Double Integrals - Short Summary (PowerPoint)

```Double Integrals
Introduction
Volume and Double Integral
z=f(x,y) ≥ 0
on rectangle
R=[a,b]×[c,d]
S={(x,y,z) in R3 | 0 ≤ z ≤ f(x,y), (x,y) in R}
Volume of S = ?
ij’s column:
y
z
(xi, yj)
f (xij*, yij*)
Rij
Sample point (xij*, yij*)
x
y
x
Area of Rij is Δ A = Δ x Δ y
*
*
f
(
x
,
y
Volume of ij’s column:
ij
ij )A
m
n
Total volume of all columns:  f ( xij* , yij* ) A
i 1 j 1
m
n
V   f ( xij* , yij* ) A
i 1 j 1
Definition
m
n
V  lim  f ( xij* , yij* ) A
m, n  i 1 j 1
Definition:
 f ( x, y)dA
The double integral R
of f over the rectangle R is
m
n
 f ( x, y)dA  lim  f ( x , y
R
m, n  i 1 j 1
*
ij
if the limit exists
m
Double Riemann sum:
n
*
*
f
(
x
,
y
 ij ij )A
i 1 j 1
*
ij
)A
Note 1. If f is continuous then the limit
exists and the integral is defined
Note 2. The definition of double integral
does not depend on the choice of sample
points
If the sample points are upper right-hand corners then
m
n
 f ( x, y)dA  lim  f ( x , y )A
R
m, n  i 1 j 1
i
j
Example 1
z=16-x2-2y2
0≤x≤2
0≤y≤2
Estimate the volume
of the solid above
the square and
below the graph
m=n=4
V≈41.5
m=n=8
V≈44.875
Exact volume?
V=48
m=n=16
V≈46.46875
Example 2
R  [1,1]  [2,2]

1  x 2 dA  ?
R
z
Integrals over arbitrary regions
f (x,y)
0
A
R

A
• A is a bounded plane
region
• f (x,y) is defined on A
• Find a rectangle R
containing A
• Define new function on R:
 f ( x, y ) if ( x, y )  A
f ( x, y )  
0, otherwise
f ( x, y)dA  f ( x, y)dA
R
Properties
Linearity
 [ f ( x, y)  g ( x, y)]dA   f ( x, y)dA   g ( x, y)dA
A
A
A
 cf ( x, y)dA  c f ( x, y)dA
A
A
Comparison
If f(x,y)≥g(x,y) for all (x,y) in R, then
 f ( x, y)dA   g ( x, y)dA
A
A
A2
A1
If A1 and A2 are non-overlapping regions then

f ( x, y )dA   f ( x, y )dA   f ( x, y )dA
A1  A2
A1
Area
1dA   dA  area of A
A
A
A2
Computation
• If f (x,y) is continuous on rectangle R=[a,b]×[c,d]
then double integral is equal to iterated integral
d b
b d
c a
a c
 f ( x, y)dA   f ( x, y)dxdy   f ( x, y)dydx
y
R
d
y
fixed
c
x
a
x
b
fixed
More general case
• If f (x,y) is continuous on
A={(x,y) | x in [a,b] and h (x) ≤ y ≤ g (x)} then
double integral is equal to iterated integral
y
b g ( x)
g(x)
 f ( x, y)dA   f ( x, y)dydx
A
A
h(x)
a
x
x
b
a h( x)
Similarly
• If f (x,y) is continuous on
A={(x,y) | y in [c,d] and h (y) ≤ x ≤ g (y)} then
double integral is equal to iterated integral
y
d g ( y)
 f ( x, y)dA   f ( x, y)dxdy
d
A
y
h(y)
c
R
c h( y )
g(y)
x
Note
If f (x, y) = φ (x) ψ(y) then

R
b
 d

f ( x, y )dA     ( x) ( y )dxdy     ( x)dx     ( y )dy 
c a
a
 c

d b
Examples
y
sin(
xy
)
dA
,
A

[
1
/
2
,
1
]

[

/
2
,

]

R
e

A
 x2
dA
where A is a triangle with vertices
(0,0), (1,0) and (1,1)
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