14.6 average value of a function 14.7 integration in ...fstitl/calculus2012/2013-lecture-09a.pdf ·...
TRANSCRIPT
. . . . . .
.Chapter 14 Multiple Integrals..
......
...1 Double Integrals, Iterated Integrals, Cross-sections
...2 Double Integrals over more general regions, Definition,Evaluation of Double Integrals, Properties of Double Integrals
...3 Area and Volume by Double Integration, Volume by IteratedIntegrals, Volume between Two surfaces
...4 Double Integrals in Polar Coordinates, More general Regions
...5 Applications of Double Integrals, Volume and First Theorem ofPappus, Surface Area and Second Theorem of Pappus,Moments of Inertia
...6 Triple Integrals, Iterated Triple Integrals
...7 Integration in Cylindrical and Spherical Coordinates
...8 Surface Area, Surface Area of Parametric Surfaces, SurfacesArea in Cylindrical Coordinates
...9 Change of Variables in Multiple Integrals, Jacobian
Math 200 in 2011
. . . . . .
.Chapter 14 Multiple Integrals..
......
14.6 Average value of a function
14.7 Cylindrical coordinates, and Spherical Coordinates
14.7 Integration in cylindrical coordinates, and spherical coordinates.
14.7 Mass, Moments, Centroid, Moment of Inertia
Math 200 in 2011
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.. Applications of Triple Integrals.
......
Suppose an object, in region D in R3, is made of different material inwhich the density (mass per unit volume) is given by δ(x, y, z),depending on the location (x, y, z). Then the total mass of D is givenapproximately by the Riemann sum ∑i δ(xi, yi, zi)∆Vi, which will
converges to the triple integral m =∫∫∫
Dδ(x, y, z) dV. We call it the
mass of the object.
Similarly, one define.
......
the center of mass (centroid) of the object by (x, y, z) =(1m
∫∫∫D
xδ(x, y, z) dV,1m
∫∫∫D
yδ(x, y, z) dV,1m
∫∫∫D
zδ(x, y, z) dV)
.
Math 200 in 2011
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Let D be a solid object in R3, and ℓ be a straight line in R3. Then the
moment of inertia I of D around the axis ℓ is∫∫∫
Dp2δ(x, y, z) dV,
where p = p(x, y, z) is the shortest distance from the point (x, y, z) ofD to the line ℓ, and δ(x, y, z) is the density of D at the point (x, y, z).
For the coordinate axii, we have.
......
Ix = Ix-axis =∫∫∫
D(y2 + z2)δ(x, y, z) dV,
Iy = Iy-axis =∫∫∫
D(x2 + z2)δ(x, y, z) dV and
Iz = Iz-axis =∫∫∫
D(x2 + y2)δ(x, y, z)dV.
.
......
For any solid region D in R3, Define the center (x, y, z) of gyration by
x =√
Ixm , y =
√Iym , z =
√Izm .
Math 200 in 2011
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.Average Value of Function of Several Variables..
......
The average value of a function of several variables defined on a
region D in R3, to be [f ]average =
∫∫∫D f dV∫∫∫D dV
.
Math 200 in 2011
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.. Cylindrical Coordinates.
......
In the cylindrical coordinate system, a point P(x, y, z) inthree-dimensional space is represented by the ordered triple (r, θ, z),where r and θ are polar coordinates of the projection Q(x, y, 0) ofP(x, y, z) onto the xy-plane and z is the directed distance from thexy-plane to P.
To convert from cylindrical to rectangular coordinates, we use theequations x = r cos θ, y = r sin θ, z = z.To convert from rectangular to cylindrical coordinates, we use theequations r =
√x2 + y2, tan θ = y
x , z = z.Math 200 in 2011
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z = r r = c (c > 0)
Math 200 in 2011
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Theorem. If D is a solid described in the cylindrical coordinates as{ ( r, θ, z ) | α ≤ θ ≤ β, r1(θ) ≤ r ≤ r2(θ), zmin(r, θ) ≤ z ≤ zmax(r, θ) },and f (x, y, z) is a continuous function defined in D,
then∫∫∫
Df (x, y, z) dV =
∫ β
α
∫ ϕ2
ϕ1
∫ r2(θ)
r1(θ)f (r cos θ, r sin θ, z) r dz dr dθ.
