Wissam Karam

Chapter VIII Triple integrals

1 TRIPLE INTEGRALS .................................................................................................................... 2

1.1 DEFINITION ...................................................................................................................................... 2 1.2 PROPRIETIES.................................................................................................................................... 2 Remark...................................................................................................................................................... 3 1.2.1 Computation techniques .............................................................................................................. 3 1.3 VOLUME EVALUATION ..................................................................................................................... 5 1.3.1 Any domain .................................................................................................................................. 5 1.3.2 Volume of domains with known base surface .............................................................................. 6 1.3.3 Volume of revolution.................................................................................................................... 7

2 CYLINDRICAL COORDINATES ................................................................................................ 7

3 SPHERICAL COORDINATES...................................................................................................... 8

4 MASS, CENTRE OF INERTIA, MOMENT OF INERTIA ...................................................... 10

4.1 MASS OF A SOLID........................................................................................................................... 10 4.1.1 Example ..................................................................................................................................... 10 4.1.2 Example ..................................................................................................................................... 11 4.2 CENTER OF INERTIA OF A SOLID..................................................................................................... 11 4.2.1 Example ..................................................................................................................................... 12 4.2.2 Example ..................................................................................................................................... 12 4.3 MOMENT OF INERTIA OF A SOLID................................................................................................... 13

C H A P T E R 8

Wissam KARAM

2222 //// 1 41 41 41 4

PART 1

1 Triple integrals

1.1 Definition

Let f be a continuous function in a 3D rectangular box B of R3 .

[ ] [ ] [ ]1 1 2 2 3 3, , ,B a b a b a b= × ×

The volume of B is given by:

( ) ( ) ( )1 2 2 3 31V b a b a b a= − × − × −

A partition P of B is the determined by partitions P1, P2, P3

1 1[ , ]a b , 2 2[ , ]a b and 3 3[ , ]a b .

This partitions B into 3D subrectangles, which we denote by S. Like it has been done for the double integral, we shall define:

( ) ( ) ( ), minSS

I P f f Vol S=∑

( ) ( ) ( ), maxSS

K P f f Vol S=∑

f being a bounded function on B. f is called integrable if there exists a unique number which greater than I P f( , ) and smaller than K P f( , ) . If this is the case, this number is called the integral of f and is denoted by:

( ), ,B B

f f x y z dxdydz=∫∫∫ ∫∫∫

1.2 Proprieties

The same theorems of “double integral” chapter are still valid here. We repeat them:

( )B B B

f g f g+ = +∫∫∫ ∫∫∫ ∫∫∫ ; B B

kf k f=∫∫∫ ∫∫∫

Let B be a 3D rectangular box, and let f be a function defined on B, bounded and continuous except possibly at the points lying on a finite number of smooth surfaces. Then f is integrable on B.

If A denotes a 3D region and f is a function on A, we define:

*

*

( ) ( ) if

( ) 0 if

f X f X x A

f X x B A

= ∈

= ∈ −

C H A P T E R 8

Wissam KARAM

3333 //// 1 41 41 41 4

Then

*

A B

f f=∫∫∫ ∫∫∫

1 2

1 2 1 2; ; Vol( ) 0V V V

f f f V V V V V= + = ∪ ∩ =∫∫∫ ∫∫∫ ∫∫∫

Remark

We shall divide the 3D space into 8 octants. The annotation is done in the trigonometrical direction.

Figure 1

1.2.1 Computation techniques

C A S E O F A R E C T A N G U L AC A S E O F A R E C T A N G U L AC A S E O F A R E C T A N G U L AC A S E O F A R E C T A N G U L A R B O XR B O XR B O XR B O X

Let 1 1 2 2 3 3[ , ] [ , ] [ , ]B a b a b a b= × ×

⇒ 31 2

1 2 3

( , , )bb b

B a a a

f f x y z dz dy dx

=

∫∫∫ ∫ ∫ ∫

C H A P T E R 8

Wissam KARAM

4444 //// 1 41 41 41 4

Figure 2

G E N E R A L G E N E R A L G E N E R A L G E N E R A L C A S E C A S E C A S E C A S E

Let a function defined over a domain B as follows:

