HUT – DEPARTMENT OF MATH.

APPLIED
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INTERNATIONAL PROGRAM
GIAÛI TÍCH 2

LEC 4: TRIPLE INTEGRAL

STEWART: PAGES 990 – 1010

• PhD. NGUYEÃN QUOÁC LAÂN (October, 2014)

 TRIPLE INTEGRAL ON BOX = ITERATED INTEGRAL
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B is some box in Oxyz: B = {(x,y,z) | a ≤ x ≤ b, c ≤ y ≤ d, r ≤ z ≤ s}

bd s

∫∫∫ f ( x, y, z )dV = ∫ ∫ ∫ f ( x, y, z )dzdydx : Inside first z → y → x

B acr
Remember : 3 = 1 + 1 + 1 (Triple Integral = Single + Single + Single).

Example : Evaluate ∫∫∫ xyz 2 dV , B = {0 ≤ x ≤ 1, − 1 ≤ y ≤ 2, 0 ≤ z ≤ 3}

B
Choose any of the 6 possible orders of integration: x – y – z (or …)
x =1 y =2
3 2
⎡ x yz 2 2⎤
yz ⎡y z 3 2 3 2
3z 2 27
3 2 2⎤
I = ∫ ∫⎢ ⎥ dydz = ∫ ∫ dydz = ∫ ⎢ ⎥ dz = ∫ dz =
0 −1⎣
2 ⎦ −
2 ⎣ 4 ⎦ 4 4
x =0 0 1 0 y = −1 0
1 2 3
27
Remark : f ( x, y, z ) and B are separable ⇒ I = ∫ xdx ⋅ ∫ ydy ⋅ ∫ z 2 dz =
0 −1 0
4
 TRIPLE INTEGRAL: DOUBLE + SINGLE (3 = 2 + 1)
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“Cylinder” E = {(x,y,z) | u1(x, y) ≤ z ≤ u1(x, y), (x, y) ∈ D ⊆ Oxz }

⎡u 2 ( x , y ) ⎤
∫∫∫ f ( x, y, z )dV = ∫∫ ⎢⎢ ∫ f ( x, y, z )dz ⎥⎥ dA : Inside first z → Double
E D ⎣ u1 ( x , y ) ⎦
Remember : 3 = 2 + 1 (Triple Integral = Double (out) + Single (in) ).

Example : Evaluate ∫∫∫ xyzdV , E = 0 ≤ z ≤ x 2 + y 2 , 0 ≤ x, y ≤ 1}{

E
Answer: D = {0 ≤ x ≤ 1, 0 ≤ y ≤ 1} (square) ⇒
x2 + y 2
[ ] ( )
11
xy 2 z = x2 + y 2 1 2 2 7
I = ∫∫ ∫ xyzdzdA = ∫∫ 2 z z =0 dA = ∫ ∫ xy x + y dydx =
2

D 0 D
200 48
1 1 x2 + y2
⎧ x = r cos θ
Remark : 1/ Direct I = ∫ ∫ ∫ xyzdzdydx 2/ D : disk ⇒ Polar ⎨
00 0 ⎩ y = r sin θ
 SOLID E: “RIGHT CYLINDER” RESPECT TO Z
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“Right cylinder” respect to z: E

= {(x, y, z) | (x,y) ∈ D, u1(x,y) ≤ z
≤ u2(x,y)}. Key: 1/ Recognize 2
equations z = u1(x,y), z = u2(x,y).
2/ Add intercep. u1(x,y) = u2(x,y).

⎡u 2 ( x , y ) ⎤
∫∫∫ f ( x, y, z )dV = ∫∫ ⎢⎢ ∫ fdz ⎥⎥ dA
E D ⎣ u1 ( x , y ) ⎦
D : a ≤ x ≤ b, g1 ( x ) ≤ y ≤ g 2 ( x ) ⇒
b g 2 ( x ) u2 ( x , y )

∫∫∫ f ( x, y, z )dV = ∫ ∫ ∫ fdzdydx

E a g1 ( x ) u1 ( x , y )
 EXAMPLE
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Example : ∫∫∫ zdV , E : bounded by x = 0, y = 0, z = 0, x + y + z = 1

E

Two equations for z: z = 0 & z = 1 – x – y. Domain D: x = 0, y = 0

is unfinite! Add interception : 1 – x – y = 0 ⇒ x + y = 1 → D.
⎧( x, y ) ∈ D
E:⎨
⎩0 ≤ z ≤ 1 − x − y
⎧0 ≤ x ≤ 1
D:⎨
⎩0 ≤ y ≤ 1 − x
1 1− x 1− x − y
⇒I=∫ ∫ ∫ zdzdydx
0 0 0
1 1− x z =1− x − y 1 1− x
⎡z 2⎤
1 1 1
1
=∫ ∫ dydx = ∫ ∫ (1 − x − y ) dydx = 6 ∫ (1 − x ) dx = 24
2 3
⎢ ⎥
0 0 ⎣ 2 ⎦ z =0 20 0 0
 OTHER CASES
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Cylinder respect to y, ( x, z ) ∈ D ⊆ Oxz

