Transcript DOUBLE & ITERATED INTEGRALS Presented by Emily To & Alvin Wong Recall Partial Derivatives… To find fx(x,y) of f(x,y), hold y constant To find.

DOUBLE & ITERATED
INTEGRALS
Presented by Emily To & Alvin Wong
Recall Partial Derivatives…
To find fx(x,y) of f(x,y), hold y constant
To find fy(x,y) of f(x,y), hold x constant
Now apply the concepts to integrals!

Integrating with respect to x (hold y constant!)
h2 ( y )
 f ( x, y )dx  f ( x, y )
x
h2 ( y )
h1 ( y )
 f (h2 ( y ), y )  f (h1 ( y ), y )
h1 ( y )

Integrating with respect to y (hold x constant!)
g2 ( x )
 f ( x, y )dy  f ( x, y )
y
g2 ( x )
g1 ( x )
 f ( x, g2 ( x ))  f ( x, g1 ( x ))
g1 ( x )

Note that the variable of integration cannot appear in the bounds
y
set by the integral (ex:
) xydy

0
An Example

Integrating with respect to y (treat x as a
constant!):
x
 (2x  y)dy
0
2 x
y
 2 xy 
2
2
x
2
 2x 
2
0
Integral of an Integral
2
2 y y2
0
3 y 2 6 y

2
2
3 ydxdy 0
2


2 y y2
3
ydx
dy



3
yx
2
dy
2
3
y

6
y


 3 y 6 y

0
2 y y2


  6 y 2  3 y 3  9 y 3  18y 2 dy
0
2
  [24y 2  12y 3 ]dy
0
 [8 y  3 y ]  8(2)  3(2)  16
3
4 2
0
3
4
Area of a Plane Region
b
 [ f ( x)  g ( x)]dx
a

f ( x)
g ( x)
dy  y 
Formula for area
between two curves
with respect to x
f ( x)
g ( x)
FTOC-I
 f ( x)  g ( x)
b

a
f ( x)
g ( x)
b
b
f ( x)
a
g ( x)
dydx   y
dx
  [ f ( x)  g ( x)]dx
a
you get
the same
equation!
***Same concept applies
when finding the area
between two curves with
respect to y!
Therefore…

If R={(x,y)│a ≤ x ≤ b, g1(x)
≤ y ≤ g2(x)}, where g1 and
g2 are continuous on [a,b],
then the area of R is given
by:
b
A
a

g2 ( x)
g1 ( x )
dydx

If R={(x,y)│c ≤ y ≤ d, h1(y)
≤ x ≤h2(y)}, where h1 and
h2 are continuous on [c,d],
then the area of R is given
by:
A 
d
c

h2 ( y )
h1 ( y )
dxdy
Finding Area Using an Iterated Integral
Example 1:
Find the area under the curve bounded by the xaxis using iterated integrals.
y
1
,
x 1
2 x5
Cont’d
5

2 0
1
x 1
5
dydx   [ y]
2

5
2
0
1
x 1
dx
1
dx
x 1
 [2 x 1]
5
2
 2 5 1  2 2 1  2
Double Integrals
Now what do iterated
integrals have to do with
double integrals?
Recall…

b
a
f ( x)dx 
n
lim  f ( x *) x
n 
i 1
i
Definition


Iterated integral: a method of evaluating a double
integral by evaluating two separate single integrals
Double integral:
 R represents rectangle R of [a,b] ·[c,d]
n
V   f ( xi , yi )  Ai
i 1
V  lim
n 
n
 f ( x , y )  A   f ( x, y)dA
i 1
i
i
i
R
V  lim
n 
n
 f ( x , y )  A   f ( x, y)dA
i 1
i
i
Ai  xi  yi
f ( xi , yi )
(xi,yi)
f ( xi , yi )  Ai
i
R
Subrectangle base area
Height
Volume of one rectangular
prism
Properties of Double Integrals

Let f and g be continuous over a closed, bounded plane region R.
 cf ( x, y)dA  c f ( x, y)dA where c is a constant
2.  [ f ( x, y) g ( x, y)]dA   f ( x, y)dA  g ( x, y)dA
R
R
R
1.
R
R
f ( x, y)dA   f ( x, y)dA  f ( x, y)dA

3.
if R can be split into
R
R1
R2
two subregions
These are the same properties that apply to single integrals
!
Volume of a Solid Region
If f is integrable over a plane region R and f(x,y)≥0
for all (x,y) in R, then the volume of the solid region
that lies above R and below the graph of f is defined
as:
V   f ( x, y)dA
R
Fubini’s Theorem

