General Double Integrals and Volumes

1. General Double Integrals

2. General Double Integrals
* Type 1 and Type 2 domains descriptions
* Integration of z = f(x, y) over both types of domains
* Changing the orders of integration z = f(x, y)

3. General Double Integrals
We will define double integrals over two types of
nonrectangular domains:

4. General Double Integrals
We will define double integrals over two types of
nonrectangular domains:
Type1: A domain that is bounded
up and down by continuous
functions y=f(x) and y=g(x)
where f(x) > g(x) with a < x < b.

5. General Double Integrals
We will define double integrals over two types of
nonrectangular domains:
Type1: A domain that is bounded
up and down by continuous
functions y=f(x) and y=g(x)
where f(x) > g(x) with a < x < b. a b
Type1
y=f(x)
y=g(x)

6. General Double Integrals
We will define double integrals over two types of
nonrectangular domains:
Type1: A domain that is bounded
up and down by continuous
functions y=f(x) and y=g(x)
where f(x) > g(x) with a < x < b.
Type2: A domain that is bounded
left and right by continuous
functions x= f(y) and x=g(y)
where f(y) > g(y) with c < y < d.
a b
Type1
y=f(x)
y=g(x)

7. General Double Integrals
We will define double integrals over two types of
nonrectangular domains:
Type1: A domain that is bounded
up and down by continuous
functions y=f(x) and y=g(x)
where f(x) > g(x) with a < x < b.
Type2: A domain that is bounded
left and right by continuous
functions x= f(y) and x=g(y)
where f(y) > g(y) with c < y < d.
a b
x=f(y)
x=g(y)
d
c
Type1
Type2
y=f(x)
y=g(x)

8. General Double Integrals
Given a type 1 domain D, a fixed x defines a
vertical cross-section with the y changes from
g(x) to f(x).
a b
y=f(x)
y=g(x)
D (Type1)
x

9. General Double Integrals
Given a type 1 domain D, a fixed x defines a
vertical cross-section with the y changes from
g(x) to f(x).
a b
y=f(x)
y=g(x)
D (Type1)
x
We decribe D as
{(x, y)| a < x < b; g(x) < y < f(x)}.

10. General Double Integrals
Given a type 1 domain D, a fixed x defines a
vertical cross-section with the y changes from
g(x) to f(x).
a b
y=f(x)
y=g(x)
D (Type1)
x
We decribe D as
{(x, y)| a < x < b; g(x) < y < f(x)}.
Example: Describe D as shown
in the figure.
2y + x = 4
0 4
2
x

11. General Double Integrals
Given a type 1 domain D, a fixed x defines a
vertical cross-section with the y changes from
g(x) to f(x).
a b
y=f(x)
y=g(x)
D (Type1)
x
We decribe D as
{(x, y)| a < x < b; g(x) < y < f(x)}.
Example: Describe D as shown
in the figure.
The x runs from 0 to 4. 2y + x = 4
0 4
2
x

12. General Double Integrals
Given a type 1 domain D, a fixed x defines a
vertical cross-section with the y changes from
g(x) to f(x).
a b
y=f(x)
y=g(x)
D (Type1)
x
We decribe D as
{(x, y)| a < x < b; g(x) < y < f(x)}.
Example: Describe D as shown
in the figure.
The x runs from 0 to 4.
For a fixed x, the y-coordinate of
(x, y)'s run from 0 to (4 – x)/2.
2y + x = 4
0 4
2
x

13. General Double Integrals
Given a type 1 domain D, a fixed x defines a
vertical cross-section with the y changes from
g(x) to f(x).
a b
y=f(x)
y=g(x)
D (Type1)
x
We decribe D as
{(x, y)| a < x < b; g(x) < y < f(x)}.
Example: Describe D as shown
in the figure.
The x runs from 0 to 4.
For a fixed x, the y-coordinate of
(x, y)'s run from 0 to (4 – x)/2.
2y + x = 4
0 4
2
So D ={(x, y)| 0 < x < 4; 0 < y < (4 – x)/2}
x