Math 200 in 2011
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Example. A solid T is bounded by the conez =
√x2 + y2 and the plane z = 2. The
mass density at any point of the solid T isproportional to the distance between theaxis of the cone and the point. Find themass of T.Solution. Since the density of the solid at a point P(x, y, z) isproportional to the distance from the z-axis to the point P(x, y, z), wesee that the density function δ(x, y, z) = k
√x2 + y2 for some constant
k. Moreover, T can be described as{ (x, y, z) | 0 ≤ x2 + y2 ≤ 4,
√x2 + y2 ≤ z ≤ 2 }.
In terms of cylindrical coordinates, T may be given as{ (r, θ, z) | 0 ≤ r ≤ 2, 0 ≤ θ ≤ 2π, r ≤ z ≤ 2 }.
So the mass of the solid T is∫∫∫T
δ(x, y, z) dV =∫∫∫
Tk√
x2 + y2 dV = k∫ 2π
0
∫ 2
0
∫ 2
rr · r dz dr dθ
= 2kπ∫ 2
0(2 − r)r2 dr = 2kπ
[2r3
3− r4
4
]2
0= 2kπ(
163
− 4) =8kπ
3.
Math 200 in 2011
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......Example. Evaluate
∫ 2
−2
∫ √4−x2
−√
4−x2
∫ 2√
x2+y2(x2 + y2) dV.
Solution. The domain D of integral in rectan-gular coordinates, and R be its shadow in xy-plane. Then in cylindrical coordinates, D canbe described as { (r, θ, z) | 0 ≤ r ≤ 2, 0 ≤ θ ≤2π, r ≤ z ≤ 2 }. Then we can simplify integral∫ 2
−2
∫ √4−x2
−√
4−x2
∫ 2√
x2+y2(x2 + y2) dV
=∫ 2
−2
∫ √4−x2
−√
4−x2
∫ 2√
x2+y2(x2 + y2) dV =
∫∫∫D(x2 + y2) dV
=∫ 2π
0
∫ 2
0
∫ 2
rr2 · r dr dθ = 2π
[r4
4
]2
0= 8π.
Math 200 in 2011
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Example. A tank D is a solid in the shape of a half-cylinder of radius 2and height 3. It is situated in R3, given by the inequalities√
x2 + y2 ≤ 2, y ≥ 0, 0 ≤ z ≤ 3. The temperature (in ◦C) at the point(x, y, z) is given by T(x, y, z) = 2yz2
√x2 + y2. Find the average
temperature in the tank.
Solution. We describe the tank D in cylindrical coordinates as{ (r, θ, z ) | 0 ≤ θ ≤ π, 0 ≤ r ≤ 2, 0 ≤ z ≤ 3 }.
Recall the formula [T]average =∫∫∫
D T dV∫∫∫D dV . Then∫∫∫
DT dV =
∫ 2
0
∫ π
0
∫ 3
02(r sin θ) · z2 · r · r dz dθ dr
= 2(∫ 2
0r3 dr
)(∫ π
0sin θ dθ
)(∫ 3
0z2 dz
)= 2 × 2 × 4 × 9 = 144.
Similarly,∫∫∫
DdV =
∫ 2
0
∫ π
0
∫ 3
0r dz dθ dr
=
(∫ 2
0r dr)(∫ π
0dθ
)(∫ 3
0dz)= 2 × π × 3 = 6π.
Hence, the average temperature in the tank is 1446π = 24
π .
Math 200 in 2011
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Example. Find the volume of the solid region D bounded below by theparaboloid z = x2 + y2, and above by the plane z = 2x.
