VC 2.05
Double Integrals, Volume Calculations,
and the Gauss-Green Formula
Example 1: A Double Integral Over a
Rectangular Region
Let f(x, y)  sin(y)  cos(x)  1 and let R be the region in the xy-plane bounded
by x  0, x   , y  2 , and y  . Calculate  f(x, y) dA :


   sin(y)  cos(x)  1  dy dx
0 2 
R
2  3  3 cos(x) is a function that sweeps out
vertical area cross sections from x  0 to x  .
If we integrate these area cross sections,
we accumulate the volume of the solid:

y
    cos(y)  y cos(x)  y  y  2  dx
0

  2  3  3 cos(x ) dx
0

  2x  3x  3 sin(x)  0
 2  3
2
Example 2: Changing the Order of
Integration
Calculate f(x, y) dA by computing  f(x, y) dy dx instead of
R

R
2  0


2 


R

   sin(y)  cos(x)  1  dx dy

 f(x, y) dx dy.

x
 x sin(y)  sin(x)  x 
dy
x0
   sin(y ) dy
2 

  y   cos(y)  2 
 2  3
2
   sin(y) is a function that sweeps out
horizontal area cross sections from y  2
to y  . If we integrate these area cross
sections, we accumulate the volume of the solid:
Computing Double Integrals Over a
Rectangular Region
Let z  f(x, y) be a surface and let R be the rectangular region in the
xy-plane bounded by a  x  b and c  y  d. Calculate  f(x, y) dA :
R
b d
d b
1) Set up   f(x, y) dx dy.
1) Set up   f(x, y) dy dx.
2) Hold y constant and
2) Hold x constant and
a c
c a
integrate with respect to x:
integrate with respect to y:
d
b
 F(x, y)
c
x b
xa
d
dy   F(b, y)  F(a, y)  dy
c
3) F(b, y)  F(a, y) is a function of y
that measures the area of horizontal
cross sections between y  c and y  d.
So we can integrate these area cross
sections to get our volume:
d
 F(b, y)  F(a, y) dy
c
 G(x, y)
a
xd
xc
b
dx    G(x,d)  G(x,c)  dx
a
3) G(x, d)  G(x, c) is a function of x
that measures the area of vertical
cross sections between x  a and x  b.
So we can integrate these area cross
sections to get our volume:
b
  G(x, d)  G(x, c) dx
a
Example 3: Computing Another Double
Integral Over a Rectangular Region
Let f(x, y)  2sin(xy)  1 and let R be the region in the xy-plane bounded by
x  4, x  4, y  3, and y  3. Calculate  f(x, y) dA :
R
3
x4
 2cos(xy)

3 4  2sin(xy)  1  dx dy  3  y  x  x 4 dy
3 4
 2cos(4y)
  2cos( 4y)

   
 4   
 ( 4)   dy
y
y
 

3 
3
3

 8 dy
3
y 3
  8y  y  3
 48
How did this end up negative??
Example 4: A Double Integral Over a
Non-Rectangular Region
Let f(x, y)  x 2  y 2  3 and let R be the region in the xy-plane bounded above by
y   x 2  4 and below by y  5. Calculate  f(x, y) dA :
 f(x, y) dA
R
R
xright yhigh ( x)

  x
2

 y2  3 dy dx
xleft ylow ( x)
3  x2  4

  x
3
3
5
x6
90  10x  3x 
is a function that sweeps
3
out vertical area cross sections from  3 to 3.
2
2
4
yhigh (x)  x2  4

 y 2  3 dy dx
y   x2  4
 2 y3

   yx 
 3y 
3
3 
 y  5
dx
6
xleft  3
xright  3


x
2
4
   90  10x  3x 
 dx
3 
ylow (x)  5
3 
3
Example 4: A Double Integral Over a
Non-Rectangular Region
Let f(x, y)  x 2  y 2  3 and let R be the region in the xy-plane bounded above by
y   x 2  4 and below by y  5. Calculate  f(x, y) dA :
If we integrate these cross sections given by
R
x6
90  10x  3x 
from x  3 to x  3,
3
we accumulate the volume of the solid:
2
4
3
6

x 
2
4
   90  10x  3x 
 dx
3 
3 
3

10 3 3 5 x 
 90x 
x  x 

3
5
21  3

15516

35
7
Example 5: Change the Order of
Integration
Calculate f(x, y) dA again, but change the order of integration :
R
26 4  y
2
y 4  y  2y 2 4  y is a function
3
3
that sweeps out horizontal area cross sections
 f(x, y) dA
R
y top


xhigh ( y)
 
