Multiple_Integrals.pdf

Multiple Integrals
Dr. Xuan Dieu Bui
School of Applied Mathematics and Informatics,
Hanoi University of Science and Technology
Dr. Xuan Dieu Bui Multiple Integrals I ♥ HUST 1 / 89

Bibliography
James Stewart, Calculus Early Transcendentals, Brooks Cole Cengage
Learning, 2012.
1 Multiple Integrals: Chapter 15,
2 Integrals depending on a parameter:
3 Line Integrals: Chapter 16,
4 Surface Integrals: Chapter 16,
5 Vector Calculus: Chapter 16,
6 Series: Chapter 11.
http://bit.ly/bai-giang

Multiple Integrals
1 Double Integrals
Double Integrals over Rectangles
Double Integrals over General Regions
Double Integrals in Polar Coordinates
Change of Variables in Double Integrals
Applications of Double Integrals
2 Triple Integrals
Triple Integrals over Rectangular Box
Triple Integrals over General Regions
Change of Variables in Triple Integrals
Triple Integrals in Cylindrical Coordinates
Triple Integrals in Spherical Coordinates
Applications of Triple Integrals

Double Integrals
Multiple Integals
1 Double Integrals
2 Triple Integrals

Double Integrals Double Integrals over Rectangles
Double Integrals
Review of the Definite Integral
Z b
a
f (x)dx = lim
n→∞
n
X
i=1
f (x∗
i )∆x.

Double Integrals
1 divide [a, b] into n subintervals [xi−1, xi ] of equal width ∆x = b−a
n
2 choose sample points x∗
i in these subintervals,
3 form the Riemann sum
n
P
i=1
f (x∗
i )∆x
4 take the limit V =
b
R
a
f (x)dx = lim
n→∞
n
P
i=1
f (x∗
i )∆x

Volumes and Double Integrals
Goal: find the volume of S := {(x, y, z) ∈ R3|0 ≤ z ≤ f (x, y), (x, y) ∈ R.}

1 divide [a, b] into m subintervals and [c, d] into n subintervals, each of
equal width.
2 choose sample points (x∗
ij , y∗
ij ) ∈ Rij = [xi−1, xi ] × [yi−1, yi ],
V (Rij) ≈ f (x∗
ij , y∗
ij )∆x∆y.
3 Riemann Sum V =
m
P
i=1
n
P
j=1
Rij ≈
m
P
i=1
n
P
j=1
f (x∗
ij , y∗
ij )∆x∆y = Sm,n.
4 take the limit lim
m,n→∞
Smn.

Double Integrals
Definition
The double integral of f over the rectangle R is
ZZ
R
f (x, y)dxdy = lim
m,n→∞
m
X
i=1
n
X
j=1
f (x∗
ij , y∗
ij )∆x∆y
if this limit exists.

Fubini’s Theorem
Theorem (Fubini’s Theorem)
If f is continuous on the rectangle R = {(x, y)|a ≤ x ≤ b, c ≤ x ≤ d},
then
ZZ
R
f (x, y)dxdy =
Z b
a
Z d
c
f (x, y)dy
!
dx =
Z d
c
Z b
a
f (x, y)dx
!
dy.
RR
R
f (x, y)dxdy = V =
b
R
a
A(x)dx, where
A(x) =
d
R
c
f (x, y)dy.

Triple Integrals
Properties
1
RR
R
[f (x, y) + g (x, y)] dxdy =
RR
R
f (x, y) dxdy +
RR
R
g (x, y) dxdy.
2
RR
R
αf (x, y) dxdy = α
RR
R
f (x, y)dxdy.
3
RR
D
f (x, y) dxdy =
RR
D1
f (x, y) dxdy +
RR
D2
f (x, y) dxdy.
x
y
O
D1 D2

Double Integrals Double Integrals over General Regions
Double Integrals over general regions
F(x, y) =
(
f (x, y), if (x, y) ∈ D,
0, if (x, y) ∈ R D.
and ZZ
D
f (x, y)dxdy =
ZZ
R
F(x, y)dxdy.

