)
α
ρ(θ)
Z P (θ)
ρ(θ)
rdrdθ =
rdr
ZZ
r drdθ.
Γ
Now let’s suppose that f is some function continuous at each point (x, y) of Ω.
Then the composition
F (r, θ) = f (r cos θ, r sin θ)
is continuous at each point (r, θ) of Γ. We will show that
ZZ
f (x, y) dxdy =
Ω
ZZ
f (r cos θ, r sin θ)r drdθ.
Γ
Proof. Our first step is to place a grid on Ω by using a finite number of rays θ = θj
and a finite number of continuous curves r = ρk (θ). This grid decomposes Ω into a
finite number of little regions
Ω1 , · · · , ΩN
with polar coordinates in sets Γ1 , · · · , ΓN . Note that
the area of each Ωi =
ZZ
r drdθ.
Γi
Writing F (r, θ) for f (r cos θ, r sin θ), we have
ZZ
F (r, θ)r drdθ =
=
i=1
F (ri∗ , θi∗ )
F (r, θ)r drdθ (additivity)
i=1 Γ
i
Γ
N
X
N ZZ
X
ZZ
Γi
r drdθ with some (ri∗ , θi∗ ) ∈ Γi (Mean-value Theorem)
Evaluating Double Integrals Using Polar Coordinates
March 17, 2001
THE DOUBLE INTEGRAL AS THE LIMIT OF RIEMANN SUMS; POLAR COORDINATES
=
N
X
F (ri∗ , θi∗ )(area
of Ωi ) =
N
X
3
f (xi∗ , yi∗ )(area of Ωi ).
i=1
i=1
This last expression is a Riemann sum for the double integral
ZZ
f (x, y) dxdy
Ω
and, as such, differs from that integral by less than any preassigned positive  provided
only that the diameters of all the Ωi are sufficiently small. This we can guarantee by
making our grid sufficiently fine.
Problem 1. Use polar coordinates to evaluate
ZZ
xy dxdy
Ω
where Ω is the portion of the unit disc that lies in the first quadrant.
Solution. Here
1
Γ : 0 ≤ θ ≤ π, 0 ≤ r ≤ 1.
2
Therefore
ZZ
ZZ
xy dxdy =
(r cos θ)(r sin θ)r drdθ
Ω
Γ
Z π/2 Z 1
1
r3 cos θ sin θ drdθ = .
8
0
0
Problem 2. Use polar coordinates to calculate the volume of a sphere of radius
=
R.
Solution. In rectangular coordinates
V =2
ZZ q
Ω
R2 − (x2 + y 2 ) dxdy
where Ω is the disc of radius R centered at the origin. Here
Γ : 0 ≤ θ ≤ 2π, 0 ≤ r ≤ R.
Therefore
V =2
ZZ q
Γ
R2 − (x2 + y 2 ) rdrdθ
Z 2π Z R √
4
R2 − r2 rdrdθ = πR3 .
3
0
0
volume of the solid bounded above by the lower nappe
Problem 3. Calculate
√ 2 the
2
of the cone z = 2 − x + y and bounded below by the disc Ω : (x − 1)2 + y 2 ≤ 1.
Solution
ZZ
q
2 − (x2 + y 2 ) dxdy
V =
=2
=
ZZ
Ω
Ω
2 dxdy −
ZZ q
(x2 + y 2 ) dxdy.
Ω
Evaluating Double Integrals Using Polar Coordinates
March 17, 2001
THE DOUBLE INTEGRAL AS THE LIMIT OF RIEMANN SUMS; POLAR COORDINATES
4
The first integral is 2(area of Ω ) = 2π. We evaluate the second integral by changing
to polar coordinates.
The equation (x − 1)2 + y 2 = 1 simplifies to x2 + y 2 = 2x. In polar coordinates
this becomes r2 = 2r cos θ, which simplifies to r = 2 cos θ. The disc Ω is the set of all
points with polar coordinates in the set
1
1
Γ : − π ≤ θ ≤ π, 0 ≤ r ≤ 2 cos θ.
2
2
Therefore
ZZ q
(x2
+
y 2 ) dxdy
=
ZZ
2
r drdθ =
Γ
Ω
We then have V = 2π − 32
.
9
Problem 4. Evaluate
ZZ
Ω
Z π/2 Z 2 cos θ
−π/2
0
r2 drdθ =
32
.