4

4 y
  
5  4  y
4

ytop  4
xlow (y)   4  y

xhigh (y)  4  y
x 2  y 2  3 dx dy
x 4 y
x

2
    y x  3x 
3
5 
 x
3
from y  5 to y  4.
x2  y2  3 dx dy
ybottom xlow ( y)

dy
4 y
 26 4  y 2

2
 
 y 4  y  2y 4  y  dy ybottom  5


3
3
5 

4
Example 5: Change the Order of
Integration
Let f(x, y)  x 2  y 2  3 and let R be the region in the xy-plane bounded above by
y   x 2  4 and below by y  5. Calculate  f(x, y) dA :
26 4  y
R
2
y 4  y  2y 2 4  y is a function
3
3
that sweeps out horizontal area cross sections

from y  5 to y  4.
 26 4  y 2

2
 
 y 4  y  2y 4  y  dy


3
3
5 

4
 4

3/2
2
 
(4  y)
15y  41y  261 
 105
 5
4

15516

35

Summary of Example 4 & 5
3  x2  4
  x
3
2

 y2  3 dy dx and
5
4
4 y
  x
5  4  y
2

 y 2  3 dx dy compute
the exact same thing!!!
We have changed the order of integration. This works whenever
we can bound the curve with top/bottom bounding functions
or left/right bounding functions.
yhigh (x)  x2  4
ytop  4
xlow (y)   4  y
xhigh (y)  4  y
xright  3
xleft  3
ylow (x)  5
ybottom  5
Computing Double Integrals Over a
Non-Rectangular Region
Let z  f(x, y) be a surface and let R be the non-rectangular region in the
xy-plane bounded by y  f(x) on top and y  g(x) on bottom. Calculate  f(x, y) dA :
xright yhigh ( x)
1) Set up
 
R
f(x, y) dy dx
xleft ylow ( x)
2) Hold x constant and
integrate with respect to y:
xright

xleft
y  yhigh ( x )
F(x, y) 
dx 
y  ylow ( x )
xright

xleft
F(x, y high (x))  F(x, y low (x))  dx


3) F(x, y high (x))  F(x, y low (x))  is a function
of x that measures the area of vertical
cross sections between x  x left and x  x right .
So we can integrate these area cross
sections to get our volume:
xright

xleft
F(x, y high (x))  F(x , y low (x))  dx


Likewise for the order
integration being reversed.
Example 6: Using the Double Integral
to Find the Area of a Region
Let R be the region in the xy-plane from Example 4 and 5 that is bounded above
by y   x 2  4 and below by y  5. Find the area of R.
Claim: For a region R, AreaR 
  x
2
3
3

  x


 4   5  dx
2
R
New Way:
Old Way:
3
 1 dA

 9 dx
3
3
 x3

 
 9x 
 3
 3
 36
R
 1 dA
R

3  x2  4
 
3
3

3

5
  y 
3
1 dy dx
y   x2  4
y  5
  x
3
2
dx
It works!!!


 4   5  dx
The Gauss-Green Formula
Let R be a region in the xy-plane whose boundary is parameterized
by (x(t),y(t)) for tlow  t  thigh . Then the following formula holds :
 n m 
R  x  y  dA 
thigh
  m  x(t), y(t) x'(t)  n  x(t), y(t) y '(t) dt
tlow
Main Idea: You can compute a double integral
as a single integral, as long as you have
a reasonable parameterization of the boundary
curve.
Main Rule: For this formula to hold:
1) Your parameterization must be counterclockwise.
2) Your tlow  t  thigh must bring you around the
boundary curve exactly ONCE.
3) Your boundary curve must be a simple closed curve.
Example 6 : Let f(x, y)  y  xy  9 and let R by the region in the xy-plane
2
2
x y
whose boundary is described by the ellipse       1.
3 4
Find  f(x, y) dx dy :
R
 n m 
Gauss-Green:  

 dx dy 
x y 
R 
thigh
  m  x(t), y(t) x'(t)  n  x(t), y(t) y '(t) dt
tlow
Let  x(t), y(t)  3cos(t),4sin(t) for 0  t  2
x
Set n  x , y    f(s, y)ds 
0
x
  y  sy  9  ds
0
x


sy
 sy 
 9s
2

0
2
x y
 xy 
 9x
2
So n  x(t), y(t)  12sin(t)cos(t)  18cos2 (t)sin(t)  27 cos(t)
2
Further,  x'(t), y'(t)   3sin(t), 4cos(t) .
thigh
Plug in :
  m  x(t), y(t) x'(t)  n  x(t), y(t) y'(t) dt
tlow
Example 6 : Let f(x, y)  y  xy  9 and let R by the region in the xy-plane
2
2
x y
whose boundary is described by the ellipse       1.
3  4
m  x(t), y(t) 
Plug into Gauss-Green :
thigh

tlow

m  x(t), y(t)  x'(t)  n  x(t), y(t)  y'(t) dt
2

0
n  x(t), y(t) 
12sin(t)cos(t)  18 cos 2 (t)sin(t)  27 cos(t)
[tlow , thigh ]
[0 ,2 ]
x '(t)
y '(t)
3sin(t)
4 cos(t)
  0  ( 3sin( t))   12sin(t)cos(t)  18 cos (t)sin(t)  27 cos(t) (4 cos(t)) dt
2
0
2