14. General Double Integrals
Given a type 1 domain D, a fixed x defines a
vertical cross-section with the y changes from
g(x) to f(x).
a b
y=f(x)
y=g(x)
D (Type1)
x
We decribe D as
{(x, y)| a < x < b; g(x) < y < f(x)}.
Example: Describe D as shown
in the figure.
The x runs from 0 to 4.
For a fixed x, the y-coordinate of
(x, y)'s run from 0 to (4 – x)/2.
2y + x = 4
0 4
2
So D ={(x, y)| 0 < x < 4; 0 < y < (4 – x)/2}
x

15. General Double Integrals
Theorem: Let z = z(x, y) over the domain
D = {(x, y)| a < x < b; g(x) < y < f(x)}.
a b
y=f(x)
y=g(x)
D (Type1)
x

16. General Double Integrals
Theorem: Let z = z(x, y) over the domain
D = {(x, y)| a < x < b; g(x) < y < f(x)}.
a b
y=f(x)
y=g(x)
D (Type1)
x
Then
∫ z(x, y)dA =
∫ ∫
x=a
b
D
z(x, y)dy ]dx.

17. General Double Integrals
Theorem: Let z = z(x, y) over the domain
D = {(x, y)| a < x < b; g(x) < y < f(x)}.
a b
y=f(x)
y=g(x)
D (Type1)
x
Then
∫ z(x, y)dA =
∫ ∫ z(x, y)dy ]dx.
x=a
b
∫y=g(x)
y=f(x)
[
D

18. General Double Integrals
Theorem: Let z = z(x, y) over the domain
D = {(x, y)| a < x < b; g(x) < y < f(x)}.
a b
y=f(x)
y=g(x)
D (Type1)
x
Then
∫ z(x, y)dA =
∫ ∫ z(x, y)dy ]dx.
x=a
b
∫y=g(x)
y=f(x)
[
The important thing here is the order of the
integration. It is set according to the the description
of the domain D.
D

19. General Double Integrals
Theorem: Let z = z(x, y) over the domain
D = {(x, y)| a < x < b; g(x) < y < f(x)}.
a b
y=f(x)
y=g(x)
D (Type1)
x
Then
∫ z(x, y)dA =
∫ ∫ z(x, y)dy ]dx.
x=a
b
∫y=g(x)
y=f(x)
[
The important thing here is the order of the
integration. It is set according to the the description
of the domain D.
D

20. General Double Integrals
Theorem: Let z = z(x, y) over the domain
D = {(x, y)| a < x < b; g(x) < y < f(x)}.
a b
y=f(x)
y=g(x)
D (Type1)
x
Then
∫ z(x, y)dA =
∫ ∫ z(x, y)dy ]dx.
x=a
b
∫y=g(x)
y=f(x)
[
The important thing here is the order of the
integration. It is set according to the the description
of the domain D.
Note that if D is the rectangle [a, b] x [c, d], then
D = {a < x < b; c < y < d} and the double integral is
∫ z(x, y)dA =
∫ ∫ z(x, y)dy ]dx.
x=a
b
∫y=c
d
[
D
D

21. General Double Integrals
Example: Draw the domain D that is bounded by y = x, and
y = 2x – x2. Find the volume of z = xy over D.

22. General Double Integrals
Example: Draw the domain D that is bounded by y = x, and
y = 2x – x2. Find the volume of z = xy over D.
0 1
D
y = x
y = 2x – x2

23. General Double Integrals
Example: Draw the domain D that is bounded by y = x, and
y = 2x – x2. Find the volume of z = xy over D.
To solve for intersection, set x = 2x – x2
and get two solutions x = 0, and 1.
0 1
D
y = x
y = 2x – x2

24. General Double Integrals
Example: Draw the domain D that is bounded by y = x, and
y = 2x – x2. Find the volume of z = xy over D.
So D is bounded above by f(x) = 2x – x2
below by g(x) =x, from x = 0 to x = 1.
To solve for intersection, set x = 2x – x2
and get two solutions x = 0, and 1.
0 1
D
y = x
y = 2x – x2

25. General Double Integrals
Example: Draw the domain D that is bounded by y = x, and
y = 2x – x2. Find the volume of z = xy over D.
So D is bounded above by f(x) = 2x – x2
below by g(x) =x, from x = 0 to x = 1.
To solve for intersection, set x = 2x – x2
and get two solutions x = 0, and 1.
0 1
D
y = 2x – x2
y = x
So D ={(x, y)| 0 < x < 1; x < y < 2x – x2}