Double Integrals over plane regions of type I
D = {(x, y)|a ≤ x ≤ b, g1(x) ≤ y ≤ g2(x)}.
x
y
O
D
y = g1(x)
y = g2(x)
a b
R = [a, b] × [c, d]
c
d
By Fubini’s Theorem,
ZZ
D
f (x, y)dxdy =
ZZ
R
F(x, y)dxdy

D = {(x, y)|a ≤ x ≤ b, g1(x) ≤ y ≤ g2(x)}.
x
y
O
D
y = g1(x)
y = g2(x)
a b
R = [a, b] × [c, d]
c
d
By Fubini’s Theorem,
ZZ
D
f (x, y)dxdy =
ZZ
R
F(x, y)dxdy
=
Z b
a
dx
Z d
c
F(x, y)dy =
Z b
a
dx
Z g2(x)
g1(x)
f (x, y)dy.

If f is continuous on a type I region D such that
D = {(x, y)|a ≤ x ≤ b, g1(x) ≤ y ≤ g2(x)}
then ZZ
D
f (x, y)dxdy =
Z b
a
dx
Z g2(x)
g1(x)
f (x, y)dy.
x
y
O
D
a b
c
d
x
y
O
D
a b
c
d
x
y
O
D
a b
c
d

Double Integrals over plane regions of type II
If f is continuous on a type II region D such that
D = {(x, y)|c ≤ y ≤ d, h1(y) ≤ x ≤ h2(y)}
then ZZ
D
f (x, y)dxdy =
Z d
c
dy
Z h2(y)
h1(y)
f (x, y)dx.
y
D
x = h1(y)
x = h2(y)
d
c

Example
Evaluate
RR
D
x2 (y − x) dxdy where D is the region bounded by y = x2 and
x = y2.
x
y
O 1
1
y = x2
x = y2

Change The Order of Integration
Example
Evaluate I =
RR
D
xey2
dx, where D = {(x, y) : 0 ≤ x ≤ 1, x2 ≤ y ≤ 1}.
x
1
y
1
O
y = x2

Exercise
Change the order of integration I =
b
R
a
dx
f2(x)
R
f1(x)
f (x, y)dy.
1 From the iterated integral, sketch the region of integration,
2 Divide it into regions of type II, for instance,
Di = {(x, y)|ci ≤ y ≤ di , gi (y) ≤ x ≤ hi (x)},
3
b
Z
a
dx
f2(x)
Z
f1(x)
f (x, y)dy =
X
i
di
Z
ci
dy
hi (y)
Z
gi (y)
f (x, y)dx.

Exercise
Change the order of integration I =
b
R
a
dx
f2(x)
R
f1(x)
f (x, y)dy.
1 From the iterated integral, sketch the region of integration,
2 Divide it into regions of type II, for instance,
Di = {(x, y)|ci ≤ y ≤ di , gi (y) ≤ x ≤ hi (x)},
3
b
Z
a
dx
f2(x)
Z
f1(x)
f (x, y)dy =
X
i
di
Z
ci
dy
hi (y)
Z
gi (y)
f (x, y)dx.
Similar for
d
R
c
dy
f2(y)
R
f1(y)
f (x, y)dx.

Example
Change the order of integration
1
R
−1
dx
1−x2
R
−
√
1−x2
f (x, y) dy.
x
1
y
1
O
D1
D2

Example
1
R
0
dy
1+
√
1−y2
R
2−y
f (x, y) dx.
x
2
1
y
2
1
O

Example
2
R
0
dx
√
2x
R
√
2x−x2
f (x, y) dy.
x
2
1
y
2
1
O

Example
√
2
R
0
dy
y
R
0
f (x, y) dx+
2
R
√
2
dy
√
4−y2
R
0
f (x, y) dx.
x
√
2
y
√
2
O

Double Integrals involving Absolute Value Functions
Evaluate
RR
D
|f (x, y)| dxdy. The curve f (x, y) = 0 divides D into two
parts,
D+
= D ∩ {f (x, y) ≥ 0}, D−
= D ∩ {f (x, y) ≤ 0}.
ZZ
D
|f (x, y)| dxdy =
ZZ
D+
f (x, y) dxdy −
ZZ
D−
f (x, y) dxdy (1)

Evaluate
RR
D
|f (x, y)| dxdy. The curve f (x, y) = 0 divides D into two
parts,
D+
= D ∩ {f (x, y) ≥ 0}, D−
= D ∩ {f (x, y) ≤ 0}.
ZZ
D
|f (x, y)| dxdy =
ZZ
D+
f (x, y) dxdy −
ZZ
D−
f (x, y) dxdy (1)
Algorithm
1 Sketch the curve f (x, y) = 0 to find D+, D−.
2 Apply formula (1).

