§ 12.1 (15.1) Double Integrals over Rectangles (矩形區域上的二重積分) ( )

§ 12.1 (15.1) Double Integrals over Rectangles (矩形區域上的二重積分) ( )

y = f x

*

1

( ) lim ( )

b n

a n i

i

f x dx f x x

→∞ =

= ∑ ∆

∫ ^{x b} n ^a

 ∆ = − 

 

 

For a function f (x, y) , where f is continuous and f (x, y) 0 on the rectangle R = = {(x, y) | a x b, c y d}.

Let S be the solid

lying below the surface z = f (x, y) and above the rectangle R, i.e.

We want to find the volume V of the solid S.

[ , ] [ , ] a b × c d _{≤ ≤ ≤ ≤}

≥

{ ^{( , , )}

⁰ ( , ), ( , ) }

S = x y z ∈ ℝ ≤ ≤ z f x y x y ∈ R

* *

1 1

( , )

m n

ij ij

i j

V f x y A

= =

≈ ∑∑ ∆

( ) ( )

, where b a and d c

A x y x y

m n

− −

∆ = ∆ ∆ ∆ = ∆ =

Def

The double integral

of a function f of two variables over a rectangle R is

if the limit exits.

Def

If f (x, y) 0 on the rectangle R = {(x, y) | a x b, c y d}.

* *

, 1 1

( , ) lim ( , )

m n

ij ij

m n i j

R

f x y dA f x y A

→∞ = =

= ∑∑ ∆

∫∫

≥ ≤ ≤ ≤ ≤

If f (x, y) 0 on the rectangle R = {(x, y) | a x b, c y d}.

The volume V of the solid lying below the surface z = f (x, y) and above the rectangle R is

≥ ≤ ≤ ≤ ≤

( , )

R

V = ∫∫ f x y dA

1 1

* *

( , )

m n

i j

ij ij

f x y A

= =

∑∑ ∆

The sum is called a double Riemann sum

(

)

, 1 1

lim 16 ( ) 2( )

m n

ij ij

m n i j

x y A

→∞ = =

= ∑∑ − − ∆

2 2

( , ) 16 2

f x y = − x − y

{ }

[0, 2] [0, 2] ( , ) 0 x y x 2, 0 y 2

= × = ≤ ≤ ≤ ≤

R

( ¹⁶

²

)

R

V = ∫∫ − x − y dA

2 2

, where and

A x y x y

m n

∆ = ∆ ∆ ∆ = ∆ =

Properties of Double Integrals (二重積分的性質)

[ ]

(1) ( , ) ( , ) ( , ) ( , )

R R R

f x y + g x y dA = f x y dA + g x y dA

∫∫ ∫∫ ∫∫

(2) ( , ) ( , ) , where is a constant

R R

cf x y dA = c f x y dA c

∫∫ ∫∫

(3) If ( , ) ( , ) on , then ( , ) ( , )

R R

f x y ≥ g x y R ∫∫ f x y dA ≥ ∫∫ g x y dA

§(15.2)Iterated Integrals (逐次積分) ( )

( , )

A x = ∫

f x y dy ( partial integration with respect to y ) ( )

( , )

B y = ∫

f x y dx ( partial integration with respect to x )

The integrals

( , ) ( , )

b d b d

f x y dydx =    f x y dy dx   

∫ ∫ ∫ ∫

and

are called iterated integrals.

( , ) ( , )

a c

f x y dydx =

  

f x y dy dx   

∫ ∫ ∫ ∫

( , ) ( , )

d b d b

c a

f x y dxdy =

  

f x y dx dy   

∫ ∫ ∫ ∫

[Ex4][Ex1]

Evaluate the iterated integrals.