9
1
dxdy
(1 + x2 + y 2 )3/2
where Ω is the triangle bounded by
y = 0, x = y, x = 1.
Solution. The vertical side of the triangle is part of the line x = 1. In polar
coordinates this is r cos θ = 1, which can be written r = sec θ. Therefore
ZZ
Ω
ZZ
1
r
dxdy
drdθ
=
(1 + x2 + y 2 )3/2
(1 + r2 )3/2
Γ
where
Γ : 0 ≤ θ ≤ π/4, 0 ≤ r ≤ sec θ.
The double integral over Γ reduces to
Z π/4 Z sec θ
0
0
"
Z π/4
r
1
√
drdθ =
2
3/2
(1 + r )
0
1 + r2
=
For θ ∈ [0, π/4]
Z π/4 0
#0
dθ
sec θ
!
1
dθ.
1− √
1 + sec 2 θ
1
cos θ
cos θ
.
=√ 2
=√
2
2 − sin 2 θ
1 + sec θ
cos θ + 1
Therefore the integral can be written
Z π/4 0
√
"
!
cos θ
−1 sin θ
√
−
sin
θ
1− √
dθ
=
2 − sin 2 θ
2
!#π/4
0
=
π π
π
− = .
4
6
12
2
The function f (x) = e−x has no elementary antiderivative. Nevertheless, by taking
a circuitous route and then using polar coordinates, we can show that
Z ∞
√
2
e−x dx = π.
−∞
Evaluating Double Integrals Using Polar Coordinates
March 17, 2001
5
SOME APPLICATIONS OF DOUBLE INTEGRATION
Proof. The circular disc Db : x2 +y 2 ≤ b2 is the set of all (x, y) with polar coordinates
(r, θ) in the set Γ : 0 ≤ θ ≤ 2π, 0 ≤ r ≤ b. Therefore
ZZ
e−(x
2 +y 2 )
dxdy =
Db
ZZ
2
e−r r drdθ =
Γ
=
Z 2π
0
Z 2π Z b
0
0
2
e−r rdrdθ
1
2
2
(1 − e−b )dθ = π(1 − e−b ).
2
Let Sa be the square −a ≤ x ≤ a, −a ≤ y ≤ a. Since Da ⊂ Sa ⊂ D2a and e−(x
positive,
ZZ
ZZ
ZZ
2
2
2
2
2
2
e−(x +y ) dxdy.
e−(x +y ) dxdy ≤
e−(x +y ) dxdy ≤
It follows that
ZZ
2
π(1 − e−a ) ≤
−a2
As a → ∞, π(1 − e
is
D2a
Sa
Da
2 +y 2 )
2 +y 2 )
e−(x
Sa
2
dxdy ≤ π(1 − e−4a ).
2
) → π and π(1 − e−4a ) → π. Therefore
lim
a→∞
ZZ
2 +y 2 )
e−(x
dxdy = π.
Sa
But
ZZ
2 +y 2 )
e−(x
Sa
=
=
Therefore
lim
a→∞
’Z a
−a
Z a
−a
e
−x2
Z a Z a
dx
2
dxdy =
−a
−a
“ ’Z a
−a
e−x dx = a→∞
lim
Z a Z a
−a
−a
2
e−(x
2 +y 2 )
dxdy
2
e−x · e−y dxdy
e
−y 2
’ZZ
Sa
“
dy =
e−(x
’Z a
2 +y 2 )
−a
−x2
e
dxdy
“2
.
=
√
π.
dx
“1/2
This integral comes up frequently in probability theory and plays an important role
in what is called “statistical mechanics.”
SOME APPLICATIONS OF DOUBLE INTEGRATION
The Mass of a Plate
Suppose that a thin distribution of matter, what we call a plate, is laid out in the
xy-plane in the form of a basic region Ω. If the mass density of the plate (the mass
per unit area) is constant, then the total mass M of the plate is simply the density λ
times the area of the plate:
M = λ×the area of Ω.
March 17, 2001
6
SOME APPLICATIONS OF DOUBLE INTEGRATION
If the density varies continuously from point to point, say λ = λ(x, y), then the
mass of the plate is the average density of the plate times the area of the plate:
M = average density × the area of Ω.
This is an integral:
ZZ
M=
λ(x, y) dxdy.
Ω
The Center of Mass of a Plate
The center of mass of a rod xM is a density-weighted average taken over the interval
occupied by the rod:
Z
b
xM M =
a
xλ(x) dx.