Solution. First we sketch the graphs of the paraboloid S1 and theplane S2, and we want to find the common intersection of S1 and S2.Let P(x, y, z) be any point of the common intersection, i.e. they satisfyboth 2x = z = x2 + y2, so (x − 1)2 + y2 = 1, which represents acylinder in space. In fact, the intersection is a curve obtained by theintersection of the plane and the cylinder above, which is a boundedclosed curve. Inside the cylinder, we have (x − 1)2 + y2 ≤ 1, i.e.x2 + y2 ≤ 2x, or equivalently the graph of S1 is below the graph of theplane S2. Hence the solid region D = { (x, y, z) | x2 + y2 ≤ 2x, andx2 + y2 ≤ z ≤ 2x }. Then one can switch to cylindrical coordinates todescribe D as { (r, θ, z) | 0 ≤ r ≤ 2 cos θ, − π
2 ≤ θ ≤ π2 and
r2 ≤ z ≤ 2r cos θ }. The volume of D =∫∫∫
D1 dV =∫∫
x2+y2≤2x
(∫ 2x
x2+y21 dz
)dA =
∫ π/2
−π/2
∫ 2 cos θ
0
(∫ 2r cos θ
r21 dz
)r drdθ.
Math 200 in 2011
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Example. Find the volume of the region D that lies inside both thesphere x2 + y2 + z2 = 4 and the cylinder x2 + y2 − 2x = 0.
Solution. The cylinder S is given by (x − 1)2 + y2 = 1, so any pointP(x, y, z) inside S is given by (x − 1)2 + y2 ≤ 1. In terms of cylindricalcoordinates, it can be described by (r, θ, z) byr2 cos2 θ + r2 sin2 θ − 2r cos θ ≤ 0, i.e. r ≤ 2 cos θ, it follows that theregion can be described in terms of cylindrical coordinates as{ (r, θ, z) | − π
2 ≤ θ ≤ π2 , r ≤ 2 cos θ and −
√4 − r2 ≤ z ≤
√4 − r2 }.
It follows from the definition of triple integral that the volume of the
region D =∫∫∫
D1 dV =
∫ π/2
−π/2
∫ 2 cos θ
0
(∫ √4−r2
−√
4−r2dz
)r drdθ =
2∫ π/2
−π/2
∫ 2 cos θ
0
√4 − r2 · rdrdθ = 2
∫ π/2
−π/2
[(4 − r2)3/2
3
]2 cos θ
0
dθ = · · · .Math 200 in 2011
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Example. Find the volume and the centroid of the solid D, boundedby the paraboloid z = b(x2 + y2) (b > 0) and the plane z = h (h > 0).
Solution. The intersection of the plane z = h and the paraboloidz = b(x2 + y2) is { (x, y, h) | x2 + y2 =
(√h/b
)2 }. D is described as{ (x, y, z) | 0 ≤ x2 + y2 ≤ h
b , b(x2 + y2) ≤ z ≤ b }, in polar coordinates
as { (r, θ, z) | 0 ≤ θ ≤ 2π, 0 ≤ r ≤√
hb , br2 ≤ z ≤ b }. The volume of
the solid D is given by∫∫∫
DdV =
∫ 2π
0
∫ √ hb
0
∫ h
br2r dz dr dθ =
2π∫ √ h
b
0(hr − br3) dz dr = 2π
(12
ha2 − 14
ba4)=
πh2
2b. Assume that
δ(x, y, z) ≡ 1. By symmetry, (x, y) = (0, 0). Then z =1V
∫∫∫D
z dV
=2b
πh2
∫ 2π
0
∫ √ hb
0
∫ h
br2rz dz dr dθ =
4bh2
∫ √ hb
0
(h2r2
− b2r5
2
)dr
=4bh2
(h2
2× 1
2(
√hb)2 − b2
2× 1
6× (
√hb)6
)=
23
h.
Math 200 in 2011
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.. Spherical Coordinatesρ =
√x2 + y2 + z2 = ∥−→OP∥ =
length of vector−→OP. ϕ = angle
between the vector−→OP and the
z-axis, 0 ≤ ϕ ≤ π.θ = angle between the shadowof vector
−→OP onto xy-plane and
the x-axis, 0 ≤ θ ≤ 2π.x = ρ sin ϕ cos θ,y = ρ sin ϕ sin θ,z = ρ cos ϕ.
Math 200 in 2011
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Math 200 in 2011
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Example. Find a spherical coordinate equation for the conez =
√x2 + y2.