26. General Double Integrals
Example: Draw the domain D that is bounded by y = x, and
y = 2x – x2. Find the volume of z = xy over D.
So D is bounded above by f(x) = 2x – x2
below by g(x) =x, from x = 0 to x = 1.
To solve for intersection, set x = 2x – x2
and get two solutions x = 0, and 1.
0 1
D
y = 2x – x2
y = x
So D ={(x, y)| 0 < x < 1; x < y < 2x – x2}
Set the iterated integral in the given
order:

27. General Double Integrals
Example: Draw the domain D that is bounded by y = x, and
y = 2x – x2. Find the volume of z = xy over D.
So D is bounded above by f(x) = 2x – x2
below by g(x) =x, from x = 0 to x = 1.
To solve for intersection, set x = 2x – x2
and get two solutions x = 0, and 1.
xy dy dx
∫
∫ y=x
x=0
1 2x–x2
0 1
D
y = 2x – x2
y = x
So D ={(x, y)| 0 < x < 1; x < y < 2x – x2}
Set the iterated integral in the given
order:

28. General Double Integrals
xy dy dx
∫
∫ y=x
x=0
1 y=2x–x2
0 1
D
y = x
y = 2x – x2

29. General Double Integrals
=
xy2
∫ y=x
x=0
1
2
| dx
y=2x–x2
0 1
D
y = x
y = 2x – x2
xy dy dx
∫
∫ y=x
x=0
1 y=2x–x2

30. General Double Integrals
=
xy2
∫ y=x
x=0
1
2
| dx
= x(2x – x2)2 – x3
∫
x=0
1
dx
1
2
y=2x–x2
0 1
D
y = x
y = 2x – x2
xy dy dx
∫
∫ y=x
x=0
1 y=2x–x2

31. General Double Integrals
=
xy2
∫ y=x
x=0
1
2
| dx
= x(2x – x2)2 – x3
∫
x=0
1
dx
1
2
= 4x3 – 4x4 + x5 – x3 dx
∫
x=0
1
1
2
y=2x–x2
0 1
D
y = x
y = 2x – x2
xy dy dx
∫
∫ y=x
x=0
1 y=2x–x2

32. General Double Integrals
=
xy2
∫ y=x
x=0
1
2
| dx
= x(2x – x2)2 – x3
∫
x=0
1
dx
1
2
= 4x3 – 4x4 + x5 – x3 dx
∫
x=0
1
1
2
= x5 – 4x4 + 3x3 dx
∫
x=0
1
1
2
y=2x–x2
0 1
D
y = x
y = 2x – x2
xy dy dx
∫
∫ y=x
x=0
1 y=2x–x2

33. General Double Integrals
=
xy2
∫ y=x
x=0
1
2
| dx
= x(2x – x2)2 – x3
∫
x=0
1
dx
1
2
= 4x3 – 4x4 + x5 – x3 dx
∫
x=0
1
1
2
= x5 – 4x4 + 3x3 dx
∫
x=0
1
1
2
=
1
2 6
( x6
+
5
4x5
)|
1
0
y=2x–x2
–
4
3x4
0 1
D
=
120
7
y = x
y = 2x – x2
xy dy dx
∫
∫ y=x
x=0
1 y=2x–x2

34. General Double Integrals
Given a type 2 domain D,
x=f(y)
x=g(y)
d
c
D Type2
x

35. General Double Integrals
Given a type 2 domain D, a fixed y defines a
vertical cross-section with the x changes from
g(y) to f(y).
x=f(y)
x=g(y)
d
c
D Type2
x
y

36. General Double Integrals
Given a type 2 domain D, a fixed y defines a
vertical cross-section with the x changes from
g(y) to f(y).
We decribe D as
{(x, y)| c < y < d; g(y) < x < f(y)}.
x=f(y)
x=g(y)
d
c
D Type2
x
y