The center of mass of a plate (xM , yM ) is determined by two density-weighted
averages, each taken over the region occupied by the plate:
ZZ
xM M =
xλ(x, y) dxdy, yM M =
Ω
ZZ
yλ(x, y) dxdy.
Ω
Problem 1. A plate is in the form of a half-disc of radius a. Find the mass of the
plate and the center of mass given that the mass density of the plate varies directly as
the distance from the center of the straight edge of the plate.
Solution.
the plate over the x-axis. The mass density can then be written
√ Place
2
2
λ(x, y) = k x + y . Here
M=
ZZ
Ω
xM M =
k
q
ZZ
x2
y2
+
’ q
x k
Ω
dxdy =
Z πZ a
0
0
(kr)rdrdθ = k
’Z π
0
1 dθ
“ ’Z a
0
2
r dr
“
1
1
= k(π)( a3 ) = ka3 π.
3
3
x2
+
“
dxdy = 0. (the integrand is odd w.r.t. x)
+
y2
y2
Thus xM = 0.
yM M =
ZZ
’ q
y k
Ω
=k
’Z π
0
x2
sin θ dθ
Since M = 13 ka3 π, we have yM =
The Center of Mass of a Plate
“
“ ’Z a

0
1
ka4
2
dxdy =
Z πZ a
“
0
0
(r sin θ)(kr)rdrdθ
1
1
r3 dr = k(2)( a4 ) = ka4 .
4
2
‘ 
/
1
ka3 π
3
‘
= 3a/2π.
March 17, 2001
7
SOME APPLICATIONS OF DOUBLE INTEGRATION
Centroids
If the plate is homogeneous, then the mass density λ is constantly M/A where A is
the area of the base region Ω. In this case the center of mass of the plate falls on the
centroid of the base region . The centroid (x̄, ȳ) depends only on the geometry of Ω:
x̄M =
ZZ
x(M/A) dxdy = (M/A)
x dxdy
Ω
ZZ
ZZ
y(M/A) dxdy = (M/A)
ZZ
ȳM =
y dxdy.
Ω
Ω
Ω
Dividing by M and multiplying through by A we have
x̄A =
ZZ
x dxdy, ȳA =
ZZ
y dxdy.
Ω
Ω
Thus x̄ is the average x-coordinate on Ω and ȳ is the average y-coordinate on Ω. The
mass of the plate does not enter into this at all.
Problem 2. Find the centroid of the region
Ω : a ≤ x ≤ b, φ1 (x) ≤ y ≤ φ2 (x).
Solution.
x̄A =
ZZ
x dxdy =
Ω
ȳA =
ZZ
Ω
y dxdy =
Z b Z φ2 (x)
a
φ1 (x)
Z b Z φ2 (x)
a
φ1 (x)
x dydx =
y dydx =
Z b
a
Z b
a
x[φ2 (x) − φ1 (x)] dx;
1
([φ2 (x)]2 − [φ1 (x)]2 ) dx.
2
Kinetic Energy and Moment of Inertia
A particle of mass m at a distance r from a given line rotates about that line with
angular speed ω. The speed v of the particle is then rω and the kinetic energy is given
by the formula
1
1
KE = mv 2 = mr2 ω 2 .
2
2
Imagine now a rigid body composed of a finite number of point masses mi located
at distances ri from some fixed line. If the rigid body rotates about that line with
angular speed ω, then all the point masses rotate about that same line with that same
angular speed ω. The kinetic energy of the body can be obtained by adding up the
kinetic energies of all the individual particles:
KE =
X
i
!
1 X
1
mi ri2 ω 2 =
mi ri2 ω 2 .
2
2 i
The expression in parentheses is called the moment of inertia (or rotational inertia)
of the body and is denoted by the letter I:
I=
X
mi ri2 .
i
Centroids
March 17, 2001
8
SOME APPLICATIONS OF DOUBLE INTEGRATION
For a rigid body in straight-line motion
1
KE = M v 2
2
where v is the speed of the body.
For a rigid body in rotational motion
1
KE = Iω 2
2
where ω is the angular speed of the body.
The Moment of Inertia of a Plate
Suppose that a plate in the shape of a basic region Ω rotates about an axis. The
moment of inertia of the plate about that axis is given by the formula
I=
ZZ
λ(x, y)[r(x, y)]2 dxdy
Ω
where λ = λ(x, y) is the mass density function and r(x, y) is the distance from the axis
to the point (x, y).