Solution. Substitute for x, y, and z with spherical coordinates:ρ cos ϕ = z =
√x2 + y2 =
√(ρ sin ϕ cos θ)2 + (ρ sin ϕ sin θ)2 = ρ sin ϕ
i.e. cos ϕ = sin ϕ, so tan ϕ = 1. As 0 ≤ ϕ ≤ π, we have ϕ = π/4..
......
Example. Find a spherical coordinate equation for the spherex2 + y2 + (z − a)2 = a2.
Solution. Substitute for x, y, and z with spherical coordinates:a2 = x2 + y2 + (z − a)2 = (ρ sin ϕ cos θ)2 + (ρ sin ϕ sin θ)2 + (ρ cos ϕ −a)2 = (ρ sin ϕ)2 + (ρ cos ϕ)2 − 2aρ cos ϕ + a2 = ρ2 − 2aρ cos ϕ + a2, i.e.ρ2 = 2aρ cos ϕ, hence ρ = 2a cos ϕ.Remark. One should note that x2 + y2 = ρ2 sin2 ϕ..
......Example. Rewrite the sphere x2 + y2 + z2 = 2az (a > 0) in sphericalcoordinates.
Solution. ρ2 = x2 + y2 + z2 = 2az = 2aρ cos ϕ, soρ = 2a cos ϕ (0 ≤ ϕ ≤ π).
Math 200 in 2011
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Theorem. If D is a solid described in the spherical coordinates as{ (ρ, ϕ, θ) | α ≤ θ ≤ β, ϕ1 ≤ ϕ ≤ ϕ2, ρ1(ϕ, θ) ≤ ρ ≤ ρ2(ϕ, θ) },
and f (x, y, z) is a continuous function defined in D,
then∫∫∫
Df (x, y, z) dV =∫ β
α
∫ ϕ2
ϕ1
∫ ρ2(ϕ, θ)
ρ1(ϕ, θ)f (ρ sin ϕ cos θ, ρ sin ϕ sin θ, ρ cos ϕ) ρ2 sin ϕ dρ dϕ dθ.
Math 200 in 2011
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Example. Prove that the volume of a solid sphere D of radius a is43 πa3.
Solution. Let D be described in spherical coordinates as{ (ρ, ϕ, θ) | 0 ≤ θ ≤ 2π, 0 ≤ ϕ ≤ π, 0 ≤ ρ ≤ a }, then the volume of
the solid sphere D is∫ 2π
0
∫ π
0
∫ a
0ρ2 sin ϕ dρ dϕ dθ =
2π∫ π
0
[ρ3
3
]a
0sin ϕ dϕ = 2π × a3
3×∫ π
0sin ϕ dϕ =
43
πa3.
Math 200 in 2011
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Example. Evaluate∫∫∫
De(x
2+y2+z2)3/2dV, where
D = { (x, y, z) | x2 + y2 + z2 ≤ a } is the solid ball of radius a.
Solution. We use spherical coordinates to describe B as as{ (ρ, ϕ, θ) | 0 ≤ θ ≤ 2π, 0 ≤ ϕ ≤ π, 0 ≤ ρ ≤ a }, thene(x
2+y2+z2)3/2= e(ρ
2)3/2= eρ3
, and∫∫∫D
e(x2+y2+z2)3/2
dV =∫ 2π
0
∫ π
0
∫ a
0eρ3
ρ2 sin ϕ dρ dϕ dθ =
2π
[∫ π
0sin ϕ dϕ
] [∫ a
0eρ3
ρ2dρ
]= 2π[1− (1)]×
[eρ3
3
]a
0
=4π
3(ea3 − 1).
Math 200 in 2011
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Example. A solid D is cut from a solid ball of radius 1 centered at theorigin by the inequalities z ≥ 0 and y ≥ x. The mass density of D at apoint (x, y, z) is given by the function δ(x, y, z) = 30z2 kg /m3.(a) Find the total mass of S. (b) Find the average mass density of S.