  48 cos (t)sin(t)  72cos (t)sin(t)  108 cos (t)  dt
2
3
2
0
 108 
k 1
k 1
sin
(b)
sin
(a)
Hint:  sink (t)cos(t)dt 

a
k 1
k 1
k 1
k 1
b

cos
(b)
cos
(a)
and  cosk (t)sin(t)dt 

a
k 1
k 1
1  cos(2t)
2
and cos (t) 
2
b
Process for Using Gauss-Green to
Compute Integrals
Let z  f(x, y) be a surface and let R be a region in the
xy-plane parameterized by (x(t),y(t)) for tlow  t  thigh . Calculate  f(x, y) dA :
R
t
high
 n m 
1) Gauss-Green:  

 dx dy   m  x(t), y(t) x'(t)  n  x(t), y(t) y'(t) dt
x y 
R 
tlow
2) Substitute as follows :
m x(t), y( t)
0




x
n  x(t), y(t) 
 f(s, y )ds
[tlow , thigh ]
Bounds for t in (x(t), y(t))
x '(t)
y'(t)
Derivative of x( t)
Derivative of y(t)
0
3) Crunch the integral! Don't forget your trig identities.
Proof of Gauss-Green Theorem
 n m 
Prove:  

 dA 

x

y

R 
thigh
  m  x(t), y(t) x'(t)  n  x(t), y(t) y '(t) dt
tlow
 m 
Let's just prove this for one piece: Show   
 dA 
y 
R 
Let y high (x ) be a function of x that traces the
boundary of the top half of R and let y low ( x)
be trace the boundary of the bottom half of R.
Let (x(t), y(t)) be a counterclockwise
tmid
parameterization of the boundary of R for
tlow <t<thigh that traverses the boundary only once.
Then x(t) for tlow  t  tmid traces out yhigh (x )
from right to left.
Also x(t) for tmid  t  thigh traces out ylow (x)
from left to right.
thigh
  m  x(t), y(t) x '(t) dt
tlow
yhigh (x)
R
ylow (x)
tlow
thigh
Proof of Gauss-Green Theorem
 n m 
Prove:  

 dA 

x

y

R 
m
R y dA 
b

thigh
  m  x(t), y(t) x'(t)  n  x(t), y(t) y '(t) dt
tlow
b yhigh ( x )
m
a y ( x ) y dy dx
low

yhigh (x)
tmid
  m x, y high (x)  m  x, y low (x)   dx


a
b


b
  m x, y high (x) dx   m  x, ylow (x)  dx
a
tlow
R
ylow (x)
a
thigh
dx
dx
  m  x(t), y(t) 
dt   m  x(t), y(t) 
dt
dt
dt
tmid
tmid
tlow
thigh
Proof of Gauss-Green Theorem
 n m 
Prove:  

 dA 

x

y

R 
tlow


tmid
m  x(t), y(t) 
thigh
  m  x(t), y(t) x'(t)  n  x(t), y(t) y '(t) dt
tlow
thigh
dx
dx
dt   m  x(t), y(t) 
dt
dt
dt
tmid
thigh
tlow

yhigh (x)
tmid
 m  x(t), y(t) x'(t)dt   m  x(t), y(t)  x'(t)dt
tmid


tmid
tmid
thigh
tlow
thigh
tmid
 m  x(t), y(t) x'(t)dt   m  x(t), y(t) x'(t)dt
tlow
R
ylow (x)
 m  x(t), y(t) x'(t)dt
tlow
 m 
Hence   
 dA 
y 
R 
thigh
 m  x(t), y(t) x'(t)dt
tlow
thigh
Proof of Gauss-Green Theorem
 n m 
Prove:  

 dA 

x

y

R 
 m 
Hence   
 dA 
y 
R 
thigh
  m  x(t), y(t) x'(t)  n  x(t), y(t) y '(t) dt
tlow
thigh
 m  x(t), y(t) x'(t)dt
tlow
 n 
We can likewise prove...    dA 
x 
R 
thigh
  n  x(t), y(t) y'(t) dt
tlow
Put it Together :
 n m 
R  x  y  dA 

n
m
dA

R x
R y dA
thigh
thigh
tlow
tlow
 n  x(t), y(t) y '(t)dt   m  x(t), y(t) x'(t)dt
thigh

  m  x(t), y(t) x '(t)  n  x(t), y(t) y'(t) dt
tlow
When Should I Use What??
y top

xright yhigh ( x)
xhigh ( y)

f(x, y) dy dx
 
thigh
f(x, y) dy dx
 n  x(t), y(t) y'(t) dt
ybottom xlow ( y)
xleft ylow ( x)
tlow
Integrate with respect to x
first then y when you have a
region R that is bounded on
the left and right by two
functions of y:
Integrate with respect to y
first then x when you have a
region R that is bounded on
the top and bottom by two
functions of x:
Use Gauss-Green when
you have a region R whose
boundary you can
parameterize with a single
function of t, (x(t),y(t)) for
tlow ≤ t ≤ thigh.
yhigh (x)
xlow (y)
R
xhigh (y)
R
(x(t), y(t))
R
tlow  t  thigh