Derivation. Decompose the plate into N pieces in the form of basic regions Ω1 , · · · , ΩN .
From each Ωi , choose a point (x∗i , yi∗ ) and view all the mass of the plate as concentrated
there. The moment of inertia of this piece is then approximately
[λ(x∗i , yi∗ )(area of Ωi )][r(x∗i , yi∗ )]2 = λ(x∗i , yi∗ )[r(x∗i , yi∗ )]2 (area of Ωi )
The sum of these approximations,
N
X
λ(x∗i , yi∗ )[r(x∗i , yi∗ )]2 (area of Ωi )
i=1
is a Riemann sum for the double integral
ZZ
λ(x, y)[r(x, y)]2 dxdy.
Ω
As the maximum diameter of the Ωi tends to zero, the Riemann sum tends to this
integral.
Problem 3. A rectangular plate of mass M , length L, and width W with lower
left-corner situated at (0, 0) rotates about the y-axis. Find the moment of inertia of
the plate about that axis (a) given that the plate has uniform mass density. (b) given
that the mass density of the plate varies directly as the square of the distance from the
rightmost side.
Solution. Let the plate be placed in the rectangle R with corners (0, 0), (L, 0), (L, W ), (0, W ).
(a) Here λ(x, y) = M/LW and r(x, y) = x. Thus
I=
ZZ
R
M 2
M ZWZL 2
x dxdy =
x dxdy
LW
LW 0 0
=
The Moment of Inertia of a Plate
M
W
LW
Z L
0
1
x2 dx = M L2 .
3
March 17, 2001
9
SOME APPLICATIONS OF DOUBLE INTEGRATION
(b) In this case λ(x, y) = k(L − x)2 but we still have r(x, y) = x. Therefore
I=
ZZ
R
2 2
k(L − x) x dxdy = k
Z WZ L
0
0
(L − x)2 x2 dxdy =
1
kL5 W.
30
We can eliminate the constant of proportionality k by noting that
M=
ZZ
R
k(L − x)2 dxdy = k
Z WZ L
0
Therefore k = 3M/L3 W and
I=
0
1
(L2 − 2Lx + x2 ) dxdy = kL3 W.
3
1
M L2 .
10
Radius of Gyration
If the mass M of an object is all concentrated at a distance r from a given axis, then
the moment of inertia about that axis is given by the product M r2 .
Suppose now that we have a plate of mass M (actually any object of mass M will
do here), and suppose that l is some line. The object has some moment of inertia I
about l. Its radius of gyration about l is the distance K for which
I = M K 2.
Namely, the radius of gyration about I is the distance from l at which all the mass of
the object would have to be concentrated to effect the same moment of inertia. The
formula for radius of gyration is usually written
K=
q
I/M .
Problem 4. A homogeneous circular plate of mass M and radius R rotates about
an axle that passes through the center of the plate and is perpendicular to the plate.
Calculate the moment of inertia and the radius of gyration.
Solution. Take the axle as the z-axis and let the plate rest on the circular
√ region
Ω : x2 + y 2 ≤ R2 . The density of the plate is M/A = M/πR2 and r(x, y) = x2 + y 2 .
Hence
ZZ
M 2
1
M Z 2π Z R 3
2
y
r drdθ = M R2 .
I=
(x
+
)
dxdy
=
2
2
πR
πR 0 0
2
Ω
q
√
The radius of gyration K is I/M = R/ 2. The circular plate of radius R has the
same moment
√ of inertia about the central axle as a circular wire of the same mass with
radius R/ 2. The circular wire is a more efficient carrier of moment of inertia than
the circular plate.
The Parallel Axis Theorem
Suppose we have an object of mass M and a line lM that passes through the center of
mass of the object. The object has some moment of inertia about that line; call it IM .
Radius of Gyration
March 17, 2001
10
TRIPLE INTEGRALS
If l is any line parallel to lM , then the object has a certain moment of inertia about l;
call that I. The parallel axis theorem states that
I = IM + d2 M
where d is the distance between the axes.
We will prove the theorem under somewhat restrictive assumptions. Assume that
the object is a plate of mass M in the shape of a basic region Ω, and assume that lM is
perpendicular to the plate. Call l the z-axis. Call the plane of the plate the xy-plane.
