Download Triple Integrals: Applications and Coordinate Transformations and more Lecture notes Engineering in PDF only on Docsity!
Contiune on 16.7 Triple Integrals
Figure 1:
E
f (x, y, z)dV =
D
[∫
u 2 (x,y)
u 1 (x,y)
f (x, y, z)dz
]
dA
Applications of Triple Integrals Let E be a solid region with a density function ρ(x, y, z).
Volume: V (E) =
E 1 dV
Mass: m =
E ρ(x, y, z)dV
Moments about the coordinate planes:
Mxy =
E
zρ(x, y, z)dV
Mxz =
E
yρ(x, y, z)dV
Myz =
E
xρ(x, y, z)dV
Center of mass: (¯x, y,¯ z¯)
x¯ = Myz /m , y¯ = Mxz /m , z¯ = Mxy/m.
Remark: The center of mass is just the weighted average of the coordinate functions over the solid region. If ρ(x, y, z) = 1, the mass of the solid equals its volume and the center of mass is also called the centroid of the solid.
Example Find the volume of the solid region E between y = 4 − x^2 − z^2 and y = x^2 + z^2.
Soln: E is described by x^2 + z^2 ≤ y ≤ 4 − x^2 − z^2 over a disk D in the xz-plane whose radius is given by the intersection of the two surfaces: y = 4 − x^2 − z^2 and y = x^2 + z^2. 4 − x^2 − z^2 = x^2 + z^2 ⇒ x^2 + z^2 = 2. So the radius is
Therefore
V (E) =
E
1 dV =
D
[∫
4 −x^2 −z^2
x^2 +z^2
1 dy
]
dA =
D
4 − 2(x^2 + z^2 )dA
∫ (^2) π
0
0
(4 − 2 r^2 )rdrdθ =
∫ (^2) π
0
[
2 r^2 −
r^4
]√ 2
0
= 4π
Example Find the mass of the solid region bounded by the sheet z = 1 − x^2 and the planes z = 0, y = − 1 , y = 1 with a density function ρ(x, y, z) = z(y + 2).
Figure 2:
Soln: The top surface of the solid is z = 1 − x^2 and the bottom surface is z = 0 over the region D in the xy-plane which is bounded by the other equations in the xy-plane and the intersection of the top and bottom surfaces. The intersection gives 1 − x^2 = 0 ⇒ x = ±1. Therefore D is a square [− 1 , 1] × [− 1 , 1].
m =
E
ρ(x, y, z)dV =
E
z(y + 2)dV =
D
[∫
1 −x^2
0
z(y + 2)dz
]
dA
− 1
− 1
∫ (^1) −x 2
0
z(y + 2)dzdxdy =
− 1
− 1
(1 − x^2 )^2 (y + 2)dxdy =
8 15
− 1
(y + 2)dy = 32/ 15
Example Find the centroid of the solid above the paraboloid z = x^2 + y^2 and below the plane z = 4. Soln: The top surface of the solid is z = 4 and the bottom surface is z = x^2 + y^2 over the region D defined in the xy-plane by the intersection of the top and bottom surfaces.
Figure 4:
in x = r cos(θ), y = r sin(θ)
∫ ∫ ∫
E
f (x, y, z)dV =
D
[∫
u 2 (x,y)
u 1 (x,y)
f (x, y, z)dz
]
dA = ∫ (^) β
α
∫ (^) h 2 (θ)
h 1 (θ)
∫ (^) u 2 (r cos θ,r sin θ)
u 1 (r cos θ,r sin θ)
f (r cos θ, r sin θ, z)rdzdrdθ
Note: dV → rdzdrdθ
Example Evaluate
E zdV^ where^ E^ is the portion of the solid sphere^ x
(^2) + y (^2) + z (^2) ≤ 9
that is inside the cylinder x^2 + y^2 = 1 and above the cone x^2 + y^2 = z^2.
Figure 5:
Soln: The top surface is z = u 2 (x, y) =
9 − x^2 − y^2 =
9 − r^2 and the bottom surface is z = u 1 (x, y) =
x^2 + y^2 = r over the region D defined by the intersection of the top (or
bottom) and the cylinder which is a disk x^2 + y^2 ≤ 1 or 0 ≤ r ≤ 1 in the xy-plane.
∫ ∫ ∫
E
zdV =
D
[∫ √
9 −r^2
r
zdz
]
dA =
∫ (^2) π
0
0
∫ √ 9 −r 2
r
zrdzdrdθ = ∫ (^2) π
0
0
[9 − 2 r^2 ]rdrdθ =
∫ (^2) π
0
0
[9r − 2 r^3 ]drdθ =
∫ (^2) π
0
[9/ 4 − 1 /4]dθ = 4π
Example Find the volume of the portion of the sphere x^2 +y^2 +z^2 = 4 inside the cylinder (y − 1)^2 + x^2 = 1.
