f(x, y) over the region R

중앙대학교 건설환경플랜트공학과 교수

김 진 홍

- 10주차 강의 내용 -

10.3 Calculus Review : Double Integrals

 The double integral of

f(x, y) over the region R

is denoted by





R

f ( x , y ) d x dy

_or R

f ( x , y ) dA

 Double integrals have properties quite similar to those of definite integrals. For any functions

f

and

g

of (x, y), defined and continuous in a region

R

,



Rkfdxdy ^ k



R fdxdy ₍

k

; constant) (1)



R(f ^g)dxdy ^



R f dxdy^



Rgdxdy



R f dxdy ^



R f dxdy^



R f dxdy

1 2

Chap. 10 Vector Integral Calculus. Integral Theorems

Formula(1)

 If R is simply connected, then there exists at least one point (x₀, y₀) in

R

such that

(2) f x y dx dy f x y A

R ( , ) ( ₀, ₀)



^

where

A

is the area of

R

. This is called the mean value theorem

for

double integrals.

Evaluation of a double integral

Evaluation of Double Integrals by Two Successive Integrations

Double integrals over a region R may be evaluated by two successive integrations.

We may integrate first over y and then over x. Then the formula is

(3)



R ^

 

b a

x h

x

g f x y dy dx

dy dx y x

f ( , ) [ ^{( )} ( , ) ]

) (

Here y=g(x) and y=h(x) represent the boundary curve of R and, keeping x constant, we integrate f(x, y) over y from g(x) to h(x). The result is a function of x, and we integrate it from x=a to x=b.

Chap. 10 Vector Integral Calculus. Integral Theorems

Similarly, for integrating first over

x

and then

y

the formula is

(4)



R ^

 

d c

y q

y

p f x y dx dy

dy dx y x

f ( , ) [ ^{( )} ( , ) ]

) (

The boundary curve of

R

is now represented by x=p(y) and x=q(y). Treating y as a constant, we integrate

f(x,y)

over x from p(y) to q(y) and then the resulting function of

y

from y=c to y=d.

Ex. 1) Evaluate the integrals (a)

 

1²

3 2

0 x ydydx _(b)

 

0³

2 2

1 x ydxdy

(a) Regarding x as a constant,

2 2

1 2 2 2

1 2

2 3

2 x

x y dy y x

y

y

 



 









 ^

 

^



0^{3 }

2 2 1 3 2

0 2

27 2

3x dx dx

dy y x

(b)

 

^



_^



_^{ }



^_^{ }_^{ }



^





2 1 3

0 2 3

1 3

0 2 2 1 3

0 2 2

1 2

9 27 3 y dy ydy dy x

dx y x dy

dx y x

x

x

Chap. 10 Vector Integral Calculus. Integral Theorems

Ex. 2) Evaluate the integrals



0^



1²ysin(xy)dxdy



xy



dy y y dy

dy dx xy

ysin( ) cos( ) ^x_x ( cos2 cos )

0 2 0 1

2 1

0







 



 



^{   }

0 sin

2 2sin 1

0

 

 







y y Ex. 3) Evaluate the integrals

 

1^{2 }

2 2

0 (x 3y )dydx





1^{2 } 2 2

0 (x 3y )dydx ² xy y^{3 y}_{y 1}²dx

0 [  ] ^_





^



^{02 (x7)dx 12}

Application of Double Integrals

The area

A

of a region

R

in the

xy

-plane is given by the double integral





R

dx dy A

The volume V beneath the surface z = f(x,y) (>0) and above a region R in the

xy

-plane is





R

f x y dx dy

V ( , )

Chap. 10 Vector Integral Calculus. Integral Theorems

Double integral as volume

Ex. 4) Evaluate x y dA



R^{(  )2} over the region 2x2

y y1x².

The parabolas intersect when 2x² 1x², x1

The region R is, R {(x,y)/1 x 1, 2x²  y 1x²} bounded by parabolas and

A d y

R x



^{(  )2 x y dydx}



^{xy y}



^yy x^{x dx} x

x

2 2 2

2

1 2 1 2

1 1

2 1

1 ( 2 ) ^_^





   



  

15 ) 32 1 2

3

( ^{4 3 2}

1

1      





 x x x x dx

Ex. 5) Find the volume V beneath the paraboloid z x² y² and above the

2. x y  region bounded by line y  2x and parabola The region R is, R {(x,y)/0 x  2,x²  y 2x}

A d y x

V ^



R^{( 2 2) x y dydx}



^{x y y}



^yy x^xdx x

x

3 2 2 2

0

2 2 2

2 0

2(  )   /3 ^_ 2



  

x dx x

x x x

x )

3 ) ( 3

) 2 ) ( 2 ( (

3 2 2 2 3 2 2

0   





x dx x x

3 ) 14 ( 3

3 4

2 6

0   





_{21 5} ⁷₆ ²¹⁶₃₅

2

0 4 5

7  



 





 x x x

Chap. 10 Vector Integral Calculus. Integral Theorems

Ex. 6) Evaluate integral



0¹



¹sin ²

x y dydx

The region R is,R {0 x 1, x  y 1} R {0 y 1, 0 x  y}





0^{1 1}sin ²

x y dydx ^

 

0^{y y dxdy} 1 2

0 sin

dy y

x ^x_x^_^y



 ^{1 2} ₀

0 [ sin ] ¹ [ysiny²]dy



0



1 0 2)]

cos(

5 .