⎧u1 ( x, z ) ≤ y ≤ u2 ( x, z )
Solid E : ⎨
⎩( x, z ) ∈ D ⊆ Oxz
⎡u 2 ( x , z ) ⎤
⇒ ∫∫∫ fdV = ∫∫ ⎢ ∫ f ( x, y, z )dy ⎥ dA
D⎣ ⎢ u ( x, z ) ⎥
E 1 ⎦

Cylinder respect to x, ( y, z ) ∈ D ⊆ Oyz

⎧u1 ( y, z ) ≤ x ≤ u2 ( y, z )
Solid E : ⎨
⎩( y, z ) ∈ D ⊆ Oyz
⎡u 2 ( y , z ) ⎤
⇒ ∫∫∫ fdV = ∫∫ ⎢ ∫ f ( x, y, z )dx ⎥ dA
E D⎢ ⎣ u1 ( y , z ) ⎥⎦
 EXAMPLE
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Evaluate ∫∫∫ x 2 + z 2 dV , E : bounded by y = x 2 + z 2 and y = 4.

E

z= y − x2
E : z = − y − x2
⎧− 2 ≤ x ≤ 2
D:⎨ 2
E : Cylindre respect z ⎩x ≤ y ≤ 4
⇒ Complicated.
y = x2 + z 2
E : cylinder respect y ⇒ E : ⇒ D : x 2 + y 2 = 4 → D : Disk
y=4
( )
4
128π
I = ∫∫ ∫ x + z dydA = ∫∫ 4 − x − z
2 2 2 2
x + z dA. Polar ⇒ I =
2 2
15
D 2 2x +z D
 APPLICATION: VOLUME OF SOLID E
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Let f ( x, y, z ) ≡ 1 ∀ M ∈ E ⇒ ∫∫∫ dV = VE : volume of E

E

Example: Find the volume of the solid bounded by x = y2, x = z,

z = 0, x = 1.
⎧z = x ⎧x = y 2 ⎧− 1 ≤ y ≤ 1
Answer : E : ⎨ ⇒ D:⎨ : Finite. y = 1 ⇒ D : ⎨ 2
2
.
⎩z = 0 ⎩x = 1 ⎩y ≤ x ≤1
VE = ∫∫∫ dV =
E
x
= ∫∫∫ dzdA = ∫∫ xdA
D 0 D
1 1
4
∫ ∫ xdxdy = 5
−1 y 2
 CYLINDRICAL COORDINATES (PAGE 1000)
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A point P(x, y, z) ∈ R3 is represented

by the triple (r, θ, z), where r and θ
are polar coord. of P’ = Projec.OxyP

P/ ⎧ x = r cos θ y
⎨ ⇒ r = x 2
+ y 2
, tan θ =
⎩ y = r sin θ x

P Example: (a) Plot the point with cylin.

coor. (2, 2π/3, 1) and find its rec. coor.
P/ (b) Find cylin. coor. of M(3, –3, –7)
∈ Oxy 2π
Answer : (a) x = 2 cos L ⇒ P(− 1, 3, 1)
3
π
(b) r = 3 2, tanθ = −1 ⇒ M⎛⎜ 3 2,− ,−7 ⎞⎟
⎝ 4 ⎠
 POLAR IN Oxy & z ⇒ CYLINDRICAL COORDINATES
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Triple Integral ( x, y, z ) : 3 = 2 + 1 & D : disk, circle ⇒ Polar (r , θ )

u2 ( x , y )
I = ∫∫∫ f ( x, y, z )dV = ∫∫ ∫ fdzdA = ∫∫ g ( x, y )dA = ∫∫ g (r , θ )rdrdθ .
E D u1 ( x , y ) D Drθ

⎧ x = r cos θ , = r sin θ
Combine 2 steps into 1 step : ⎨ ⇒ Cylindrical.
⎩z = z

E = {(x,y) ∈ D, u1(x,y) ≤ z ≤ u2(x,y)},

where D in polar: {α ≤ θ ≤ β, h1(θ) ≤ r
≤ h2(θ)} ⇒ Triple Integral I = ∫∫∫ f dV:
β h2 (θ )u2 (r ,θ )
I=∫ ∫ ∫ f (r cosθ , r sin θ , z )rdzdrdθ
α h1 (θ ) u1 (r ,θ )
 EXAMPLE
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( )
Evaluate I = ∫∫∫ x 2 + y 2 dV , E = x 2 + y 2 ≤ 4, x 2 + y 2 ≤ z ≤ 2 { }
E