If f(x,y) is continuous on rectangle R [a,b]·[c,d], then
b
d
a
c
 f ( x, y)dA  
f ( x, y )dydx  
d
c

b
a
f ( x, y )dxdy
R

1.
2.
Basically…
Hold x constant and integrate with respect to y from c
to d, then integrate with respect to x from a to b
-ORHold y constant and integrate with respect to x from a
to b, then integrate with respect to y from c to d
Note that…

If f(x,y)=g(x)h(y) and we are integrating over the
rectangle R=[a,b] ∙ [c,d], then
 f ( x, y)dA   g ( x)h( y)dA  (
b
a
R

d
g ( x)dx)( h( y )dy)
c
R
Meaning that…
If the function can be broken into separate x and y
components, then we can integrate the components
individually and multiply the results!
Example
3

1
2
0
3
2
x y dydx
2
3
 (  y dy)( x dx)
2
0
3
1
2
3
y  x 
   
 3  0  4 1
3
4
160
 8  81 1   8 
       20 
3
 3  4 4   3 
Switching the Order of Integration
Integrating wrt y first
3

1

0
3
1
2
3
1

x y dydx
3
0 1
2
 3 y 
 x   dx
3 0


2
2
3
3
Integrating wrt x first
3
8x
dx
3
0
3
x
2
  y  dy
4
1
4
2
  20y 2 dy
0
3
 2 x  160

 
3
 3 1
4
VERSUS

2
x 3 y 2 dxdy
SAME RESULTS!!
2
 20 y  160

 
3
 3 0
3
Double Integrals over General Regions
***Since most regions are not rectangular, D represents
any region.

If D={(x,y)│a ≤ x ≤ b,
g1(x) ≤ y ≤ g2(x)},
where g1 and g2 are
continuous on [a,b],
then V   f ( x, y)dA
D
is:
b
V 
a

g2 ( x)
g1 ( x )
f ( x, y)dydx

If D={(x,y)│c ≤ y ≤ d,
h1(y) ≤ x ≤h2(y)},
where h1 and h2 are
continuous on [c,d],
then V   f ( x, y)dA
D
is:
V 
d
c

h2 ( y )
h1 ( y )
f ( x, y)dxdy
Example

Evaluate to find the volume over general region D
with the given information:
y
D 1  x2 dA D:{(x,y)|0≤x≤4, 0 ≤y ≤√x)}
Cont’d
y
4
x
y
dA
D 1  x 2  0 0
dydx
2
1 x
x
 y

 
dx

2
0 2(1  x )

0
4
x
u=x2

dx
du=2xdx
0 2(1  x 2 )
4
2
1 16 1
1
16
 
du  ln 1  u 0  1 ln 17
4 0 1 u
4
4
Finding Volume Using a Double Integral
Evaluate
2
2
[
x

y
]dydx

R
where R: [1,3] ∙ [2,4]
2
2
where R: [1,3] ∙ [2,4]
[
x

y
]
dydx

Evaluate
R
2
2
[
x

y
]dydx

R

3 4

1 2

3
1

4
 2
y 
 x y   dx
3 2

3
1
[ x 2  y 2 ]dydx
3
 2 56
2 x  3 dx


3
58 280
 2 x 56x 


  74  3  3
3 1
 3
3
Order matters when…
Evaluate the integral by reversing the order of
integration.
2 8
2
2

0
x
3
x sin( y )dydx
y=x3x=y1/3
Reversing the order of integration

2 8
0
x
sin(
y
)
dydx
3
2
2
x
Old
New
dx
0≤x≤2
0≤x≤y1/3
dy
x3≤y≤8
0≤y≤8
Cont’d
8

y1 / 3
0 0
2
2
x sin( y )dxdy
y1 / 3
1 64
1
64
sin udu   cos u 0

6 0
6
1
  cos 64  cos 0
6
x
2 
    sin( y ) dy
0
3
0
8 y
2 
    sin( y )dy
0 3


1
1
1
2
  (cos 64  1)   cos 64 
u=y
6
6
6
du=2ydy
8
3
QUIZ TIME!!! No calculator (:
1. Find the volume under the surface
f(x,y)=ylnx,R  y ln xdA,where R is a
rectangular region [1,2]∙[0,3]. (10
pts.)
2. Find the volume under the surface
f(x,y)=8xy-2y,  [8xy  2 y]dA , where
D
D is the region between y=x3 and
y=-2x(x-4) in the first quadrant.
Provide a graph. (12 pts.)
Quiz Corrections
Cont’d