Example
Evaluate
RR
D
|x + y|dxdy, D :

(x, y) ∈ R2 ||x ≤ 1| , |y| ≤ 1 .
O
x
1
D+
y
1
D−

Example
Evaluate
RR
D
q
|y − x2|dxdy, D :

(x, y) ∈ R2 ||x| ≤ 1, 0 ≤ y ≤ 1 .
O x
1
D+
y
1
D−

Solving Double Integrals Using Symmetry
Theorem
If
1 f (x, y) is an odd function with respect to y and,
2 D is symmetric with respect to x− axis
then ZZ
D
f (x, y) dxdy = 0.

Theorem
If
1 f (x, y) is an odd function with respect to y and,
then ZZ
D
f (x, y) dxdy = 0.
Theorem
If
1 f (x, y) is an even function with respect to y and,
ZZ
D
f (x, y) dxdy = 2
ZZ
D+
f (x, y) dxdy.

Theorem
If
1 f (−x, −y) = −f (x, y),
2 D is symmetric with respect to the origin,
then
RR
D
f (x, y) dxdy = 0.

Theorem
If
1 f (−x, −y) = −f (x, y),
2 D is symmetric with respect to the origin,
then
RR
D
f (x, y) dxdy = 0.
Example
Evaluate
RR
|x|+|y|≤1
|x| + |y|dxdy.
O x
1
y
1

Double Integrals Double Integrals in Polar Coordinates
The polar coordinate of a point M is a pair (r, θ), where



r =
−
−
→
OM
θ =

−
−
→
OM,Ox.
O
y
x
O
M
θ
r = |
−
−
→
OM|
x
y
Polar coordinates vs rectangular coordinates:
(
x = r cos θ,
y = r sin θ.

∆Ai =
1
2
r2
i ∆θ −
1
2
r2
i−1∆θ =
1
2
(r2
i − r2
i−1)∆θ
=
1
2
(ri + ri−1)(ri − ri−1)∆θ = r∗
i ∆r∆θ.

ZZ
R
f (x, y)∆A = lim
m,n→∞
m
X
i=1
n
X
j=1
f (r∗
i cos θ∗
j , r∗
i sin θ∗
j )∆Ai
= lim
m,n→∞
m
X
i=1
n
X
j=1
f (r∗
i cos θ∗
j , r∗
i sin θ∗
j )r∗
i ∆r∆θ
=
ZZ
R
f (r cos θ, r sin θ) r drdθ.

If f is continuous on a polar region of the form
(
θ1 ≤ θ ≤ θ2
r1 (θ) ≤ r ≤ r2 (θ) ,
then
I =
θ2
Z
θ1
dθ
r2(θ)
Z
r1(θ)
f (r cos θ, r sin θ) r dr

Example
Evaluate I =
RR
D
dxdy, where D :
(
4x ≤ x2 + y2 ≤ 8x,
x ≤ y ≤
√
3x.
x
4
2 8
y
O

Example
Evaluate
RR
D
dxdy
(x2+y2)2 , where D :
(
4y ≤ x2 + y2 ≤ 8y
x ≤ y ≤ x
√
3.
x
y
4
8
O
y = x
y = x
√
3

Example
Evaluate
RR
D
xy2dxdy where D is bounded by
(
x2 + (y − 1)2
= 1
x2 + y2 − 4y = 0.
x
y
2
4
O

Example
Evaluate
RR
D
xy
x2+y2 dxdy, where D :
(
x2 + y2 ≤ 12, x2 + y2 ≥ 2x
x2 + y2 ≥ 2
√
3y, x ≥ 0, y ≥ 0.
x
2 2
√
3
y
2
√
3
O
D1
D2

Double Integrals Change of Variables in Double Integrals
Change of variables in double integrals
Back in Calculus I,
b
R
a
f (x)dx =
d
R
c
f (x(t)) x′
(t) dt, where x = x(t).
Similarly,
RR
R
f (x, y)dxdy =
RR
S
f (x(u, v), y(u, v)) factor dudv
Example
Consider the transformation T :
(
x = x(u, v) = u2 − v2,
y = y(u, v) = 2uv.
Find the image of the square S = {(u, v) : 0 ≤ u ≤ 1, 0 ≤ v ≤ 1}.

∆A ≈ |ru × rv |∆u∆v =
x′
u x′
v
y′
u y′
v
∆u∆v,

∆A ≈ |ru × rv |∆u∆v =
x′
u x′
v
y′
u y′
v
∆u∆v, J =
D(x, y)
D(u, v)
=
x′
u x′
v
y′
u y′
v
.