(a) (b)

[Sol]:

(a)

3 2 2

0 1

x y dy dx

∫ ∫ ∫ ∫

x y dxdy

3 2 3 3 3

2

0 1 0 0

2 2 2 2 2 2

0 1

2

1

1 2

2 1 x y 2

x y dy dx =    dy    dx =   x y   dx =    x − x   dx

  

 

∫ ∫ ∫ ∫ ∫ ∫

3 2 3

1

27 3 x d x x

= ∫ = =

(b)

3 0

2

0

1 27

2 2

3

2 x d x x

= ∫ = =

2 3 2 3

3 2 3

0 0

2 2

2

1 0 1 1 1

1

3 9

x y dxdy =    x y dx    dy =     x y     dx = y dy

∫ ∫ ∫ ∫ ∫ ∫

2 2

1

9 27

2 2

= y =

Fubini’s Theorem

If f (x, y) is continuous on the rectangle R = {(x, y) | a x b, c y d}, then

More generally, this is true if we assume that f is bounded on R, f is discontinuous only on a finite

number of smooth curves, and the iterated integrals exist.

≤ ≤ ≤ ≤

( , )

( , )

( , )

a c c a

R

f x y dA = f x y dydx = f x y dxdy

∫∫ ∫ ∫ ∫ ∫

If f (x, y) 0 , ≥

, where

( ) ( )

( , )

b

A x A x f x

V = ∫ d x = ∫ y dy

So,

, where

( ) ( ) ( , )

a

A x A x

f x

V = ∫ d x = ∫ y dy

( ) ( , )

( , )

a a

d c b

R

f x y dA = V = A x dx = f x y dy dx

∫ ∫ ∫ ∫ ∫

Similarly ,

So,

, where

( ) ( )

( , )

d

c

B y B y

f x

V = ∫ d y = ∫ y dx

( , )

( )

( , )

c c

R

b

B y

f x y dx

f x y dA = V = dy = dy

∫ ∫ ∫

∫ ∫

B(y)

[Ex5] [Ex2]

Evaluate ^{( 3}

) , where

R

x − y dA

∫∫

[Sol]:

[Ex6] [Ex3]

2 2

0

2

0

( 7) 7 12

2

x − dx = x − x = −

= ∫

{( , ) | 0 2, 1 2}

R = x y ≤ ≤ x ≤ ≤ y

( ³

)

( ³

)

(

)

R

dA dy dx dx

x − y = x − y = xy y −

∫∫ ∫ ∫ ∫

[Ex6] [Ex3]

Evaluate sin , where R = .

R

y xydA

∫∫ [1, 2] [0, ] × π

[Sol]:

[ ]

2 2

0 1 0 =1

sin sin cos

x R

y xydA =

y xy dx dy =

− xy

dy

∫∫ ∫ ∫ ∫

0^π

( cos 2 − y + cos ) y dy

= ∫

0

1 sin 2 sin 0

2 y y

π

− + =

 

=  

 

[Ex7] [Ex4]

Find the volume of the solid S that is bounded by the elliptic paraboloid

the planes x = 2, y = 2, and the three coordinate planes

.

2 2

2 16

x + y + = z [Sol]:

2

2

1

16 2

3

x

x x y x dy

=

− −

 

=  

 

∫

( ¹⁶

²

)

R

V = ∫∫ − x − y dA

( )

2 2

0 0

2 2

16 x 2 y ^dx dy

= ∫ ∫ − −

0

0

2 2

0

3 2

0

16 2

3 88 4

3

88 4

3 3

48

x

x x y x dy

y dy

y y

=

− −

−

−

=  

 

 

=  

 

 

=  

 

=

∫

∫

[ ] [ ]

( ) ( )

( )

( ) , where , ,

a c

R

g x h y dA = g x dx h y dy R = a b × c d

∫∫ ∫ ∫

[Ex8] [Ex5]

Evaluate sin cos , where R = .

R

x y dA

∫∫ [0, ] [0, ] ^π ₂ × ^π ₂

2 2

0 0

sin cos sin cos

R

x ydA =

x y dxdy

∫∫ ∫ ∫

[Sol]:

R

⁼ ( ^∫

^{sin xdx} )( ^∫

^{cos y dy} )

( ^{− cos x}

)( ^{sin y}

)

=

= ⋅ 1 1

= 1

Type Ι : D

has the form:

D

= {(x, y)| a x b and g

(x) y g

(x)}.

Type Π : D

has the form:

D

= {(x, y)| c y ≤ ≤ d and h

(y) x ≤ ≤ h

(y)} .

