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Chapter 14 Multiple Integrals
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1. Double Integrals, Iterated Integrals, Cross-sections
2. Double Integrals over more general regions, Definition,
Evaluation of Double Integrals, Properties of Double Integrals
. Area and Volume by Double Integration, Volume by Iterated
Integrals, Volume between Two surfaces
.4 Double Integrals in Polar Coordinates, More general Regions
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. Applications of Double Integrals, Volume and First Theorem of
Pappus, Surface Area and Second Theorem of Pappus,
Moments of Inertia
6. Triple Integrals, Iterated Triple Integrals
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. Integration in Cylindrical and Spherical Coordinates
8. Surface Area, Surface Area of Parametric Surfaces, Surfaces
7
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Area in Cylindrical Coordinates
.9 Change of Variables in Multiple Integrals, Jacobian
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Math 200 in 2011
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Chapter 14 Multiple Integrals
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14.2 Properties of double integrals
14.3 Area and volume of double integration
14.4 Double integrals and iterated integral in polar coordinates
14.4 Gaussian probability distribution
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Math 200 in 2011
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∫∫
∫∫
1
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∫ ∫D
2
D
( f (x, y) + g(x, y) )dA =
∫∫
cf (x, y) dA = c
D
∫∫
D
f (x, y) dA +
D
g(x, y) dA.
f (x, y) dA.
. If f (x, y) ≥ g(x, y) for all (x, y) ∈ D, then
∫∫
∫∫
f (x, y) dA ≥
g(x, y) dA.
3
4.
∫ ∫D
D
∫ ∫D
f (x, y) dA =
D1
∫∫
f (x, y) dA +
D2
f (x, y) dA, where
D = D1 ∪ D2 , and D1 and D2 do
not overlap except perhaps on their
boundary.
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∫∫
∫∫
5
D
dA =
D
1 dA = Area of D = A(D).
. If m ≤ f (x, y) ≤ M for all (x, y) ∈ D, then
∫∫
6
mA(D) ≤
.
D
f (x, y) dA ≤ MA(D).
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Math 200 in 2011
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. Double Integrals in Polar Coordinates.
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Proposition. If f (x, y) is continuous on a region D in xy-plane, and D
can be described in polar coordinates in the following form
{ (r, θ ) | α ≤ θ ≤ β, g1 (θ ) ≤ r ≤ g2 (θ ) }, then
∫∫
.
D
f (x, y) dA =
∫ β ∫ g2 (θ )
α
g1 ( θ )
f (r cos θ, r sin θ ) r dr dθ.
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Math 200 in 2011
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Given a region R in xy-plane, sometimes it is rather easy to describe
the boundary of R in terms of polar coordinates instead of rectangular
coordinates. An example is x2 + y2 = a2 , can be easily described as
{ (r, θ ) | 0 ≤ θ ≤ 2π, 0 ≤ r ≤ a }. It is the reason why one needs to
develop the double integral in terms of polar coordinates, just like the
one in rectangular coordinates.
Without really getting into the details, one can subdivide the plane
region R in terms of polar parameters, just like defining double
integral, so that we have a sectorial area ∆A = r ∆r ∆θ, so the
Riemann sum is written ∑ f (ri cos θj , ri sin θj ) ri ∆r ∆θ, which will
ij
converges to the double integral
∫∫
R′
f (r cos θ, r sin θ ) r dr dθ, where R′
is the domain for θ and r representing the domain R in xy-plane.
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Math 200 in 2011
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Example. Use a double integral to find the
area enclosed by one loop of the
three-leaved
rose r = sin 3θ.
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Solution. The region, in polar coordinates, is
{ (r, θ ) | 0 ≤ θ ≤ π/3, 0 ≤ r ≤ sin 3θ },
hence the area of the region D is
∫ π/3 ∫ sin 3θ
∫ π/3
∫
1 π/3
1
2
r dr dθ =
sin (3θ )dθ =
(1 − cos(6θ ) ) dθ =
2
4 0
0
0
[
]π/3 0
1
1
π
θ − sin(6θ )
= .
4
6
12
0
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Math 200 in 2011
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2 ≤ 1, y ≥ 0 }
Example. D = { (x, y) | x2 + y√
√
2
2
. = { (x, y) | 0 ≤ y ≤ 1, − 1 − y ≤ x ≤ 1 − y }.
One can describe D in terms of polar coordinates as follows
{ (r, θ ) | − π2 ≤ θ ≤ π2 , 0 ≤ r ≤ 1 }.
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Example. For any a > 0, prove that the equation r = 2a sin θ in polar
2
2
.coordinates is the circle x + y = 2ay.