37. General Double Integrals
Given a type 2 domain D, a fixed y defines a
vertical cross-section with the x changes from
g(y) to f(y).
We decribe D as
{(x, y)| c < y < d; g(y) < x < f(y)}.
Example: Describe D as shown
in the figure.
2y + x = 4
0 4
2
y
x=f(y)
x=g(y)
d
c
D Type2
x
y

38. General Double Integrals
Given a type 2 domain D, a fixed y defines a
vertical cross-section with the x changes from
g(y) to f(y).
We decribe D as
{(x, y)| c < y < d; g(y) < x < f(y)}.
Example: Describe D as shown
in the figure.
The y run from 0 to 2. 2y + x = 4
0 4
2
y
x=f(y)
x=g(y)
d
c
D Type2
x
y

39. General Double Integrals
Given a type 2 domain D, a fixed y defines a
vertical cross-section with the x changes from
g(y) to f(y).
We decribe D as
{(x, y)| c < y < d; g(y) < x < f(y)}.
Example: Describe D as shown
in the figure.
The y run from 0 to 2.
For a fixed y, the x-coordinate of
(x, y)'s run from 0 to 4 – 2y.
2y + x = 4
0 4
2
y
x=f(y)
x=g(y)
d
c
D Type2
x
y

40. General Double Integrals
Given a type 2 domain D, a fixed y defines a
vertical cross-section with the x changes from
g(y) to f(y).
We decribe D as
{(x, y)| c < y < d; g(y) < x < f(y)}.
Example: Describe D as shown
in the figure.
The y run from 0 to 2.
For a fixed y, the x-coordinate of
(x, y)'s run from 0 to 4 – 2y.
2y + x = 4
0 4
2
So D ={(x, y)| 0 < y < 2; 0 < x < 4 – 2y}
y
x=f(y)
x=g(y)
d
c
D Type2
x
y

41. General Double Integrals
Theorem: Let z = z(x, y) over the domain
D = {(x, y)| c < y < d; g(y) < x < f(y)}. x=f(y)
x=g(y)
d
c
D Type2
x
y

42. General Double Integrals
Theorem: Let z = z(x, y) over the domain
D = {(x, y)| c < y < d; g(y) < x < f(y)}.
Then
∫ f(x, y)dA =
∫ ∫ dy.
y=c
d
D
x=f(y)
x=g(y)
d
c
D Type2
x
y

43. General Double Integrals
Theorem: Let z = z(x, y) over the domain
D = {(x, y)| c < y < d; g(y) < x < f(y)}.
Then
∫ f(x, y)dA =
∫ ∫ z(x, y)dx ]dy.
y=c
d
∫
x=g(y)
f(y)
[
D
x=f(y)
x=g(y)
d
c
D Type2
x
y

44. General Double Integrals
Theorem: Let z = z(x, y) over the domain
D = {(x, y)| c < y < d; g(y) < x < f(y)}.
Then
∫ f(x, y)dA =
∫ ∫ z(x, y)dx ]dy.
y=c
d
∫
x=g(y)
f(y)
[
The order of the integration is set according to the the
description of the domain D with ∫ dy outside and ∫ dx
inside to be calculated first.
D
x=f(y)
x=g(y)
d
c
D Type2
x
y

45. General Double Integrals
Theorem: Let z = z(x, y) over the domain
D = {(x, y)| c < y < d; g(y) < x < f(y)}.
Then
∫ f(x, y)dA =
∫ ∫ z(x, y)dx ]dy.
y=c
d
∫
x=g(y)
f(y)
[
The order of the integration is set according to the the
description of the domain D with ∫ dy outside and ∫ dx
inside to be calculated first.
Again, if D is the rectangle [a, b] x [c, d], then
D = {c < y < d; a < x < b } and the double integral is
∫ f(x, y)dA =
∫ ∫ z(x, y)dx ]dy.
y=c
d
∫
x=a
b
[
D
D
x=f(y)
x=g(y)
d
c
D Type2
x
y

46. General Double Integrals
Example: D is bounded by x = 2y, and x = y2 + 1 from y = 0
to y = 1. Draw D. Find the volume of z = 2x + y over D.

47. General Double Integrals
Example: D is bounded by x = 2y, and x = y2 + 1 from y = 0
to y = 1. Draw D. Find the volume of z = 2x + y over D.
To solve for intersection, set 2y = y2 + 1
and get y =1.