⎧⎪( x, y ) ∈ D : x 2 + y 2 ≤ 4 : Circle
E:⎨ & f = x 2 + y 2 ⇒ Cylind. coor.
⎪⎩u1 ( x, y ) = x 2 + y 2 ≤ z ≤ u2 = 2
⎧ x = r cos θ , y = r sin θ
Answer : Cylind. ⎨
⎩z = z
⎧0 ≤ r ≤ 2 ⎧(r , θ ) ∈ D
⇒ D:⎨ &E:⎨
⎩0 ≤ θ ≤ 2π ⎩r ≤ z ≤ 2
⎡ 2 ⎤ 2π 2 2
( )
⇒ I = ∫∫ ⎢ ∫ x 2 + y 2 dz ⎥ dS = ∫ ∫ ∫ r 3dzdrdθ
⎢ ⎥
D ⎢ x2 + y2 ⎥⎦
⎣ 0 0r
2π 2
1 4 1 5 ⎤ 16π
2
⎡
= ∫ dθ ∫ r (2 − r )dr = 2π ⎢ r − r ⎥ =
3

0 0
⎣2 5 ⎦0 5
SIMPLIFY SPHERE: SPHERICAL COORDINATES (P. 1005)
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0 ≤φ ≤π
P(x, y, z) ∈ R3 ↔ (ρ, θ, φ), ρ = OP ≥ 0,
θ: cylindrical angle, φ: angle (Oz, OP)
ρ cos φ ⎧ x = ρ sin φ cos θ
⎪ ⎧
⎪ ρ 2
= x 2
+ y 2
+ z 2
⎨ y = ρ sin φ sin θ ⇒ ⎨ 2 2
ρ sin φ ⎪ z = ρ cos φ ⎪
⎩ ρ sin φ = x 2
+ y 2
⎩

Example: (a) Plot the point with sphe.

coor. (2, π/4, π/3) and find its rec. coor.
(b) Find cylin. coor. of M (0,2 3,−2 )
π π
(a) ρ = 2, θ = ,φ = ⇒ P( 3 2, 3 2, 1)
4 3
(b) ρ = x 2 + y 2 + z 2 = 4 ⇒ M(4, π 2 , 2π 3)
 TRIPLE INTEGRAL IN SPHERICAL COORDINATES
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x2 + y2 + z 2 = R2 Ball: x2 + y2 + z2 ≤ R2 in spherical
coordinate: 0 ≤ ρ ≤ R, 0 ≤ θ ≤ 2π, 0 ≤ φ
≤ π ⇒ Triple integral ∫∫∫fdxdydz on B:
π 2π R

ρ=R ∫∫∫ fdV = ∫ ∫∫ f ( ρ ,θ , φ )ρ 2 sin φdρdθdφ

B 0 0 0

φ = c,
π
<c
2
<π

x = y ⇔θ =π 4 φ = c, 0 < c < π 2
 EXAMPLE
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Evaluate I = ∫∫∫ e (x )
2 2 232
+ y +z
dV , B is the unit ball : x 2 + y 2 + z 2 ≤ 1
B

⎧ x = ρ sin φ cos θ ⎧0 ≤ ρ ≤ 1
⎪ ⎪
Answer : Spherical coordinate ⎨ y = ρ sin φ sin θ ⇒ B : ⎨0 ≤ θ ≤ 2π
⎪ z = ρ cos φ ⎪0 ≤ φ ≤ π
⎩ ⎩

As x 2 + y 2 + z 2 = ρ 2 ⇒ I = ∫
π 2π 1
(ρ )
232
∫ ∫ e ρ 2 sin φdρdθdφ =
0 0 0
π 2π 1 ρ =1
⎡ 1 ρ3 ⎤ 4π
= ∫ sin φdφ ∫ dθ ∫ ρ e dρ = [− cos φ ]φ =0 (2π )⎢ e ⎥
2 ρ φ =π 3
= (e − 1)
0 0 0
⎣3 ⎦ ρ =0 3

⎧⎪− 1 ≤ x ≤ 1,− 1 − x 2 ≤ y ≤ 1 − x 2
2
1 1− x L
In Oxyz, B : ⎨ ⇒∫ ∫ ∫ eL
:?
⎪⎩− 1 − x 2 − y 2 ≤ z ≤ 1 − x 2 − y 2 −1− 1− x 2 − L