Solution. The curve x2 + y2 = 2ax can be described in polar
coordinates as r2 = r2 (cos2 θ + sin2 θ ) = x2 + y2 = 2ay = 2ar sin θ, so
one has r = 2a sin θ (0 ≤ θ ≤ π ). Geometrically, it means a
right-angled triangle circumscribed in a circle of diameter 2a.
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Math 200 in 2011
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Example. Find the volume of the solid D bounded by the paraboloid
2
2
S
. : z = 25 − x − y and the xy-plane.
Solution. The paraboloid S : z = 25 − x2 − y2
intersect the xy-plane π : z = 0 in the curve C :
0 = 25 − x2 − y2 , which is a circle x2 + y2 = 52 .
So the shadow R of the solid D after projecting
onto xy-plane is given by the circular disc R =
{ (x, y) | x2 + y2 ≤ 52 }, in polar coordinates is
given by { (r, θ ) | 0 ≤ θ ≤ 2π, 0 ≤ r ≤ 5}.
Then the volume of the solid D is given by
∫∫
R
(25 − x2 − y2 )dA =
∫ 5
2π
0
∫ 2π ∫ 5
(25r − r3 ) dr = 2π
0
(25 − r2 ) r dr dθ =
]5
r4
625
−
=
π.
2
4 0
2
0
[
25r2
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Math 200 in 2011
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Example. Find the volume of the solid that lies under the paraboloid
z. = x2 + y2 , above the xy-plane, and inside the cylinder x2 + y2 = 2x.
Solution. The cylinder x2 + y2 = 2x lies over
the circular disk D which can be described as
{ (r, θ ) | − π/2 ≤ θ ≤ π/2, 0 ≤ r ≤ 2r cos θ }
in polar coordinates. The reason is that if we
write (x, y, z) = (r cos θ, r sin θ, z) for any point
in the cylinder, then r2 = x2 + y2 ≤ 2x =
2r cos θ, i.e. r ≤ 2 cos θ. As 2 cos θ ≥ r ≥ 0,
it follows that
−π/2 ≤ θ ≤ π/2. The height of the solid is the z-value of the
paraboloid from the xy-plane. Hence the volume V of the solid is
∫∫
∫ π/2 ∫ 2 cos θ
3π
(x2 + y2 ) dA =
r2 · r dr dθ =
.
2
D
−π/2 0
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Math 200 in 2011
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Example.
Find the volume of the solid that lies below the hemisphere
√
z = 9 − x2 − y2 , above the xy-plane, and inside the cylinder
2
2
.x + y = 1.
Solution. Let R be the shadow of D after projecting on xy-plane, then R is the circular disk centered at the origin with radius 1, in polar coordinates { (r, θ ) | 0 ≤ √
r ≤ 1, 0 ≤ θ ≤ √
2π }. Moreover, the top ztop = 9 − x2 − y2 = 9 − r2 , and
zbottom = 0.
Hence
volume of the solid D is
∫ 2π ∫the
∫
1√
2π 1 √
2
9 − r · r dr dθ =
9 − r2 d(8 − r2 )
−2 0
0
]1
[0
]
√
2π [ 3/2
2π
(9 − r2 )3/2
=
9 − 83/2 =
(27 − 16 2).
= −π
3/2
3
3
0
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Math 200 in 2011
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Example. Find the volume of the solid bounded above by the
paraboloid
z = 8 − x2 − y2 , and below by the paraboloid z = x2 + y2 .
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Solution. Let P(x, y, z) be the intersection of two
paraboloids, then one has 8 − x2 − y2 = z = x2 +
y2 , so x2 + y2 = 4 = 22 , which is a circle. The
shadow R of the solid D is then the circular disc,
in polar coordinates { (r, θ ) | 0 ≤ r ≤ 2, 0 ≤ θ ≤
2π }. As 0 ≤ r ≤ 2, we have r2 ≤ 4, i.e. r2 ≤ 8 − r2 ,
so one knows that the top of the solid is given by
ztop = 8 − x2 − y2 = 8 − r2 , and the bottom of the
solid is given by zbottom = x2 + y2 = r2 .
Hence
volume of the solid D is ∫
∫ 2π ∫the
2
2
(8 − r2 − r2 ) · r dr dθ = 2π
(8r − r3 ) dr
0
0
0
]2
[
r4
= 24π.
= 2π 4r2 −
4 0
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Math 200 in 2011
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Example. Determine the volume of the region D common to the
interiors of the cylinders x2 + y2 = 1 and x2 + z2 = 1, in the first
octant.
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Solution. One immediately √
recognizes the solid D has a top given by
x2 + z2 = 1, i.e. zmax (x) = 1 − x2 , and xy-plane as the bottom.