0 y



 0.5(1cos1)

Let f(x, y) be the density (=mass per unit area) of a distribution of mass in the

xy

-plane. Then the total mass M in

R

is





R

f x y dx dy

M ( , )

the center of gravity of the mass in

R

has the coordinate , where x, y





^

 R Ryf x y dxdy

y M dy

dx y x M xf

x 1 ( , )

) ,

1 ( _and

the moments of inertia and of the mass about the x- and

y-

axes, respectively, are ^x

I

I_y

dy dx y x f x I

dy dx y x f y

Ix ^



R ² ( , ) , y ^



R ² ( , )

and the polar moment of inertia

I

₀ about the origin of the mass is, dy

dx y x f y x I

I

I₀  x  y 



R( ²  ²) ( , )

Chap. 10 Vector Integral Calculus. Integral Theorems

Change of Variables in Double Integrals, Jacobian

For a definite integral the formula for the change from

x

to

u

is

(5)



^



^ du

du u dx x f dx x

b f

a ( ) ( ( ))

Here we assume that x = x(u) is continuous and has a continuous derivative.

The formula for a change of variables in double integrals from

x, y

to

u, v

is

; the absolute value of the Jacobian

(7) u

y v x v y u x v

y u

y v

x u

x v

u y J x







 







 







 







 

 

) , (

) , (

(6) dudv

v u

y v x

u y v u x f dy

dx y x

R f R



^{( , ) }



^{( ( , ), ( , )) }_⁽₍ ^,_{, )}⁾

Chap. 10 Vector Integral Calculus. Integral Theorems

) , (

) , (

v u

y x





Ex. 1) Change of Variables in a Double Integral

Evaluate the following double integral over the square

R

. dy

dx y

R(x²  ²)



Sol.) The shape of

R

;x y  u, x y  v

2 1 2 / 1 2 / 1

2 / 1 2 / 1 ) , (

) ,

( 

 



  v u

y J x

R

corresponds to the square 0≤u≤ 2, 0≤v≤2 .

3 8 2

)1 2(

) 1

( ^{2 2 2}

0 2

0 2

2  

 

 



R ^{x y dxdy u v dudv}

) 2(

), 1 2(

1 u v y u v

x   



Therefore,

Region R in Example 1

In case of the polar coordinates r and θ, and . Thenxrcos y rsin r r

r r

y

J x  

 

 









 sin cos

sin cos

) , (

) ,

( and

(8) f x y dxdy f r  r  rdrd

R R



^{( , ) }



^{* ( cos , sin )}

where R* is the region in the rθ-plane corresponding to R in the xy-plane.

Chap. 10 Vector Integral Calculus. Integral Theorems

Ex. 2) Evaluate x y dA, where

R

is the region in the upper half-plane

R(3 4 ²)



Sol.) The region R is,

bounded by circles x²  y² 1 and x² y² 4.

} 4 1

, 0 / ) ,

{(   ²  ² 

 x y y x y

R A d y

R x



^{(3 4 2)} 0^{ (3rcos 4r2sin2)rdrd} 2

1 



 

^{(1r2,0}

^

^

^

⁾





 

d dr r

r cos 4 sin ) 3

( ^{2 3 2}

0 2

1 



 



^{ }



^

 r³ r^{4 2 r}_{r 1}²d

0 cos  sin ^_





^



^{ 2}



^d

0 7cos 15sin









 

 d

 

  





0 ^{7cos 15}₂ ^{(1 cos2 )}









0

2 4 sin 15 2

sin 15

7   

 2 

15

Chap. 10 Vector Integral Calculus. Integral Theorems

Ex. 3) Evaluate dydx

y x

x

 

0² x ^ _{ } 8

2 2

2

5 1

Sol.) dydx

y x

x

 

0² x ^ _{ } 8

2 2

2

5

1 ^ 

 rdrd

r²

2 /

4 /

8

0 5

1



 

 ^ drd^

r r

2 2

/

4 /

8

0 5

2 2

1



 





^



 r d



^

 ^/2

4 /

8 0 2) 5 2 ln(

1 ^ 

 d





 ^/2

4

) /

5 ln 13 2(ln 1

5 ln13 8

 

) 2 0

, 8

(x y x²  x

Ex. 4) Double Integrals in Polar Coordinates. Center of Gravity. Moment of Inertia Let be the mass density in the region(Fig. below). Find the total mass, the center of gravity, and the moments of inertia .

1 ) , (x y  f

, 0

,I I I_{x y} Sol.) The total mass,

4 2

1

2 / 0 2

/ 0

1 0

 

 ^

  







^dxdy

 

^rdrd



^d

M R

The center of gravity,

 

 



 





3 cos 4

3 1 cos 4

4 ^/2

0 2

/

0 1

0  



 

^{r rdrd}



^d

x

 3

 4

y for reasons of symmetry The moments of inertia are







 ^

 r rdrd d

dxdy y Ix R

2 2 / 0 2

/ 0

1 0

2 2

2 sin

4

sin



1

 



^{ }



) 16 2 0

8 ( ) 1

2 cos 1

8 ( 1

2 / 0

π θ π

d

π θ













16

 

Iy for reasons of symmetry,

0 8





I_x I_y I

Example 4

Chap. 10 Vector Integral Calculus. Integral Theorems

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