≤ ≤ ≤ ≤

§12.2 (15.3) Double Integrals over General Regions

Theorem

(1) If f (x, y) is continuous on the typeΙ region D

, then

(2) If f (x, y) is continuous on the type Π region D

, then

2

1 1

( )

( , )

( , )

a g x D

f x y dA = f x y dydx

∫∫ ∫ ∫

2

1 2

( ) ( )

( , )

( , )

c h x D

f x y dA = f x y dxdy

∫∫ ∫ ∫

[Ex1][Ex1]

Evaluate , where D is the region bounded by the parabolas of y = 2x

and y = 1+ x

.

[Sol]:

2

2 2

2

2 1,1 {( , ) | 1 1 2 1 }

1 .

y x

x D x y x and x y x

y x ^⇒

 =

= − ∴ = − ≤ ≤ ≤ ≤ +

 = +



∵

( ² )

D

x + y dA

∫∫

( )

Therefore 2

D

x + y dA

∫∫

4 3 2 1

5

1

3 2 32

4 3 2 15

5

x x x

x x

−

− ⋅ − + ⋅ + +

= =

D

( )

2

2 1 1

1 2 ^x

2

x

x y dy dx

+

= ∫ ∫

+

( )

1

1 2

2 ^{y x}

y x

d x

x y y

− =

= ∫ +

( ) ( ) ( ) ( )

1

1

2 2

2 2 2 2

1 1 2 2 dx

x x x x x x

−

 

= ∫   + + + − −  

( )

1 1

4 3 2

3 x x 2 x x 1 dx

= ∫

− − + + +

[Ex2][Ex2]

Find the volume of the solid S that lies under the paraboloid z = x

+ y

and above the region D in the xy-plane bounded by the line y = 2x and the parabola y = x

. [Sol 1]:

2

2 0, 2

y x

y x ^⇒ x

 =

 = =

∵ 

(

)

(

)

Therefore ∫∫ x + y dA = ∫ ∫

x + y dy dx = ⋅⋅⋅

{( , ) | 0 2

2 }.

D x y x and x y x

∴ = ≤ ≤ ≤ ≤

[Sol 2]:

( )

( )

Therefore

x D

dA dy dx

x + y = x + y = ⋅⋅⋅

∫∫ ∫ ∫

2

2 0, 2 0, 4

y x

x y

y x ^⇒

 =

= ⇒ =

 =

∵ 

(

)

(

)

Therefore

x D

dA dy dx

x + y = x + y = ⋅⋅⋅

∫∫ ∫ ∫

{( , ) | 0 4 1 }.

D x y y and 2 y x y

∴ = ≤ ≤ ≤ ≤

[Ex3][Ex3]

Evaluate , where D is the region bounded by the line of y = x− 1, and the parabola y

= 2x + 6.

D

∫∫ xydA

[Sol 1]:

2

1 2, 4

2 6

y x

y x ^⇒ y

 = −

 = + = −

∵ 

1 2

2

4 1

2 3

Therefore

y D

xydA

xy dx dy

− −

= = ⋅⋅⋅

∫∫ ∫ ∫

1

{( , ) | 2 4 2 3 1} .

D x y y and y x y

∴ = − ≤ ≤ − ≤ ≤ +

[Sol 2]:

2

1 2, 4 3,5

2 6

y x

y x

y x ^⇒

 = −

= − ⇒ = −

 = +

∵ 

2

2, 4 3,5

2 6 y x

y x ^⇒ = − ⇒ = −

 = +

∵ 

1 2 6 5 2 6

3 2 6 1 1

So

x x

D

xydA

xy dy dx

xy dy dx

− − + − −

= + = ⋅⋅⋅

∫∫ ∫ ∫ ∫ ∫

2 6 2 6 when 3 1 or

( , ) .

1 2 6 when 1 5

x y x x

D x y

x y x x

 

− + ≤ ≤ + − ≤ ≤ −

 

∴ =  

− ≤ ≤ + − ≤ ≤

 

 

[Ex4] [Ex4]

Find the volume of the solid S bounded by the planes x +2 y+ z=2, x = 2y, x = 0 and z = 0.