Moreover, the shadow R of the solid D is a circle
√ disk in the 1st
quadrant, so R =∫ ∫{ (x, y) | 0 ≤ x ≤ 1, 0 ≤ y ≤ 1 − x2 }. The volume
√
of D is given by
( 1 − x2 − 0) dA
R
[
]1
∫ 1
∫ 1 ∫ √1−x2 √
x3
2
2
2
1 − x dy dx =
(1 − x )dx = x −
=
= .
3
3
0
0
0
0
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Math 200 in 2011
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Example. Determine the volume of the region D common to the
interiors of the cylinders x2 + y2 = 1 and x2 + z2 = 1, in the first
octant.
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Solution. One immediately √
recognizes the solid D has a top given by
x2 + z2 = 1, i.e. zmax (x) = 1 − x2 , and xy-plane as the bottom.
Moreover, the shadow R of the solid D is a circle disk in the 1st
quadrant, so R can be described in terms of polar coordinates as
{ (r, θ ) | 0 ≤ r ≤ 1, 0 ≤ θ ≤ π/2 }. The volume of D is given by
∫ π/2 ∫ 1 √
∗
1 − r2 cos2 θ rdrdθ = 23 .
0
0
(*): the answer 32 is given by integration via rectangular coordinates.
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Math 200 in 2011
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Example. Show that the volume of the solid region E bounded by√the
three
cylinders x2 + y2 = 1, y2 + z2 = 1 and x2 + z2 = 1 is 16 − 8 2.
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Solution. The solid is symmetric with respect to the 3 coordinate axes. Thus, it suffices to compute the volume of the portion D
with nonnegative x, y, z-coordinates. In other
words, D lies in the first octant of the coordinate system.
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Math 200 in 2011
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Example. Show that the volume of the solid region E bounded by√the
three
cylinders x2 + y2 = 1, y2 + z2 = 1 and x2 + z2 = 1 is 16 − 8 2.
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Solution. The solid is symmetric with respect to the 3 coordinate axes. Thus, it suffices to compute the volume of the portion D
with nonnegative x, y, z-coordinates. In other
words, D lies in the first octant of the coordinate system.
√
2
For the
√ solid D, it is bounded on top by the graphs of z = 1 − x and
z = 1 − y2 . On its side and bottom, it is bounded by the cylinder
x2 + y2√= 1 and the three
√coordinate planes. Furthermore, the graphs
of z = 1 − x2 and z = 1 − y2 intersect along√a curve on the plane
x = y. Thus the solid is under the graph of z = 1 − x2 over the
region D in xy-plane described as { (r, θ ) | 0 ≤ r ≤ 1, 0 ≤ θ ≤ π/4 }
in the∫∫
polar
volume of original solid E
√ coordinates. Hence,
∫ π/4 ∫the
1√
= 16 D 1 − x2 dA = 16 0
1 − r2 cos2 θrdrdθ =
0
]
[
r
=
1
∫ π/4 1−sin3 θ
∫ π/4
2 cos2 θ )3/2
dθ = 16 0
dθ =
16 0
− (1−r cos
θ
3 cos2 θ
r=0
√
π/4
16
= 16 − 8 2.
3 [tan θ − sec θ − cos θ ]0
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Math 200 in 2011
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∫ +∞
√
π
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2
0
∫ +∞
∫ M
(
)
Solution. As
e−x dx = lim
e−x dx = lim 1 − e−M = 1,
Example. Prove with the polar-coordinates that
.
M→+∞ 0
0
e
− x2
dx =
M→+∞
it follows from comparison test with 0 ≤ e−x ≤ e−x that the improper
∫ +∞
2
∫ M
e−x dx converges. Let IM =
e−x dx. Need to prove
0
0
√
π
π
2
that lim IM =
. It suffices to show that
= lim IM
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2
4
M→+∞
M→+∞
integral
2
By Fubini
( theorem,)we(have
∫ M
2
IM
=
∫∫
RM
e
e−x dx ·
0
−x2 −y2
2
∫ M
0
e−y dy
2
)
2
∫ M∫ M
=
0
0
e−x e−y dy dx =
2
2
dA, where RM = [0, M] × [0, M] is a square of length M.
Continue...
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Math 200 in 2011
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Example. Prove with the polar-coordinates that
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∫ +∞
0
e
− x2
√
dx =
π
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2
Solution. Let DM be quarter circular disk of radius M centered at
origin. By means of polar coordinates, we have
∫∫
JM =
2 − y2
DM
∫
e−x
dA =
∫ π/2 ∫ M
0
0
e−r r dr dθ
2
2
π M − r2 − 1
π
=
e
×
d(−r2 ) = (1 − e−M ).
2 0
2
4
π
It follows that lim JM = .
