Double Integrals
over General Regions
Double Integrals over General Regions Type I
Double integrals over general regions are evaluated as iterated integrals.
Most double integrals fall into two categories determined by the region of
integration D.
Type I Region: The lower and upper boundaries are graphs of continuous
functions y = g1(x) and y = g2(x), for 𝑎 ≤ 𝑥 ≤ 𝑏.


D  ( x, y ) a  x  b, g1( x )  y  g 2 ( x )
The double integral of 𝑓 𝑥, 𝑦 over the region D
is set up as:
Double Integrals over General Regions Example 1
Evaluate the integral
  xy
2
 1 dA where D is the region bounded by
D
the curves 𝑥 = 0, 𝑦 = 𝑥 and 𝑦 = 2 − 𝑥 .
The lines 𝑦 = 𝑥 and 𝑦 = 2 − 𝑥 intersect at (1,1).
 The projection of the region D onto the x-axis determines the limits
of integration for x: 0 ≤ 𝑥 ≤ 1
 The curves determine the limits of integration for y: 𝑥 ≤ 𝑦 ≤ 2 − 𝑥
  xy
D
2

 1 dA  
  xy
1 2 x
0 x

 1 dydx
2 x

y
  x  y
0
 3
 x

1
1
2
3
dx
  x4

41
(2  x )3
   x
 (2  x )     x  dx 
0
30
3

  3

Double Integrals over General Regions Type II
Type II Region: The left and right boundaries are graphs of continuous
functions x = h1(y) and x = h2(y), for c ≤ y ≤ d.


D  ( x, y ) c  y  d , h1( y )  x  h2 ( y )
The double integral of 𝑓(𝑥, 𝑦) over the
region D is set up as:
Double Integrals over General Regions Example 2
Evaluate the integral
  2  e  dA where D is the region bounded by the
x
D
curves y =1, y = x and y = − x +10.
The given lines intersect at (1, 1), (5, 5) and (9, 1)
 The projection of the region on the y-axis
determines the limits of integration for y:
1≤𝑦≤5
 The curves determine the limits of
integration for x:
𝑦 ≤ 𝑥 ≤ 10 − 𝑦
5 10 y
D  2  e  dA  1  y  2  e dxdy  
x
5

5
10 y
2 x  e x 
y
1
x

dy
  2(10  y )  e y 10  2 y  e y dy  32  e1  e9  2e5  32.35
1
Double Integrals over General Regions Example 3
The region D is given. Set up
D f ( x, y )dA
both ways if possible:
Type I:
2
4
2 x
2
4
f ( x, y )dydx
Type I:
2 x
0 0
Type II:
f ( x, y )dydx
0 
y
y
f ( x, y )dxdy
Type II:
2 2
0  y f ( x, y )dxdy
Double Integrals over General Regions Example 4
Reverse the order of integration and evaluate the integral:
1 1
0  y
x 3  1 dxdy
The integral is set up as Type II.
D : 0  y  1,
y  x 1
As is, it is impossible to evaluate the integral.
Convert to Type I region: 0  x  1, 0  y  x 2
1 x2
0 0
x 3  1 dydx

1
0
x
3
x2
 1  y 0
1
dx   x 2 x 3  1 dx
0
(u  x 3  1, du  3x 2dx )


2
2
1
1
2
3/2
u   2 2 2  1  0.41
  udu 
1 9
3 1
3 3 
Double Integrals over General Regions Example 5
Assume 𝑓(𝑥, 𝑦) ≥ 0 on the region D in the 𝑥𝑦-plane.
Recall that
 f ( x, y )dA
represents the volume of the solid below f and
D
above the region D.
Example 5: Calculate the volume under the plane z = y and above the
region bounded by y = 9 − x2 and the x-axis.
We can look at the region D as Type I region:
V   ydydx  
D
3

9 x 2
3 0
ydydx
9 x 2
 y2 
  
2 0
3 
3

3
dx
1
2 2
   9  x  dx
2 3
648

5
Double Integrals over General Regions Example 6
Calculate the volume bounded by 𝑧 = 𝑥 2 + 𝑦 2 and 𝑧 = 18 − 𝑥 2 − 𝑦 2 .
The two surfaces intersect along a curve C:
18  x 2  y 2  x 2  y 2  x 2  y 2  9
The circle of radius 3 is the
projection of C on the xy-plane and it
is also the boundary of the region of
integration D.
Double Integrals over General Regions Example 6 continued


V    upper surface    lower surface  dA
D


 


V   18  x 2  y 2  x 2  y 2 dA  2 9  x 2  y 2 dA
D
D
The limits for D, as a type I region, are:
3  x  3,  9  x 2  y  9  x 2
V  2
3

9 x 2
3  9  x 2
9  x

 y 2 dydx
9 x 2


2


 2  9  x 2 y  1 y 3 
dx
3
3   9 x 2

3/2
3
2
2
 4  9  x
dx  81 (using trigonometric substitution)

3
3
3


Note: By symmetry of both the domain and the integrand, we can write
V  4  2
3

0 0
9 x 2


9  x 2  y 2 dydx