Denoting the points of Ω by (x, y) we have
I − IM =
ZZ
Ω
=
ZZ
= 2xM
ZZ
xλ(x, y) dxdy + 2yM
Ω
2
)
−(x2M + yM
2x2M M
+
Ω
λ(x, y)[(x − xM )2 + (y − yM )2 ] dxdy
2
2
λ(x, y)[2xM x + 2yM y − (xM
+ yM
)] dxdy
Ω
=
ZZ
λ(x, y)(x2 + y 2 ) dxdy −
2
2yM
M
−
2
(xM
ZZ
ZZ
yλ(x, y) dxdy
Ω
λ(x, y) dxdy
Ω
2
2
)M = (x2M + yM
+ yM
)M = d2 M.
An obvious consequence of the parallel axis theorem is that IM ≤ I for all lines
l parallel to lM . To minimize moment of inertia we must pass our axis through the
center of mass.
TRIPLE INTEGRALS
Once we are familiar with double integrals
ZZ
f (x, y) dxdy,
Ω
it is not hard to understand triple integrals
ZZZ
f (x, y, z) dxdydz.
T
Basically the only difference is this: instead of working with functions of two variables
continuous on a plane region Ω, we will be working with functions of three variables
continuous on some portion T of three-space.
March 17, 2001
11
TRIPLE INTEGRALS
The Triple Integral over a Box
For double integration we began with a rectangle
R : a1 ≤ x ≤ a2 , b1 ≤ y ≤ b2 .
For triple integration we begin with a box
Π : a1 ≤ x ≤ a2 , b1 ≤ y ≤ b2 , c1 ≤ z ≤ c2 .
To partition this box, we first partition the edges. Taking a partition
P1 : a1 = x0 < x1 < x2 < · · · < xm = a2 of [a1 , a2 ],
a partition
P2 : b1 = y0 < y1 < y2 < · · · < yn = b2 of [b1 , b2 ]
and a partition
P3 : c1 = z0 < z1 < z2 < · · · < zq = c2 of [c1 , c2 ]
we form the Cartesian product
P = P1 × P2 × P3 = {(xi , yj , zk ) : xi ∈ P1 , yj ∈ P2 , zk ∈ P3 }
and call this a partition of Π. P breaks up Π into m × n × q nonoverlapping boxes
Πijk : xi−1 ≤ x ≤ xi , yj−1 ≤ y ≤ yj , zk−1 ≤ z ≤ zk .
Taking
Mijk as the maximum value of f on Πijk
and
mijk as the minimum value of f on Πijk
we form the upper sum
Uf (P ) =
q
m X
n X
X
Mijk (volume of Πijk ) =
q
m X
n X
X
Mijk (xi −xi−1 )(yj −yj−1 )(zk −zk−1 )
q
m X
n X
X
mijk (xi −xi−1 )(yj −yj−1 )(zk −zk−1 )
i=1 j=1 k=1
i=1 j=1 k=1
and the lower sum
Lf (P ) =
q
m X
n X
X
mijk (volume of Πijk ) =
i=1 j=1 k=1
i=1 j=1 k=1
As in the case of functions of one and two variables, it turns out that, with f continuous
on Π, there is one and only one number I that satisfies the inequality
Lf (P ) ≤ I ≤ Uf (P ) for all partitions P of Π.
DEFINITION THE TRIPLE INTEGRAL OVER A BOX Π
The unique number I that satisfies the inequality
Lf (P ) ≤ I ≤ Uf (P ) for all partitions P of Π.
is called the triple integral of f over Π and is denoted by
ZZZ
f (x, y, z) dxdydz.
Π
The Triple Integral over a Box
March 17, 2001
12
TRIPLE INTEGRALS
The Triple Integral over a More General Solid
We start with a three-dimensional, bounded, open, connected set and adjoin to it the
boundary. We now have a three-dimensional, bounded, closed, connected set T . We
assume that T is a basic solid; that is, we assume that the boundary of T consists of a
finite number of continuous surfaces z = α(x, y), y = β(x, z), x = γ(y, z).
Now let’s suppose that f is some function continuous on T . To define the triple
integral of f over T we first encase T in a rectangular box Π. We then extend f to all
of Π by defining f to be zero outside of T . This extended function f is bounded on Π,
and it is continuous on all of Π except possibly at the boundary of T . In spite of these
possible discontinuities, f is still integrable over Π; that is, there still exists a unique
number I such that
Lf (P ) ≤ I ≤ Uf (P ) for all partitions P of Π.
The number I is by definition the triple integral
ZZZ
f (x, y, z) dxdydz.