ZZ
R
f (x, y)dA = lim
m,n→∞
m
X
i=1
n
X
j=1
f (xi , yj)∆A
= lim
m,n→∞
m
X
i=1
n
X
j=1
f (x(ui , vj), y(ui , vj))
x′
u x′
v
y′
u y′
v
∆u∆v
=
ZZ
S
f (x(u, v), y(u, v))|J|dudv.

Change of variables
Let T :
(
x = x(u, v),
y = y(u, v)
, S → R,
x(u, v), y(u, v) has continuous partial derivatives on S,
the transformation is a 1 − 1 map.
the Jacobian J =
x′
u x′
v
y′
u y′
v
6= 0 on S.
Then
RR
D
f (x, y)dxdy =
RR
S

Change of variables
Let T :
(
x = x(u, v),
y = y(u, v)
, S → R,
x(u, v), y(u, v) has continuous partial derivatives on S,
the transformation is a 1 − 1 map.
the Jacobian J =
x′
u x′
v
y′
u y′
v
6= 0 on S.
Then
RR
D
f (x, y)dxdy =
RR
S
Example
Evaluate I =
RR
D
4x2 − 2y2

dxdy, where D :
(
1 ≤ xy ≤ 4
x ≤ y ≤ 4x.

Example
Evaluate I =
RR
D
4x2 − 2y2

dxdy, where D :
(
1 ≤ xy ≤ 4
x ≤ y ≤ 4x.
x
y
O 1
1
y = 4x
y = x
xy = 4
xy = 1

Polar coordinate transformation
1
(
x = r cos θ,
y = r sin θ
2 J = D(x,y)
D(r,θ) =
x′
r x′
θ,
y′
r y′
θ
= r
3
RR
R
f (x, y)dxdy =
RR
S
f (r cos θ, r sin θ) r drdθ.

Polar coordinates
If D : x2
a2 + y2
b2 ≤ 1, then
(
x = ar cos ϕ
y = br sin ϕ
, J = abr
Example
Evaluate
RR
D
9x2 − 4y2 dxdy, where D : x2
4 + y2
9 ≤ 1.
x
2
y
3
O

Polar Coordinates
If D : (x − a)2
+ (y − b)2
≤ R2, then
(
x = a + r cos ϕ
y = b + r sin ϕ
, J = r
Example
Evaluate
2
R
0
dx
√
2x−x2
R
−
√
2x−x2
p
2x − x2 − y2dy.
x
2
y
O

Polar Coordinates
Example
Evaluate
RR
D
xydxdy, where D : (x − 2)2
+ y2 ≤ 1.
x
3
1
y
O

Polar Coordinates
Example
Evaluate
RR
D
xydxdy, where D : (x − 2)2
+ y2 ≤ 1, y ≥ 0
x
3
1
y
O

Double Integrals Applications of Double Integrals
Example
Compute the area of the domain bounded by
(
y2 = x, y2 = 2x
x2 = y, x2 = 2y.
x
y
O
y = x2 x2 = 2y
x = y2
2x = y2
1 3
√
2 3
√
4 2

Example
Compute the area of the domain D bounded by
(
y = 0, y2 = 4ax
x + y = 3a, (a 0) .
y
O x
3a
3a
−6a

Example
(
x2 + y2 = 2x, x2 + y2 = 4x
x = y, y = 0.
y = x
x
2 4
y
O

Example
Compute the area of the domain D bounded by r = 1, r = 2
√
3
cos ϕ.
x
y
O

Example
x2 + y2
2
= 2a2xy (a 0).
x
y
O
r = a
√
sin 2ϕ

Example
Compute the area of the domain D bounded by x3 + y3 = axy (a 0)
(Descartes leaf)
x
y
O 1
2
1
2
y = −x − 1
3

Example
Compute the area of the domain D bounded by r = a (1 + cos ϕ) (a 0)
(Cardioids)
x
y
O
2a
a
−a

Ω :
(
0 ≤ z ≤ f (x, y),
(x, y) ∈ D
⇒ V (Ω) =
ZZ
D
f (x, y) dxdy.
y
z
x
O
z = f (x, y)
D

Volume of cylindrical objects
V :
(
z1(x, y) ≤ z ≤ z2(x, y),
(x, y) ∈ D
⇒ V (Ω) =
ZZ
D
(z2 (x, y) − z1 (x, y))dxdy.
y
z
x
O
z = z2(x, y)
z = z1(x, y)
D
Ω