[Sol]:

{( , ) | 0 1 1 }

2 2

x x

D = x y ≤ ≤ x and ≤ ≤ − y

( )

0

2 2

So the volume = 2 2 (2 2 )

D

x

V ∫∫ − − x y dA = ∫ ∫

− − x y dy dx = ⋅⋅⋅

[Ex5][Ex5] [Evaluating by reversing

the order of integration

] Evaluate the iterated integral

.

0 1

sin ( )

x

y dy dx

∫ ∫

[Sol]:

{( , ) | 0 1 and 1}.

{( , ) | 0 1 and 0 }.

D x y x x y

x y y x y

= ≤ ≤ ≤ ≤

= ≤ ≤ ≤ ≤

∵

1 1 1 1

2 2 2 2

0 0 0 0 0

2

1 0 1

0

sin sin ( ) sin sin

1 1

2 cos 2

( ) ( ) ( )

( ) (1 cos1)

y y

x

y dy dx y dx dy x y dy y y dy

− y

∴ = = =

= = −

∫ ∫ ∫ ∫ ∫ ∫

[Ex] (Evaluating by reversing the order of integration) Evaluate the iterated integral

.

0

1

y

x + dxdy

∫ ∫

[Sol]:

2

{( , ) | 0 1 and 1}

{( , ) | 0 1 and 0 }

D x y y y x

x y x y x

= ≤ ≤ ≤ ≤

= ≤ ≤ ≤ ≤

∵

1 1 1 2

0 0 0

3 3

1

1

y

x dx dy x dydx

∴ ∫ ∫ ∫ ∫

+ = ∫ ∫ ∫ ∫

+

1 0

1 2 3

0

3

2 3

0

3 1 2

0

1

1 2 1

3 3

2 2 1

9

1

( )

2

x

dx x x

x y x

+ dx

⋅ +

−

= +

=

=

 

=  

∫

∫

Properties of Double Integrals

[ ]

(1) ( , ) ( , ) ( , ) ( , )

D D D

f x y + g x y dA = f x y dA + g x y dA

∫∫ ∫∫ ∫∫

where is a

(2) ( , ) ( , ) , constant

D D

cf x y dA = c f x y d A c

∫∫ ∫∫

If on , t

(3) ( , ) ( , ) hen ( , ) ( , )

D D

f x y ≥ g x y D ∫∫ f x y dA ≥ ∫∫ g x y d A

1

1 2 2

If , where and don't overlap except perhaps on th

(4) D = D ∪ D D D eir boundaries,

(除了邊界外都沒有重疊)

1 1

1 2 2

2

( ,

then ) ( , ) ( , )

D D D

f x y dA = f x y dA + f x y dA

∫∫ ∫∫ ∫∫

∪

D₁

: Type Ι

: Type Π

(除了邊界外都沒有重疊)

( ) ^{, where} ( ) ^is

(5) 1 the area of .

D

A D D

dA = A D

∫∫

If for all ( , ) in , th

(6) ( , )

( ) ( , ) ( )

en

D

x y D

m f x y M

m A D f x y dA M A D

≤ ≤

⋅ ≤ ∫∫ ≤ ⋅

[Ex]

Find area of the region D in the xy-plane bounded by the line y = 2 x and the parabola y = x

.

[Sol 1]:

2 2

2 0, 2 {( , ) | 0 2 and 2 }

y x

x D x y x x y x

y x

 =

⇒ = ∴ = ≤ ≤ ≤ ≤

 =

∵ 

2

2 2

1

1

( )

A D = ∫∫ dA = ∫ ∫

dydx

[Sol 2]:

2

0

2 2 2 2 3

0 0

2 2

0

1 4

3 3

(2 ) ( ) =

x D

x

y

dx x x dx x x

= =   −   − =

∫∫ ∫ ∫

∫ ∫

2 2 3

0

2

2 2

0

1 4

3 3

2 0, 2

( ) (2 ) ( )

A dx x x

y x

y x x

D x x

⇒

∴ = = − =

 =

 = =



 − 

 

∫

[Ex6][Ex6]