Solution. (a) D is described in spherical coordinates by{ (ρ, θ, ϕ) | 0 ≤ ϕ ≤ π/2, π/4 ≤ θ ≤ 5π/4 0 ≤ ρ ≤ 1 }.The total mass of D is∫∫∫
Dδ(x, y, z) dV =
∫ π/2
0
∫ 5π/4
π/4
∫ 1
030(ρ cos ϕ)2ρ2 sin ϕ dρ dθ dϕ
= 30(∫ π/2
0cos2 ϕ sin ϕ dϕ
)(∫ 5π/4
π/4dθ
)(∫ 1
0ρ3 dρ
)= 2π.
(b) The volume of D is∫∫∫
DdV =
∫ π/2
0
∫ 5π/4
π/4
∫ 1
0ρ2 sin ϕ dρ dθ dϕ
=
(∫ π/2
0sin ϕ dϕ
)(∫ 5π/4
π/4dθ
)(∫ 1
0ρ2 dρ
)= 1 × π × 1
3=
π
3.
The average mass of D is 2ππ/3 = 6.
Math 200 in 2011
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Example. Use spherical coordinates to find the volume of the solidthat lies above the cone z =
√x2 + y2 and below the sphere
x2 + y2 + z2 = z.
Solution. The sphere x2 + y2 + z2 = z can be written as ρ2 = ρ cos ϕ,i.e. ρ = cos ϕ. Since the upper solid cone bounded by z =
√x2 + y2
can be best described in spherical coordinate by 0 ≤ ϕ ≤ π4 . D can
be parameterized by{ (ρ, ϕ, θ) | 0 ≤ θ ≤ 2π, 0 ≤ ϕ ≤ π/4, 0 ≤ ρ ≤ cos ϕ }
in terms of spherical coordinates. For any point P(x, y, z) in D, thedistance from the point P to the xy-plane is z = ρ cos ϕ. The volume of
D is∫∫∫
DdV =
∫ 2π
0
∫ π/4
0
∫ cos ϕ
0ρ2 sin ϕ dρ dϕ dθ =
5π
9√
2.
Math 200 in 2011
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Example. Let D be a region bounded above by the spherex2 + y2 + z2 = 32 and below by the cone z =
√x2 + y2. The mass
density at any point in D is its distance from the xy-plane. Find thetotal mass m of D.
Solution. Since the cone z =√
x2 + y2 can be best described inspherical coordinate by 0 ≤ ρ ≤ π
4 . In terms of spherical coordinates,D can be parameterized by{ (ρ, ϕ, θ) | 0 ≤ θ ≤ 2π, 0 ≤ ϕ ≤ π/4, 0 ≤ ρ ≤ 3 }. For any pointP(x, y, z) in D, the distance from the point P to the xy-plane isz = ρ cos ϕ. The total mass
m =∫∫∫
DzdV =
∫ 2π
0
∫ π/4
0
∫ 3
0ρ cos ϕ · ρ2 sin ϕ dρdϕdθ =
2π∫ 3
0ρ3dρ ·
∫ π/4
0sin ϕ cos ϕdϕ = 2π × 34
4× sin2(π/4)
2=
81π
8.
Math 200 in 2011
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Example. Let D be ice-cream solid bounded above by the spherex2 + y2 + z2 = 2az and below by the cone z =
√3(x2 + y2). Find the
volume of D.
Solution. The sphere x2 + y2 + z2 = 2az is written in sphericalcoordinates as ρ2 = 2aρ cos ϕ, i.e. ρ = 2a cos ϕ.The cone z =
√3(x2 + y2) is written in spherical coordinates as
ρ cos ϕ =√
3ρ2 sin2 ϕ, i.e. tan ϕ = 1√3
and so ϕ = π/6.D can be described in spherical coordinates as{ (ρ, ϕ, θ) | 0 ≤ θ ≤ 2π, 0 ≤ ϕ ≤ π/6, 0 ≤ ρ ≤ 2a cos ϕ }.
The volume of D is given by∫ 2π
0
∫ π/6
0
∫ 2a cos ϕ
0ρ2 sin ϕ dρdϕdθ =
2π × 8π
3
∫ π/6
0cos3 ϕ sin ϕ dϕ =
16π
3
[−1
4cos4 ϕ
]π/6
0=
712
πa3.
Math 200 in 2011