Example
Compute the volume of the object given by
(
3x + y ≥ 1, y ≥ 0
3x + 2y ≤ 2, 0 ≤ z ≤ 1 − x − y.
y
z
x
O

Example
Compute the volume of the object bounded by the surfaces
(
z = 4 − x2 − y2
2z = 2 + x2 + y2.
y
z
x
O
2z = 2 + x2 + y2
z = 4 − x2 − y2

Example
Compute the volume of the object given by V :
(
0 ≤ z ≤ 1 − x2
− y2
y ≥ x, y ≤
√
3x
y
z
x
O 1
1

Example
Compute the volume of the object given by V :
(
x2
+ y2
+ z2
≤ 4a2
x2
+ y2
− 2ay ≤ 0
y
z
x
O
2a
2a
2a

Example







z =
x2
a2
+
y2
b2
, z = 0
x2
a2
+
y2
b2
=
2x
a
x
z
O
z = x2
a2 + y2
b2
a
1

Example
V :



az = x2
+ y2
z =
q
x2 + y2
y
z
x
O a
−a
a

Area of a curved surface
S =
RR
D
q
1 + z′2
x + z′2
y dxdy.
y
z
x
O
z = f (x, y)
D

Triple Integrals
Multiple Integals
1 Double Integrals
2 Triple Integrals

Triple Integrals Triple Integrals over Rectangular Box
Triple Integrals
Z b
a
f (x)dx = lim
n→∞
n
X
i=1
f (x∗
i )∆x.

Triple Integrals
1 divide [a, b] into n subintervals [xi−1, xi ] of equal width ∆x = b−a
n
2 choose sample points x∗
i in these subintervals,
3 form the Riemann sum
n
P
i=1
f (x∗
i )∆x
4 take the limit V =
b
R
a
f (x)dx = lim
n→∞
n
P
i=1
f (x∗
i )∆x

Goal: find the volume of S := {(x, y, z) ∈ R3|0 ≤ z ≤ f (x, y), (x, y) ∈ R.}

1 divide [a, b] into m subintervals and [c, d] into n subintervals, each of
equal width.
2 choose sample points (x∗
ij , y∗
ij ) ∈ Rij = [xi−1, xi ] × [yi−1, yi ],
V (Rij) ≈ f (x∗
ij , y∗
ij )∆x∆y.
3 Riemann Sum V =
m
P
i=1
n
P
j=1
Rij ≈
m
P
i=1
n
P
j=1
f (x∗
ij , y∗
ij )∆x∆y = Sm,n.
4 take the limit lim
m,n→∞
Smn.

1 divide B into sub-boxes by
dividing [a, b] into l subintervals [xi−1, xi ] of equal width ∆x,
dividing [c, d] into m subintervals [yj−1, yj ] of equal width ∆y,
dividing [r, s] into n subintervals [zk−1, zk ] of equal width ∆z.

2 Choose sample points (x∗
ijk, y∗
ijk, z∗
ijk) in each box Bijk. Each sub-box
has volume ∆V = ∆x∆y∆z.

2 Choose sample points (x∗
ijk, y∗
ijk, z∗
ijk) in each box Bijk. Each sub-box
has volume ∆V = ∆x∆y∆z.
3 form the triple Riemann sum
l
P
i=1
m
P
j=1
n
P
k=1
f (x∗
ijk, y∗
ijk, z∗
ijk)∆V .

Definition
ZZZ
B
f (x, y, z)dV = lim
l,m,n→∞
l
X
i=1
m
X
j=1
n
X
k=1
f (x∗
ijk, y∗
ijk, z∗
ijk)∆V .
Again, the triple integral always exists if is continuous.

Definition
ZZZ
B
f (x, y, z)dV = lim
l,m,n→∞
l
X
i=1
m
X
j=1
n
X
k=1
f (x∗
ijk, y∗
ijk, z∗
ijk)∆V .
Again, the triple integral always exists if is continuous.
Theorem (Fubini’s Theorem)
If f is continuous on the rectangular box B = [a, b] × [c, d] × [r, s] then
ZZZ
B
f (x, y, z)dxdydz =
Z b
a
dx
Z d
c
dy
Z s
r
f (x, y, z)dz.