Estimate the integral , where D is the disk with center the origin and radius 2.

sin cos D

x y

e dA

∫∫

[Sol]:

1 sin cos 1

2

1 sin cos

Since 1 sin 1 and 1 cos 1 , we have 1 sin cos 1 and therefore

( ) (2) 4

( ) ( )

x y

x y

x x x y

e e e e

A D

e A D e dA e A D

π π

−

−

− ≤ ≤ − ≤ ≤ − ≤ ≤

≤ ≤ =

= =

∴ ⋅ ≤ ∫∫ ≤ ⋅

∵

1 sin cos

1 sin cos

sin cos

( ) ( )

4 4

4 4

x y D

x y D

x y D

e A D e dA e A D

e e dA e

e dA e

e

π π

π π

−

⇒

⇒

∴ ⋅ ≤ ≤ ⋅

⋅ ≤ ≤ ⋅

≤ ≤

∫∫

∫∫

∫∫

§ 12.3 (15.4) Double Integrals in Polar Coordinates (極座標表示二重積分)

θ

r

i

( , ) ( , ) P r

θ

y

O

x

2 2 2

r = x + y cos x = r θ

sin y = r θ

tan y

θ = x

A polar rectangle is the region A polar rectangle is the region

{( , ) | , } R = r θ a ≤ ≤ r b α θ ≤ ≤ β

Theorem

If f (x, y) is continuous on a polar rectangle ,

where , then _{0 ≤ β α − ≤ 2 π}

( , )

( cos , sin )

a R

f x y dA

f r r r drd

α

θ θ θ

∫∫ = ∫ ∫

{( , ) | , }

R = r θ a ≤ ≤ r b α θ ≤ ≤ β

[Ex1][Ex1]

Evaluate , where R is the region in the upper half-plane bounded by the circles x

+ y

= 1 and x

+ y

= 4.

(3 4

)

R

x + y dA

∫∫

[Sol]:

2 2

{( , ) | 0 and 1 4 } D = x y y ≥ ≤ x + y ≤

∵

{( , ) | 1 r θ r 2 and 0 θ π }

= ≤ ≤ ≤ ≤

x²

+

1

x²

+

4

( )

2 2 2 2

0 1

(3 4

15 2

) 3 cos 4 sin

R

x y dA

r θ r θ rdrd θ

π

∴ + = +

= ⋅⋅⋅

=

∫∫ ∫ ∫

[Ex] Evaluate .

2 2 2

3 9

( )

3 0

x x y

e dydx

− − +

∫ ∫

[Sol]:

{( , ) | 0 9

and 3 3 }

D = x y ≤ ≤ y − x − ≤ ≤ x

∵

{( , ) | 0 r θ r 3 and 0 θ π }

= ≤ ≤ ≤ ≤

( ) ( )

2 2 2 2

3 9 3

3 0 0 0

x x y r

e dydx

e r d d r θ

− − + −

∴ ∫ ∫

= ∫ ∫

9 2

y = −x

( ) ( )

( )

( )

2

2

3

0 0

3

0 9

1 2

2 1

r

r

d e dr

e e

π

r

π

θ

π

−

−

−

−

=

=

= −

∫ ∫

[Ex2][Ex2]

Find the volume of the solid S bounded by the planes z =0 and the paraboloid z = 1− x

− ^y

[Sol]:

2 2

{( , ) | 1 }

D x y x y

π

∴ = + ≤

= ≤ ≤ ≤ ≤

2 2

2 2

1 1 = 0

0

z x y

x y

z ^⇒

 = − −

− −

 =



∵

= {( , ) | 0 r θ ≤ ≤ r 1 and 0 ≤ ≤ θ 2 }. π

( )

( )

2 1

0 0

2 2

2

So = 1

1

2

D

V dA

drd

x y r r

π

θ

π

− −

= −

= ⋅⋅⋅

=

∫∫

∫ ∫

Theorem

If f (x, y) is continuous on a polar region of the form , then

1 2

{( , ) | , ( ) ( ) }

D = r θ α θ ≤ ≤ β h θ ≤ ≤ r h θ

2

1

( ) ( )

( , )