� a and b are 2 real numbers such that a < b

� 1( )g x and 2 ( )g x 2 functions defined over [a, b] such that ( )1 2( )g x g x≤

� 1( , )h x y and 2 ( , )h x y 2 functions defined over [ ] [ ]1 2, ( ), ( )a b g x g x× such

that: 1 2( , ) ( , )h x y h x y≤

Then 2 2

1 1

( ) ( , )

( ) ( , )

( , , )g x h x yb

B a g x h x y

f f x y z dz dy dx

=

∫∫∫ ∫ ∫ ∫

We can simply prove that:

2

1

( , )

( , )

( , , )h x y

B D h x y

f dxdy f x y z dz=∫∫∫ ∫∫ ∫

where D is the projection of B on the plane xOy.

C H A P T E R 8

Wissam KARAM

5555 //// 1 41 41 41 4

Figure 3

1.3 Volume evaluation 1.3.1 Any domain

when 1f = the triple integral of f over a domain D is the volume of D.

( )D

V D dxdydz= ∫∫∫

E XE XE XE X AAAA M P L E 1M P L E 1M P L E 1M P L E 1

Evaluate the volume of the tetrahedral defined by x y z x y z a> > > + + <0 0 0, , et .

MethodMethodMethodMethod #1111

0.2 0.4 0.6 0.8 1

0.2

0.4

0.6

0.8

1

0.2 0.4 0.6 0.8 1

0.2

0.4

0.6

0.8

1

Figure 4

C H A P T E R 8

Wissam KARAM

6666 //// 1 41 41 41 4

( )3

00 0 6

a a xa x y

A D

aV dxdydz dxdy dz dx a x y dy

−− −

= = = − − =∫∫∫ ∫∫ ∫ ∫ ∫

MethodMethodMethodMethod #2222

( )3

2

00 0 0

1

2 6

a y za a za

A

aV dxdydz dz dy dx a z dz

− −−

= = = − =∫∫∫ ∫ ∫ ∫ ∫

We note that ( )0

aV A z dz= ∫ , where ( )A z is the area of the triangle at the altitude z.

1.3.2 Volume of domains with known base surface The volume of a solid that’s section between the planes z = a and z = b have an area

)(zA , is:

∫=b

adzzAV )(

E XE XE XE X AAAA M P L EM P L EM P L EM P L E

Figure 5

Evaluate the volume of the pyramid having a square base formed by the following points {1,1,1} ; {1,-1,1} ; {-1,-1,1} ;{1,-1,1}

C H A P T E R 8

Wissam KARAM

7777 //// 1 41 41 41 4

The base is at an altitude z=1 and it’s area equals 2×2=4. If we cut this pyramid with a plane parallel to xOy, having an altitude z, the section is also square and its area is ( ) 24A z z=

The pyramid volume is therefore: ( )1 1 2

0 0

44

3V A z dz z dz= = =∫ ∫

1.3.3 Volume of revolution

The volume obtained by turning the curve {z=f(x) a < x < b}, around the x axis is:

∫=b

adxxfV 2)(π

E XE XE XE X AAAA M P L EM P L EM P L EM P L E

The volume obtained by turning the curve sin ; 0z x x π= ≤ ≤ around the x axis.

0

1

2

3

x

-1

-0.5

0

0.5

1

y

-1

-0.5

0

0.5

1

z

0

1

2

3

x

Figure 6

0

1

2

3 -1

-0.5

0

0.5

1-1

-0.5

0

0.5

1

0

1

2

3 Figure 7

2

2 2

0( ) sin

2

b

aV f x dx xdx

π ππ π= = =∫ ∫

2 Cylindrical coordinates

A point M of the 3D space can be defined by cylindrical coordinates as follows:

0.5 1 1.5 2 2.5 3x

0.20.40.60.81

z

C H A P T E R 8

Wissam KARAM

8888 //// 1 41 41 41 4

t r

x

y

z

Figure 8

cos ; sin ; x r y r z zθ θ= = =

The Jacobean of this variables substitution is:

r z

r z

r z

x x x

J y y y r

z z z

θ

θ

θ

′ ′ ′′ ′ ′= =′ ′ ′

3 Spherical coordinates

q

j

r

xy

z

Figure 9

A point M of the 3D space can be defined by spherical coordinates as follows:

sin cos ; sin sin ; z= cosx yρ ϕ θ ρ ϕ θ ρ ϕ= =

This coordinates are given for:

0 ; 0 ; 0 2ρ ϕ π θ π≤ ≤ ≤ ≤ ≤

C H A P T E R 8

Wissam KARAM

9999 //// 1 41 41 41 4

The Jacobean of this variables substitution is:

2 sin

x x x

J y y y

z z z

ρ θ ϕ

ρ θ ϕ

ρ θ ϕ

ρ ϕ′ ′ ′′ ′ ′= =′ ′ ′

E XE XE XE X AAAA M P L E 3M P L E 3M P L E 3M P L E 3

Evaluate the volume between the cone 2 2 2z x y= + and the sphere 2 2 2x y z z+ + = .

S O L U T I O N

This sphere is centered at the point 1

0,0,2

and have a radius of 1

2;

2

2 2 2 2 2 1 1

2 4x y z z x y z

+ + = ⇒ + + − =

-0.5 -0.25 0 0.25 0.5

-0.5

-0.2500.250.5

0

0.25

0.5

0.75

1

-0.5-0.25 0 0.25 0.5

-0.5-0.25

00.25

0.5

0

0.25

0.5

0.75

1

0

0.25

0.5

Figure 10

In spherical coordinated the sphere becomes:

2 2 2 2 cos cosx y z z ρ ρ ϕ ρ ϕ+ + = ⇒ = ⇒ =

So

*

2 sinD D

V dxdydz d d dρ ϕ ρ ϕ θ= =∫∫∫ ∫∫∫

cos2 42

0 0 0

sin8

V d d d

πϕπ πθ ϕ ϕ ρ ρ= =∫ ∫ ∫

C H A P T E R 8

Wissam KARAM

1 01 01 01 0 //// 1 41 41 41 4

PART 2

4 Mass, Centre of inertia, Moment of inertia

4.1 Mass of a solid

We shall call a solid every pair ( ),S ρ where S is a part of 3R and : S Rρ +→ a

continuous application called space density of the solid ( ),S ρ .

We shall call mass of a solid ( ),S ρ the real number mmmm defined by

( ), ,S

m x y z dxdydzρ= ∫∫∫ , where ( ), ,x y z covers S

4.1.1 Example

Evaluate the mass of the solid having the following density 3rρ = and defined by:

; cos ; 03 3

r z rπ πθ θ− ≤ ≤ = ≤ ≤

S T U D Y O F T H E S O L I D ES T U D Y O F T H E S O L I D ES T U D Y O F T H E S O L I D ES T U D Y O F T H E S O L I D E ::::

� 2 2 0 0z r z x y≤ ≤ ⇔ ≤ ≤ + ⇒ the cone is its upper frontier

�

2

2 2 2 21 1 cos cos 0

2 4r r r x y x x yθ θ = ⇒ = ⇒ + − = ⇒ − + =

⇒ the

circular part 2

21 1

2 4x y − + =

with 3 3

π πθ− ≤ ≤ is its lateral frontier

0 0.2 0.4

-0.2

00.2

0

0.2

0.4

0

0 0.2 0.4

-0.2

00.2

0

0.2

0.4

0

Figure 11

C H A P T E R 8

Wissam KARAM

1 11 11 11 1 //// 1 41 41 41 4

0.2 0.4 0.6 0.8 1

-0.4

-0.2

0.2

0.4

0.2 0.4 0.6 0.8 1

-0.4

-0.2

0.2

0.4

Figure 12

S O L U T I O NS O L U T I O NS O L U T I O NS O L U T I O N

The masse of M is given by:

cos cos3 3 33 4

0 0 03 3 3

33 3 cos

4

r

M d rdr rdz d r dr d

π π πθ θ

π π πθ θ θ θ

− − −

= = =∫ ∫ ∫ ∫ ∫ ∫

But 4 3 cos2 cos4cos

8 2 8

θ θθ = + +

⇒

3

0

3 sin 4 3 2 1 4 3 7 33 2sin 2 2sin sin

16 4 16 3 4 3 16 8M

π

θ π πθ θ π π = + + = + + = +

4.1.2 Example

Evaluate the mass of a ball S, centered at the origin and having a radius R, the density being given by