Π
We define the triple integral over T by setting
ZZZ
f (x, y, z) dxdydz =
T
ZZZ
f (x, y, z) dxdydz.
Π
Volume as a Triple Integral
The simplest triple integral of interest is the triple integral of the function that is
constantly one on T. This gives the volume of T :
volume of T =
RRR
T
dxdydz.
Proof. Set f (x, y, z) = 1 for all (x, y, z) in T . Encase T in a box Π. Define f to
be zero outside of T. An arbitrary partition P of Π breaks up T into little boxes Πijk .
Note that
Lf (P ) = the sum of the volumes of all the Πijk that are contained in T
Uf (P ) = the sum of the volumes of all the Πijk that intersect T .
It follows that
Lf (P ) ≤ the volume of T ≤ Uf (P ).
The arbitrariness of P gives the formula.
The Triple Integral over a More General Solid
March 17, 2001
13
TRIPLE INTEGRALS
Some Properties of the Triple Integral
Below we give without proof the salient elementary properties of the triple integral.
They are all analogous to what you saw in the one- and two-variable cases. The T
referred to is a basic solid. The functions f and g are assumed to be continuous on T .
I. The triple integral is linear:
ZZZ
[αf (x, y, z) + βg(x, y, z)] dxdydz
T
=α
ZZZ
f (x, y, z) dxdydz + β
T
ZZZ
g(x, y, z) dxdydz.
T
II. It preserves order:
RRR
T , then T f (x, y, z) dxdydz
≥ 0;
if f ≥ 0 onRRR
RRR
if f ≤ g on T , then T f (x, y, z) dxdydz ≤ T g(x, y, z) dxdydz.
III. It is additive:
if T is broken up into a finite number of basic solids T1 , · · · , Tn , then
RRR
T
f (x, y, z) dxdydz =
RRR
T1
f (x, y, z) dxdydz + · · · +
RRR
Tn
f (x, y, z) dxdydz
IV. It satisfies a mean-value condition: namely, there is a point (x0 , y0 , z0 ) in T for
which
ZZZ
f (x, y, z) dxdydz = f (x0 , y0 , z0 )(the volume of T ).
T
We call f (x0 , y0 , z0 ) the average value of f on T .
The notion of average given above enables us to write
ZZZ
f (x, y, z) dxdydz = (the average value of f on T )(the volume of T ).
T
We can also take weighted averages: if f and g are continuous and g is nonnegative
on T , then there is a point (x0 , y0 , z0 ) in T for which
ZZZ
f (x, y, z)g(x, y, z) dxdydz = f (x0 , y0 , z0 )
T
ZZZ
g(x, y, z) dxdydz.
T
We call f (x0 , y0 , z0 ) the g-weighted average of f on T .
The formulas for mass, center of mass, and moments of inertia derived in the
previous section for two-dimensional plates are easily extended to three-dimensional
objects.
Suppose that T is an object in the form of a basic solid. If T has constant mass
density λ (here density is mass per unit volume), then the mass of T is the density λ
times the volume of T :
M = λV.
Some Properties of the Triple Integral
March 17, 2001
14
TRIPLE INTEGRALS
If the mass density varies continuously over T, say λ = λ(x, y, z), then the mass of T
is the average density of T times the volume of T :
M=
ZZZ
λ(x, y, z) dxdydz.
T
The coordinates of the center of mass (xM , yM , zM ) are density-weighted averages:
ZZZ
xλ(x, y, z) dxdydz,
yM M =
ZZZ
yλ(x, y, z) dxdydz,
zM M =
ZZZ
zλ(x, y, z) dxdydz.
xM M =
T
T
T
If the object T is homogeneous (constant mass density M/V ), then the center of
mass of T depends only on the geometry of T and falls on the centroid (x̄, ȳ, z̄) of the
space occupied by T . The density is irrelevant. The coordinates of the centroid are
simple averages over T :
x̄V =
ZZZ
x dxdydz, ȳV =
T
ZZZ
T
y dxdydz, z̄V =
ZZZ
z dxdydz.
T
The moment of inertia of T about a line is given by the formula
I=
ZZZ
λ(x, y, z)[r(x, y, z)]2 dxdydz.
T
Here λ(x, y, z) is the mass density of T at (x, y, z) and r(x, y, z) is the distance of
(x, y, z) from the line in question. The moments of inertia about the x, y, z axes are
again denoted by Ix , Iy , Iz .
Some Properties of the Triple Integral
March 17, 2001