( cos , sin )

h D

f x y dA

f r r r drd

α θ

θ θ θ

∫∫ = ∫ ∫

[Ex3]Use a double integral to find the area enclosed by one loop of the four-leaved rose r = cos2θ

[Sol]:

[Sol]:

{( , ) | 4 4 , 0 cos 2 } D = r θ − π ≤ ≤ θ π ≤ ≤ r θ

∵

cos 2 2

0

4 cos 2 4

4 0 4

1

( ) = 1

D

A D dA r d rd r d

π θ π θ

π

θ

θ

− −

∴ ∫∫ = ∫ ∫ = ∫      

( )

4 2 4

4 4

1 1

cos 2 1 cos 4

2

d 4

d

π

θ θ

θ θ

− −

+

= ∫ = ∫

4

4

1 4

1 sin 4

4 8

π

π

θ θ π

−

+ =

 

=  

 

[Ex3][Ex4]

Find the volume of the solid that lies under the paraboloid z = x

+y

above the xy-plane, and inside the cylinder

x

+ y

= 2x.

[Sol]:

2 2 2 2

2 ( i.e. ( 1) 1 ) is a circle centered (1,0) x + y = x x − + y =

2 2 2

In polar coordinates, x + y = 2 becomes x r = 2 cos , or r θ r = 2 cos θ {( , ) | 2 2 , 0 2 cos }

D = r θ − π ≤ ≤ θ π ≤ ≤ r θ

∵

(

)

( )

=

V x y dA

r r drd

π

θ

∴ ∫∫ + = ∫ ∫

π

∫∫ ∫ ∫

2 4 2cos 2 4

2 0 2

1 4 cos

4 r d d

π θ π

π

θ

θ θ

− −

 

=   =

 

∫ ∫

2 2

4

0 0

1 cos 2

8 cos 8

d 2 d

π

θ θ

^{ +} θ ^ θ

= =  

 

∫ ∫

( )

2 0

1 2

2

^ 1 2 cos 2 θ 1 cos 4 θ ^ d θ

=  + + + 

 

∫

2

0

3 1 3

2 8 2

2 sin 2 sin 4 2 3

2 2

π

θ π π

θ θ

    

=  + +  =    =

    

§12.5 (15.7) Triple Integrals (三重積分) If f is defined on a rectangular

box

{ ^{( , , ) , ,} }

B = x y z a ≤ ≤ x b c ≤ ≤ y d r ≤ ≤ z s Let ∆ = ∆ ∆ ∆ V x y z .

(

)

1 1 1

, ,

l m n

ijk ijk ijk

i j k

f x y z V

= = =

∑∑∑ ∆

The triple Riemann sum

is

1 1 1

i= j= k=

Def

The triple integral of a function f of three variables over the box B is

if the limit exits.

(

)

1 1 1

, ,

, ,

( , , ) lim

l m n

ijk ijk ijk

i j k

l m n B

f x y z V

f x y z dV

= = =

→∞

∆

= ∑∑∑

∫∫∫

Fubini’s Theorem for Triple Integrals

If f is continuous on the rectangular box B =

= , then

{ ^{( , , ) x y z a ≤ ≤ x b c , ≤ ≤ y d r , ≤ ≤ z s} }

[ , ] [ , ] [ , ] a b × c d × r s

( , , )

( , , )

( , , )

r a c r c a

B

f x y z dV = f x y z dydxdz = f x y z dxdydz

∫∫∫ ∫ ∫ ∫ ∫ ∫ ∫

[Ex1][Ex1]

Evaluate the triple integral , where B is the rectangular box given by ∫∫∫ xyz dV

Evaluate the triple integral , where B is the rectangular box given by

[Sol]:

B

xyz dV

∫∫∫

2 2 1 2

3 2 1 3 2 3 2

2 2

0 1 0 0 1 0 1

2

2

x

B x

x yz yz

xyz dV xyz dxdydz dydz dydz

=

− − −

=

= = =

∫∫∫ ∫ ∫ ∫ ∫ ∫ ∫ ∫

{ ^{( , , ) 0 1, 1 2, 0 3} }

B = x y z ≤ ≤ − ≤ ≤ x y ≤ ≤ z

2 3

2 2 2 3

3 3

0 0

1 0

3 27

4 4 4 4

y

y

y z z z

dz dz

=

=−

= ∫ = ∫ = =

(

)(

)(

)

