北邮高等数学英文课件Lecture 11-1

A1 , A2 , , An . Choose a point ( xk , yk ) in each piece Ak , then
y
A
x
Sn f ( xk , yk )Ak .
k 1
16
n
Calculating Double Integrals Over Rectangles
on the xOy-plane and its density is a function of the point ( x , y ), f ( x , y ), then, how can we find the mass of this sheet metal? To find the mass, we suppose that f ( x , y )
( )
f ( x , y )d , f ( x , y )dA
( )
or
n
( )
f ( x , y )dxdy .
Thus,
( )
f ( x , y )d lim f ( x
A 0 k 1
k
, yk )Ak
As with functions of a single variable, the sums approach this limit no matter how the intervals [a , b] and [c , d ] that determine (σ) are partitioned, as long as the norms of the partitions both go to zero.
( ) ( ) ( ) ( )
( )
2. Additivity with respect to the domain of integration
Suppose that ( ) ( 1 ) ( 2 ) and ( 1 ),( 2 ) have no common part except for their
9
Geometric meaning of the double integral
Suppose that z f ( x , y ) 0,( x , y ) , then it can be think of a cylindrical body in three dimensional space.
17
Fubini’s Theorem for Calculating Double Integrals
Suppose that we wish to calculate the volume under the plane z 4 x y over the rectangular region ( ) : 0 x 2,0 y 1 on the xy-plane. If we denote the area of the cross-section at x as A( x ) , then the volume is
boundaries. Then
( )
( 1 )
( 2 )
f ( x , y )d
(
f ( x , y )d
1)
(

f ( x , y )d .
6
1)
Properties of Double Integral
3. Domination (1) f ( x , y )d 0 , if f ( x , y ) 0 on ( ).
denoted by k . Choose any point
Summation
Pk k and form the sum
பைடு நூலகம்
is
f ( P )
k 1 k n
n
k
.
If the limit lim f ( Pk ) k exist, where d 0 k 1 Precision n d max d ( k ) , we say that f is integrable
2 x
x
O
which is the area under the curve z 4 x y
l f ( x , y )d L .
( )
4. Mean Value Theorem Suppose that f C (( )) and ( ) is a closed bounded, and connected domain. Then there exists at least one point , ( ), such that
( ) : a x b, c y d .
Then we make a network of lines parallel to x- and y-axes. These lines divide (σ) into small pieces of area A xy . We number these in some order
V (4 x y )d
( )

as the integral
x2 x 0
A( x )dx .
z 4 x y
z
For each value of x, we may calculate A( x )
A( x )
y 1 y0
4
(4 x y )dy ,
is defined on a rectangular region Ω given by
: a x b, c y d .
We divide Ω into small pieces of area A and number them in some order A1 , A2 , , An .
Section 11.1
Concept and Properties of Double Integrals
School of Science, BUPT
1
Mass of a Thin Rectangular Sheet Metal A
k
Suppose a thin rectangular sheet metal lies
k 1
over the domain
( ).
4
Riemann, Bernhard (1826-1866), German mathematician
The Notation of the Double Integral
If function f is integral over the domain ( ), then the limit of Riemann sum is called the integral of the multivariable function f on the domain
2
Mass of a Thin Rectangular Sheet Metal A
k
We choose a point ( xk , yk ) in each piece
Ak and form the sum
Sn f ( xk , yk )Ak .
k 1 n
If f is continuous throughout Ω, then, as we refine the mesh (or two-dimensional partition) width to make the “norm” of each piece go to zero, we can expect that the sum should have a limit and the limit should be the mass of the thin rectangular steel
( )
(2) (3) (4)
( )
f ( x , y )d g ( x , y )d , if
( )
f ( x , y ) g ( x , y ) on ( ).
( )
f ( x , y )d
f ( x , y ) d
( )
If l f ( x , y ) L, ( x , y ) ( ), then
If f is continuous throughout (σ), then as we refined the mesh width to make both x and y go to zero, the sums approach a limit called the double integral of f over (σ). The notation for it is
Suppose f ( x , y ) and g( x , y ) are both integrable over the domain ( ), then 1. Linearity Property (1) (2)
kf ( x , y )d k f ( x , y )d , where k is a constant. f ( x , y ) g( x , y ) d f ( x , y )d g( x , y )d
( x k , yk )
metal.
3
The Concept of the Double Integral
Definition Double Integral
Page 256/definition 11.1.1
Suppose that a scalar function f pieces of area
is defined on a closed bounded
plane region ( ). Suppose, however, ( ) can be divided into pieces of