Figure 6:
Soln: The top surface is z =
4 − x^2 − y^2 =
4 − r^2 and the bottom is z = −
4 − x^2 − y^2 = −
4 − r^2 over the region D defined by the cylinder equation in the xy-plane. So rewrite the cylinder equation x^2 + (y − 1)^2 = 1 as x^2 + y^2 − 2 y + 1 = 1 ⇒ r^2 = 2r sin(θ) ⇒ r = 2 sin(θ).
V (E) =
E
1 dV =
D
∫ √ 4 −r 2
− √ 4 −r^2
1 dzdA =
∫ (^) π
0
∫ (^) 2 sin(θ)
0
∫ √ 4 −r 2
− √ 4 −r^2
1 rdzdrdθ = ∫ (^) π
0
∫ (^) 2 sin(θ)
0
2 r
4 − r^2 drdθ (by substitution u = 4 − r^2 ) = ∫ (^) π
0
[(4 − 4 sin^2 (θ))^3 /^2 − (4)^3 /^2 ]dθ (use identity 1 = cos^2 (θ) + sin^2 (θ)) = ∫ (^) π
0
[1 − | cos(θ)|^3 ]dθ =
∫ (^) π/ 2
0
[1 − cos^3 (θ)]dθ +
∫ (^) π
π/ 2
[1 + cos^3 (θ)]dθ = ∫ (^) π/ 2
0
[1 − (1 − sin^2 θ) cos θ]dθ +
∫ (^) π
π/ 2
[1 + (1 − sin^2 θ) cos θ]dθ =
16 /3[(θ − sin θ + sin^3 θ/3)|π/ 0 2 + (θ + sin θ − sin^3 θ/3)|ππ/ 2 ] = 16π/ 3 − 64 / 9
- Triple Integrals in Spherical Coordinates
The volume element in spherical coordinates is a spherical wedge with sides dρ, ρdφ, rdθ. Replacing r with ρ sin(φ) gives:
dV = ρ^2 sin(φ)dρdφdθ
For our integrals we are going to restrict E down to a spherical wedge. This will mean a ≤ ρ ≤ b, α ≤ θ ≤ β, c ≤ φ ≤ d,
E
f (x, y, z)dV =
∫ (^) β
α
∫ (^) d
c
∫ (^) b
a
f (ρ sin(φ) cos(θ), ρ sin(φ) sin(θ), ρ cos(φ))ρ^2 sin(φ)dρdφdθ
Figure 9: One example of the sphere wedge, the lower limit for both ρ and φ are 0
The more general formula for triple integration in spherical coordinates: If a solid E is the region between g 1 (θ, φ) ≤ ρ ≤ g 2 (θ, φ), α ≤ θ ≤ β, c ≤ φ ≤ d, then
∫ ∫ ∫
E
f (x, y, z)dV =
∫ (^) β
α
∫ (^) d
c
∫ (^) g 2 (θ,φ)
g 1 (θ,φ)
f (ρ sin(φ) cos(θ), ρ sin(φ) sin(θ), ρ cos(φ))ρ^2 sin(φ)dρdφdθ
Example Find the volume of the solid region above the cone z^2 = 3(x^2 + y^2 ) (z ≥ 0) and below the sphere x^2 + y^2 + z^2 = 4. Soln: The sphere x^2 + y^2 + z^2 = 4 in spherical coordinates is ρ = 2. The cone z^2 = 3(x^2 + y^2 ) (z ≥ 0) in spherical coordinates is z =
3(x^2 + y^2 ) =
√^3 r^ ⇒^ ρ^ cos(φ) = 3 ρ sin(φ) ⇒ tan(φ) = 1/
3 ⇒ φ = π/6. Thus E is defined by 0 ≤ ρ ≤ 2, 0 ≤ φ ≤ π/6, 0 ≤ θ ≤ 2 π.
V (E) =
E
1 dV =
∫ (^2) π
0
∫ (^) π/ 6
0
0
ρ^2 sin(φ)dρdφdθ = ∫ (^2) π
0
∫ (^) π/ 6
0
sin(φ)dφdθ =
16 π 3
Figure 10:
Figure 11:
Example Find the centroid of the solid region E lying inside the sphere x^2 +y^2 +z^2 = 2z and outside the sphere x^2 + y^2 + z^2 = 1 Soln: By the symmetry principle, the centroid lies on the z axis. Thus we only need to compute ¯z The top surface is x^2 + y^2 + z^2 = 2z ⇒ ρ^2 = 2ρ cos(φ) or ρ = 2 cos(φ). The bottom surface is x^2 + y^2 + z^2 = 1 ⇒ ρ = 1. They intersect at 2 cos(φ) = 1 ⇒ φ = π/3.
m =
E
1 dV =
∫ (^2) π
0
∫ (^) π/ 3
0
∫ (^) 2 cos(φ)
1
ρ^2 sin(φ)dρdφdθ = ∫ (^2) π
0
∫ (^) π/ 3
0
cos^3 (φ) sin(φ)dφdθ −
∫ (^2) π
0
∫ (^) π/ 3
0
sin(φ)dφdθ =
11 π 12