0 1 0

3 1 27

OR 9

3 2 2 2 2 4

z y x

z dz ydy xdx

− −

     

 = =     = ⋅ ⋅ = 

   

     

 ∫ ∫ ∫ 

{

}

(1) If E = ( , , ) ( , ) x y z x y ∈ D , u x ( , ) y ≤ ≤ z u ( , ) x y ,

2

1

( , ) ( , )

then ( , , ) ( , , )

E

u x y x y

D u

z

f x y z dV =    f x y z d    dA

∫∫∫ ∫∫ ∫

{

}

(2) If E = ( , , ) ( , ) x y z y z ∈ D , u y ( , ) z ≤ ≤ x u ( , ) y z ,

2( , )

then ∫∫∫ f x y z dV ( , , ) = ∫∫ ∫   

f x y z d ( , , ) x    dA

then ( , , ) ( , , )

E y z

D u

x

f x y z dV =    f x y z d    dA

∫∫∫ ∫∫ ∫

{

}

(3) If E = ( , , ) ( , ) x y z x z ∈ D , u x ( , ) z ≤ ≤ y u ( , ) x z ,

2

1

( , ) ( , )

then ( , , ) ( , , )

E

u x z x z

D u

y

f x y z dV =    f x y z d    dA

∫∫∫ ∫∫ ∫

[Ex2][Ex2]

Evaluate , where E is the solid tetrahedron

bounded by the four planes x = 0, y = 0, z = 0, and x+y+z = 1.

∫∫∫ zdV

[Sol]:

{ ( , , ) ( , ) , 0 1 }

E = x y z x y ∈ D ≤ ≤ − − z x y

∵

{ ^{( , , ) 0 x y z x 1, 0 y 1 x , 0 z 1 x y} }

= ≤ ≤ ≤ ≤ − ≤ ≤ − −

1 1 1 1

0 0 0 0

1

24

x y x x y

E D

z dV 

z dz dA 

z dz dy dx

∴ ∫∫∫ = ∫∫ ∫     = ∫ ∫ ∫ = ⋅⋅⋅ =

Exercise

Evaluate , where E is the solid tetrahedron bounded by the coordinate planes and the planes 2x+y+z = 4.

Ans:16/3

∫∫∫ dV

[Ex3][Ex3]

Evaluate , where E is the region bounded by the paraboloid y = x

+ z

and the plane y = 4.

2 2

E

x + z dV

∫∫∫

[Sol 1]:

{ ( , , ) ( , )

,

}

E = x y z x y ∈ D

≤ ≤ z

∵

{ ^{( , , ) x y z 2 x 2, x}

^{y 4,}

^z

}

= − ≤ ≤ ≤ ≤ − ≤ ≤

2

2 1

2 2 2 2

y x

E D

x z dV

x z dz dA

− −

 

∴ + =  + 

 

∫∫∫ ∫∫ ∫

2

2 2

2 4

2 2

2

y x

x y x⁻

x z dz dy dx

− − −

= ∫ ∫ ∫ +

( It is hard to evaluate! )

[Sol 2]:

{ ( , , ) ( , )

,

4 }

E = x y z x z ∈ D x + z ≤ ≤ y

∵

2 2

3

2 2 4 2 2

x z

E D

x z dV x z dy dA

+

 

∴ ∫∫∫ + = ∫∫ ∫   +  

For the region D

, let x = r cos and θ z = r sin , then θ

{( , ) | 0 2 , 0 2}

D = r θ ≤ ≤ θ π ≤ ≤ r

3

2 2 2 2

(4 )

D

x z x z dA

 

= − − +

 

∫∫

3

{( , ) | 0 2 , 0 2}

D = r θ ≤ ≤ θ π ≤ ≤ r

3

2 2 2 2 2 2

So (4 )

E D

x + z dV =  − x − z x + z  dA

 

∫∫∫ ∫∫

( )