Triple Integrals Triple Integrals over General Regions
If E is a general bounded solid region, choose
B = [a, b] × [c, d] × [r, s] ⊃ V
and define
F(x, y, z) =
(
f (x, y, z), if (x, y, z) ∈ E,
0, if (x, y, z) ∈ B E.

If E is a general bounded solid region, choose
B = [a, b] × [c, d] × [r, s] ⊃ V
and define
F(x, y, z) =
(
f (x, y, z), if (x, y, z) ∈ E,
0, if (x, y, z) ∈ B E.
Definition
ZZZ
E
f (x, y, z)dV =
ZZZ
B
F(x, y, z)dV .

Triple Integrals over Regions of type I
If V :
(
z1(x, y) ≤ z ≤ z2(x, y),
(x, y) ∈ D
then
I =
ZZZ
V
f (x, y, z) dxdydz =
ZZ
D
dxdy
z2(x,y)
Z
z1(x,y)
f (x, y, z) dz
y
z
x
O
z = z2(x, y)
z = z1(x, y)
D
V

Triple Integrals over Regions of type I
Reduce the triple integral into a double integral
1. Find the projection of V onto Oxy.
2. Find
(
the lower boundary z = z1 (x, y) ,
the upper boundary z = z2 (x, y)
of V .
3. Apply the formula
I =
ZZZ
V
f (x, y, z) dxdydz =
ZZ
D
dxdy
z2(x,y)
Z
z1(x,y)
f (x, y, z) dz
.
The idea:
Triple Integrals ⇒ Double Integrals ⇒ Iterated integrals

Example
Evaluate
RRR
V
x2 + y2

dxdydz, where V :
p
x2 + y2 ≤ z ≤
p
1 − x2 − y2.
y
z
x
O
z =
p
1 − x2 − y2
z =
p
x2 + y2
D

By the same sort of argument,
1 Type II:
RRR
E
f (x, y, z)dxdydz =
RR
D

u2(y,z)
R
u1(y,z)
f (x, y, z)dx
#
dydz.
2 Type III:
RRR
E
f (x, y, z)dxdydz =
RR
D

u2(x,z)
R
u1(x,z)
f (x, y, z)dy
#
dxdz.

Solving triple integrals using symmetry
Theorem
If
1 V is symmetric with respect to z = 0,
2 f (x, y, z) is an odd function with respect to z
then
RRR
V
f (x, y, z) dxdydz = 0.

Solving triple integrals using symmetry
Theorem
If
2 f (x, y, z) is an odd function with respect to z
then
RRR
V
f (x, y, z) dxdydz = 0.
Theorem
If
2 f (x, y, z) is an even function with respect to z
then
RRR
V
f (x, y, z) dxdydz = 2
RRR
V +
f (x, y, z) dxdydz.
Note: The role of x, y, z can be interchangeable.

Triple Integrals Change of Variables in Triple Integrals
1 Calculus 1,
b
R
a
f (x)dx =
d
R
c
f (x(u)) x′
(u) du, where x = x(u).

1 Calculus 1,
b
R
a
f (x)dx =
d
R
c
f (x(u)) x′
2
RR
R
f (x, y)dxdy =
RR
S
f (x(u, v), y(u, v))|J|dudv,
(
x = x(u, v),
y = y(u, v).

1 Calculus 1,
b
R
a
f (x)dx =
d
R
c
f (x(u)) x′
2
RR
R
f (x, y)dxdy =
RR
S
f (x(u, v), y(u, v))|J|dudv,
(
x = x(u, v),
y = y(u, v).
3
ZZZ
B
f (x, y, z)dxdydz =
ZZ
S
f (x(u, v, w), y(u, v, w), z(u, v, w)) factor dudvdw,
where







x = x(u, v, w),
y = y(u, v, w),
z = z(u, v, w).

Consider the transformation: T :







x = x(u, v, w)
y = y(u, v, w)
z = z(u, v, w).
, V ′ → V satisfies
1 T is a 1 − 1 map.
2 x(u, v, w), y(u, v, w), z(u, v, w) are continuous and have continuous
partial derivatives on V ′.
3 The Jacobian determinant
J =
D(x, y, z)
D(u, v, w)
=
x′
u x′
v x′
w
y′
u y′
v y′
w
z′
u z′
v z′
w
6= 0 in V ′
.
Then
ZZZ
V
f (x, y, z)dxdydz =
ZZZ
V ′
f (x(u, v, w), y(u, v, w), z(u, v, w)) |J|dudvdw.
(1)

Example
Evaluate
RRR
V
(x + y + z)dxdydz, where V is bounded by







x + y + z = ±3
x + 2y − z = ±1
x + 4y + z = ±2
.