( ) 2 2 2, ,x y z x y zρ = + +

Using spherical coordinates:

( )2 4

3 4

0 0 0

, , sin 2 24

R

S

Rm x y z dxdydz d d d R

π π

ρ θ ϕ ϕ ρ ρ π π= = = × × =∫∫∫ ∫ ∫ ∫

4.2 Center of inertia of a solid

The center of inertia of a solid ( ),S ρ is the point G of 3R defined by:

C H A P T E R 8

Wissam KARAM

1 21 21 21 2 //// 1 41 41 41 4

( )

( )

( )

1, ,

1, ,

1, ,

GS

GS

GS

x x x y z dxdydzm

y y x y z dxdydzm

z z x y z dxdydzm

ρ

ρ

ρ

= =

=

∫∫∫

∫∫∫

∫∫∫

where ( ), ,x y z covers S and m is the mass of ( ),S ρ .

4.2.1 Example

Find the center of gravity of the hemisphere:

z a x y= − −2 2 2

having a constant density of 1. we can simply verify that, due to the symmetry, x y= = 0.

( )

( )

2 2 2

0

22 2 2

2 22 2 5

0 0

1 3

2

3

4

3 4

4 5

a x y

GD D

D

a

z zdxdydz dxdy zdzm

a x y dxdy

a r drd aπ

π

π

θπ

− −

= =

= − − =

= − =

∫∫∫ ∫∫ ∫

∫∫

∫ ∫

4.2.2 Example

Find the center of gravity G of the homogeneous solid ( ),S ρ where S is a part of

the sphere, centered at the origin and having a radius of 1. It is defined in spherical coordinates by:

;0 ;0 14 4

π πθ ϕ π ρ− ≤ ≤ ≤ ≤ ≤ ≤

C H A P T E R 8

Wissam KARAM

1 31 31 31 3 //// 1 41 41 41 4

00.250.50.751

-0.5

0

0.5 -1

-0.5

0

0.5

1

Figure 13

It has the form of a slice of an orange.

S O L U T I O NS O L U T I O NS O L U T I O NS O L U T I O N

Using the proportionality, the masse m is given by:

4 232 4 3S

m dxdydzπ π πρρ ρ

π= = × =∫∫∫

Due to symmetrical reasons 0G Gy z= =

Then: ( )( )12 34

0 04

3

3cos

3 2

8

GS

x xdxdydz

d sin d dπ

π

π

ρπρ

θ θ ϕ ϕ ρ ρπ −

=

=

=

∫∫∫

∫ ∫ ∫

4.3 Moment of inertia of a solid

Let H be a point of a line or a plane of 3R ; for every point M of

3R , we denote ( ),d M H

as the distance from M to H. The moment of inertia of a solid ( ),S ρ with respect to H is the real number HI defined

by:

( ) ( )( )2,H

S

I M d M H dxdydzρ= ∫∫∫

C H A P T E R 8

Wissam KARAM

1 41 41 41 4 //// 1 41 41 41 4

where ( ), ,M x y z covers S.

IN PARTICUAR:

The moments of inertia of a solid with respect to the 3 main axis (x, y , z) are given by:

I y z x y z dxdydz

I z x x y z dxdydz

I x y x y z dxdydz

x

A

y

A

z

A

= +

= +

= +

∫∫∫

∫∫∫

∫∫∫

( ) ( , , )

( ) ( , , )

( ) ( , , )

2 2

2 2

2 2

ρ

ρ

ρ