4
M→+∞
2
2
As f (x, y) = e−x −y ≥ 0, and DM ⊂ RM ⊂ DM√2 , it follows from
2 ≤ J √ . The result follows
property of double integral that JM ≤ IM
M 2
from sandwich theorem that
π
π
2
= lim JM ≤ lim IM
≤ lim JM√2 = lim JM = .
4
4
M→+∞
M→+∞
M→+∞
M→+∞
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Math 200 in 2011
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Applications of Double Integrals
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Suppose a (planar) object, in region R, is made of different material in
which the density (mass per unit area) is given by δ(x, y), depending
on the location (x, y). Then the total mass of R is given approximately
by the Riemann sum∫∑
∫ i δ(xi , yi )∆Ai , which will converges to the
double integral m =
.
D
δ(x, y) dA. We call it the mass of the object.
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Similarly,(one can define the center of mass (centroid)
of the object by
)
∫∫
∫∫
1
1
(x, y) =
xδ(x, y) dA,
yδ(x, y) dA .
m D
m D
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Math 200 in 2011
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Example. A lamina R is made of the part of circular disc of radius a in
1st quadrant. Its density is proportional to the distance from the
.origin. Determine the position of its centroid.
Solution. In terms of polar coordinates, D can be described
as
√
{ (r, θ ) | 0 ≤ r ≤ a, 0 ≤ θ ≤ π/2 }. Let δ(x, y) = k x2 + y2 = kr.
∫ π/2 ∫ a
π ka3
kπa3
Then the mass m =
kr · r dr dθ = ·
=
.
2 3
6
0
0
As all the conditions on the lamina R are symmetric in x and y, so
∫
∫
1 π/2 a
x=y=
(r sin θ ) · kr · r dr dθ =
(∫ π/2m 0
)0 (∫ a
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k
k
a4
3a
3
sin θdθ ·
r dr =
×
1
×
=
.
3
m
4
2π
kπa
/6
0
0
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Math 200 in 2011
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First Theorem of Pappus: Volume of Revolution
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Suppose that a plane region R is revolved around an axis in its plane
generating a solid of revolution with volume V. Assume that the axis
does not intersect the interior of R. Then the volume V of the solid is
V = A · d,
where A is the area of R and d is the distance traveled by the centroid
.of R.
Remark. Cut the region R into vertical strips, and each vertical strip
after rotating will form a ring, which contributes ∆Vi = 2πxi∗ ∆Ai ,
where Ai is the area of the vertical strip. It follows from Riemann sum
that
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n
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Math 200 in 2011
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Volume of a Sphere
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Prove
that the volume of a sphere is 43 πa3 .
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Solution. Let R be a region bounded by the upper semi-circle x2 + y2 = a2 , y ≥ 0 and x-axis.
In polar coordinates, R can be described as
{ (r, θ ) | 0 ≤ θ ≤ π, 0 ≤ r ≤ a }. Using polar coordinates, the area of
∫ π∫ a
a2
r dr dθ =
R is =
× π = πa2 /2. Similarly, y of R is given by
2
0
0
∫∫
∫ π∫ a
1
1
ydA
=
r sin θ · r dr dθ
πa2 /2 R(
πa2 /2
) ( ∫0 a 0 )
[ 3 ]a
∫ π
1
1
r
2
π
=
sin
θ
dθ
r
dr
=
×
[−
cos
θ
]
×
0
3 0
πa2 /2
πa2 /2
0
0
4a
2
a3
=
. By Pappus’s theorem, the volume of a sphere
=
×2×
3
3π
πa2
4a πa2
4
of radius a = 2πy× Area of R = 2π ·
·
= πa3 .
3π
2
3
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Math 200 in 2011
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Second Theorem of Pappus: Surface Area of Revolution
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Suppose that a plane curve C is revolved around an axis in its plane
that does not intersect the curve. Then the area A of the surface of
revolution generated is
A = s · d,
where s is the arc-length of C and d is the distance traveled by the
.centroid of C.
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Math 200 in 2011
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Let R be a plane lamina and ℓ be a straight line that may or may not
lie
∫ ∫ in xy-plane. Then the moment of inertia I of R around the axis ℓ is
R
p2 δ(x, y)dA, where p = p(x, y) is the shortest distance from the
point (x, y) of R to the line ℓ, and δ(x, y) is the density of R at the point
.(x, y).
For the coordinate axii, we have
.
∫∫
Ix = Ix-axis =
Iz = Iz-axis =
.
∫∫R
R
(y2 )δ(x, y)dA, Iy = Iy-axis =
∫∫
R
(x2 )δ(x, y)dA and
(x2 + y2 )δ(x, y)dA.
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For any plane lamina
R, Define the center (x̂, ŷ, ẑ) of gyration by
√
√
√
Iy
Ix
Iz
x̂ = m , ŷ = m , ẑ = m
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Math 200 in 2011
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