Example
Evaluate
RRR
V
(x + y + z)dxdydz, where V is bounded by







x + y + z = ±3
x + 2y − z = ±1
x + 4y + z = ±2
.
Consider the transformation







u = x + y + z
v = x + 2y − z
w = x + 4y + z
J−1
=
D (u, v, w)
D (x, y, z)
=
1 1 1
1 2 −1
1 4 1
= 6 ⇒ J =
1
6
.

Triple Integrals Triple Integrals in Cylindrical Coordinates
z
y
x
O
M
M′
ϕ
r = |
−
−
→
OM′|
b
b
Cylindrical Coordinate
M(r, ϕ, z) , where (r, ϕ) is the
polar coordinate of M′.

z
y
x
O
M
M′
ϕ
r = |
−
−
→
OM′|
b
b
Cylindrical vs rectangular coor.
x = r cos ϕ, y = r sin ϕ, z = z.

z
y
x
O
M
M′
ϕ
r = |
−
−
→
OM′|
b
b
Cylindrical vs rectangular coor.
x = r cos ϕ, y = r sin ϕ, z = z.
The transformation
Taking the transformation







x = r cos ϕ
y = r sin ϕ
z = z.
then J = D(x,y,z)
D(r,ϕ,z) = r and
I =
ZZZ
V
f (x, y, z)dxdydz =
ZZZ
Vrϕz
f (r cos ϕ, r sin ϕ, z)rdrdϕdz.

Example
Evaluate
RRR
V
x2 + y2

dxdydz, where V :
(
x2
+ y2
≤ 1
1 ≤ z ≤ 2
y
z
x
O
V
1
2

Example
Evaluate
RRR
V
z
p
x2 + y2dxdydz, where:
a) V is bounded by: x2 + y2 = 2x and z = 0, z = a (a 0).
b) V is a half of the sphere x2 + y2 + z2 ≤ a2, z ≥ 0 (a 0)
y
z
x
O
y
z
x
O

Example
Evaluate I =
RRR
V
ydxdydz, where V is bounded by:
(
y =
p
z2 + x2
y = h.
y
z
x
O h

Example
Evaluate I =
RRR
V
p
x2 + y2dxdydz where V is bounded by:
(
x2
+ y2
= z2
z = 1.
y
z
O

Example
Evaluate
RRR
V
dxdydz
√
x2+y2+(z−2)2
, where V :
(
x2
+ y2
≤ 1
|z| ≤ 1.
y
z
x
O

Triple Integrals Triple Integrals in Spherical Coordinates
z
y
x
O
M
M′
ϕ
θ
r = |
−
−
→
OM|
b
b
The spherical coordinate of a point M is an
ordered triple (r, θ, ϕ), where
r =
−
−
→
OM , θ =

Oz,
−
−
→
OM

, ϕ =

Ox,
−
−
→
OM′


z
y
x
O
M
M′
ϕ
θ
r = |
−
−
→
OM|
b
b
r =
−
−
→
OM , θ =

Oz,
−
−
→
OM

, ϕ =

Ox,
−
−
→
OM′

Spherical vs rectangular coor.
x = r sin θ cos ϕ, y = r sin θ sin ϕ, z = r cos θ.

z
y
x
O
M
M′
ϕ
θ
r = |
−
−
→
OM|
b
b
r =
−
−
→
OM , θ =

Oz,
−
−
→
OM

, ϕ =

Ox,
−
−
→
OM′

Spherical vs rectangular coor.
x = r sin θ cos ϕ, y = r sin θ sin ϕ, z = r cos θ.
The transformation
Let







x = r sin θ cos ϕ
y = r sin θ sin ϕ
z = r cos θ,
then J = D(x,y,z)
D(r,θ,ϕ) = −r2 sin θ and
ZZZ
V
f (x, y, z)dxdydz =
ZZZ
Vrθϕ
f (· · · , · · · , · · · )r2
sin θdrdθdϕ.