2 2

2

0^π 0

4 r r r dr d θ

= ∫ ∫ −

( )

2 2

2 4

0^π

d θ

4 r r dr

= ∫ ∫ −

3 5 2

0

4 128

2 3 5 15

r r π

π ^{ }

=  −  =

 

§ 12.6 (15.8) Triple Integrals in Cylindrical Coordinates

(圓柱座標或柱面座標)

2 2 2

cos

sin tan

r x y

x r y r y

z z x

z z θ

θ θ =

 = + 

 =  

 =  

  

 =  

  = 

{

}

1 2

If ( , , ) ( , ) , ( , ) ( , ) {( ,

w ) | , ( ) ( ) }

here u x y z u x y

D r h r

D

h E x y z x y

θ α θ β θ θ

≤ ≤

= ≤ ≤ ≤ ≤

= ∈

2

1

( , ) ( , )

( , , ) ( , ,

then

)

u x y

E D

f x y z dV =    f x y z dz dA   

∫∫∫ ∫∫ ∫

( )

2 2

1 1

( ) ( cos , sin )

( )^{h u}( cos , sin )^{r r}

cos , sin ,

h u r r

f r r z r dzdrd

β θ θ θ

α θ θ θ

θ θ θ

= ∫ ∫ ∫

[Ex3][Ex3]

Find the volume of the solid E that lies within the cylinder x

+ y

= 1, below the plane z = 4, and above the paraboloid z = 1− x

− y

.

,

{( , z ) | 0 2 , 0 1, 1 4 } E = r θ ≤ ≤ θ π ≤ ≤ r − r ≤ ≤ z

∵

[Sol 1]: (Use triple integral )

2

2 1 4 0 0 1

1

E r

V dV

r d zdrd θ

∴ = ∫∫∫ = ∫ ∫ ∫

2

7

(4-(1 r )) r drd

π

θ π

= ∫ ∫ − = ⋅⋅⋅ =

[Sol 2]: (Use double integral )

2 2

{( , ) | 1 } {( , ) | 0 1 and 0 2 }.

D x y x y r θ r θ π

∴ = + ≤ = ≤ ≤ ≤ ≤

0 0

(4-(1 )) r 2 r drd θ

= ∫ ∫ − = ⋅⋅⋅ =

(

) (

)

¹

⁷

= 4 1 = 4 (1 ) (4 ( ))

D

2

D D

V dA x y dA x y dA

r r drd θ π

= − =⋅⋅⋅ =

− − − ∫∫ − − − ∫ ∫ −

∫∫ ∫∫

∵

[Ex4][Ex4]

Evaluate

2

2 2 2

2 4 2

2 2

2 4

x

x x y

x y dz dy dx

−

− − − +

+

∫ ∫ ∫

[Sol]:

{( , r θ , ^z ) | 0 θ 2 , 0 π r 2, r z 2 }

= ≤ ≤ ≤ ≤ ≤ ≤

{ ^{( , , ) 2 2, 4 x}

^{4 x}

^{, x}

² }

E = x y z − ≤ ≤ x − − ≤ ≤ y − + y ≤ ≤ z

2

2 2 2

2 4 2

2 2 2 2

2 4

x

x x y

E

x y dz dy dx x y dV

−

− − − +

+ = +

∫ ∫ ∫ ∫∫∫

E

2 2 2 2 2

2 3

0 0 0 0

(2 )

r

r r dz dr d d r r dr

π π

θ θ

= ∫ ∫ ∫ = ∫ ∫ −

2

4 5

0

1 1

2 5

2 16

r r 5 π

π ^{ }

=   −   =

Exercise Evaluate , where E is the solid enclosed by the paraboloid cylinder , and the xy-plane . ^Ans:

e dV

∫∫∫

2 2

1

z = + x + y x

+ y

= 5 π ( e

− − e 5)

§ 12.7 (15.8) Triple Integrals in Spherical Coordinates (球面座標)

sin cos

sin sin cos

x y z

θ θ

ρ φ ρ φ

ρ φ

 =

 =

  =



2 2 2 2

( ρ = x + y + z )

( ρ = x + y + z )