Example
Evaluate
RRR
V
x2 + y2 + z2

dxdydz, where V :
(
1 ≤ x2
+ y2
+ z2
≤ 4
x2
+ y2
≤ z2
.
y
z
x
O
V1

Example
Evaluate
RRR
V
p
x2 + y2 + z2dxdydz, where V : x2 + y2 + z2 ≤ z.
y
z
x
O

Spherical and Cylindrical Coordinates - Special Cases
1 V : x2
a2 + y2
b2 + z2
c2 ≤ 1 ⇒







x = ar sin θ cos ϕ
y = br sin θ sin ϕ
z = cr cos θ
, J = −abcr2 sin θ

1 V : x2
a2 + y2
b2 + z2
c2 ≤ 1 ⇒







z = cr cos θ
2 V : (x − a)2
+ (y − b)2
+ (z − c)2
≤ R2
⇒







x = a + r sin θ cos ϕ
y = b + r sin θ sin ϕ
z = c + r cos θ
, J = −r2
sin θ

1 V : x2
a2 + y2
b2 + z2
c2 ≤ 1 ⇒







z = cr cos θ
2 V : (x − a)2
+ (y − b)2
+ (z − c)2
≤ R2
⇒







x = a + r sin θ cos ϕ
y = b + r sin θ sin ϕ
z = c + r cos θ
, J = −r2
sin θ
3 V : x2+y2
a2 + z2
b2 ≤ 1 ⇒







z = bz′
x = ar cos ϕ
y = ar sin ϕ

Example
Evaluate
RRR
V
z
p
x2 + y2dxdydz, where V : x2+y2
a2 + z2
b2 ≤ 1, z ≥ 0.

Example
Evaluate
RRR
V
z
p
a2 + z2
b2 ≤ 1, z ≥ 0.
Cylindrical Coordinate.
Let







z = bz′
x = ar cos ϕ
y = ar sin ϕ
, then
Spherical Coordinate.
Let







y = ar sin θ sin ϕ
z = br cos θ
, then

Example
Evaluate
RRR
V
z
p
a2 + z2
b2 ≤ 1, z ≥ 0.
Let







z = bz′
x = ar cos ϕ
y = ar sin ϕ
, then
Let







z = br cos θ
, then
J = a2br and







0 ≤ ϕ ≤ 2π,
0 ≤ r ≤ 1,
0 ≤ z′ ≤
√
1 − r2.
J = a2br2 sin θ and







0 ≤ ϕ ≤ 2π,
0 ≤ θ ≤ π
2 ,
0 ≤ r ≤ 1.

Example
Evaluate
RRR
V
z
p
a2 + z2
b2 ≤ 1, z ≥ 0.
Let







z = bz′
x = ar cos ϕ
y = ar sin ϕ
, then
Let







z = br cos θ
, then
J = a2br and







0 ≤ ϕ ≤ 2π,
0 ≤ r ≤ 1,
0 ≤ z′ ≤
√
1 − r2.
J = a2br2 sin θ and







0 ≤ ϕ ≤ 2π,
0 ≤ θ ≤ π
2 ,
0 ≤ r ≤ 1.
I =
2πa3b2
15
.

Triple Integrals Applications of Triple Integrals
Volume of a solid Object
V =
ZZZ
V
dxdydz.
Example
Compute the volume of the domain V bounded by







x + y + z = ±3
x + 2y − z = ±1
x + 4y + z = ±2.

V =
ZZZ
V
dxdydz.
Example







x + y + z = ±3
x + 2y − z = ±1
x + 4y + z = ±2.
Change of variables







u = x + y + z
v = x + 2y − z
w = x + 4y + z

V =
ZZZ
V
dxdydz.
Example







x + y + z = ±3
x + 2y − z = ±1
x + 4y + z = ±2.
Change of variables







u = x + y + z
v = x + 2y − z
w = x + 4y + z
J−1
=
D (u, v, w)
D (x, y, z)
=
1 1 1
1 2 −1
1 4 1
= 6

V =
ZZZ
V
dxdydz.
Example







x + y + z = ±3
x + 2y − z = ±1
x + 4y + z = ±2.
Change of variables







u = x + y + z
v = x + 2y − z
w = x + 4y + z
J−1
=
D (u, v, w)
D (x, y, z)
=
1 1 1
1 2 −1
1 4 1
= 6 ⇒ J =
1
6
,

V =
ZZZ
V
dxdydz.
Example







x + y + z = ±3
x + 2y − z = ±1
x + 4y + z = ±2.
Change of variables







u = x + y + z
v = x + 2y − z
w = x + 4y + z
J−1
=
D (u, v, w)
D (x, y, z)
=
1 1 1
1 2 −1
1 4 1
= 6 ⇒ J =
1
6
, V =
1
6
ZZZ
Vuvw
dudvdw = 8.

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