48. General Double Integrals
Example: D is bounded by x = 2y, and x = y2 + 1 from y = 0
to y = 1. Draw D. Find the volume of z = 2x + y over D.
To solve for intersection, set 2y = y2 + 1
and get y =1.
1 x = y2 + 1
x = 2y

49. General Double Integrals
Example: D is bounded by x = 2y, and x = y2 + 1 from y = 0
to y = 1. Draw D. Find the volume of z = 2x + y over D.
So D is bounded to the right by y2 + 1,
to the left 2y, from y = 0 to y = 1.
To solve for intersection, set 2y = y2 + 1
and get y =1.
1 x = y2 + 1
x = 2y

50. General Double Integrals
Example: D is bounded by x = 2y, and x = y2 + 1 from y = 0
to y = 1. Draw D. Find the volume of z = 2x + y over D.
So D is bounded to the right by y2 + 1,
to the left 2y, from y = 0 to y = 1.
To solve for intersection, set 2y = y2 + 1
and get y =1.
To set the iterated integral, write D in
correct order as
{(x, y)| 0 < y < 1;
1 x = y2 + 1
x = 2y

51. General Double Integrals
Example: D is bounded by x = 2y, and x = y2 + 1 from y = 0
to y = 1. Draw D. Find the volume of z = 2x + y over D.
So D is bounded to the right by y2 + 1,
to the left 2y, from y = 0 to y = 1.
To solve for intersection, set 2y = y2 + 1
and get y =1.
To set the iterated integral, write D in
correct order as
{(x, y)| 0 < y < 1; 2y < x < y2+1 }
1 x = y2 + 1
x = 2y

52. General Double Integrals
Example: D is bounded by x = 2y, and x = y2 + 1 from y = 0
to y = 1. Draw D. Find the volume of z = 2x + y over D.
So D is bounded to the right by y2 + 1,
to the left 2y, from y = 0 to y = 1.
To solve for intersection, set 2y = y2 + 1
and get y =1.
To set the iterated integral, write D in
correct order as
{(x, y)| 0 < y < 1; 2y < x < y2+1 }
then set the integral in this order:
dy
∫
y=0
1
1 x = y2 + 1
x = 2y
2x + y dx
x=y2+1
∫x=2y

53. General Double Integrals
dy
∫
y=0
1
2x + y dx
x=y2+1
∫x=2y
1 x = y2 + 1
x = 2y

54. General Double Integrals
=∫y=0
1
dy
∫
y=0
1
2x + y dx
x=y2+1
∫x=2y
x2 + yx| dy
x=2y
x=y2+1
1 x = y2 + 1
x = 2y

55. General Double Integrals
=∫y=0
1
= (y2 + 1)2 + y(y2 +1) – (2y)2 – y(2y)dy
∫
y=0
1
dy
∫
y=0
1
2x + y dx
x=y2+1
∫x=2y
x2 + yx| dy
x=2y
x=y2+1
1 x = y2 + 1
x = 2y

56. General Double Integrals
=∫y=0
1
= (y2 + 1)2 + y(y2 +1) – (2y)2 – y(2y)dy
∫
y=0
1
=∫y=0
1
dy
∫
y=0
1
2x + y dx
x=y2+1
∫x=2y
x2 + yx| dy
x=2y
x=y2+1
1 x = y2 + 1
x = 2y
y4 + y3 – 4y2 + y + 1 dy

57. General Double Integrals
=∫y=0
1
= (y2 + 1)2 + y(y2 +1) – (2y)2 – y(2y)dy
∫
y=0
1
=∫y=0
1
dy
∫
y=0
1
2x + y dx
x=y2+1
∫x=2y
x2 + yx| dy
x=2y
x=y2+1
1 x = y2 + 1
x = 2y
y4 + y3 – 4y2 + y + 1 dy
=
37
60

58. General Double Integrals
Its possible that a
domain is both type 1
and type 2:
y=2x or x=y/2
1
2
D

59. General Double Integrals
Its possible that a
domain is both type 1
and type 2:
y=2x or x=y/2
1
Let f(x, y) = e .
2
D
x2

60. General Double Integrals
Its possible that a
domain is both type 1
and type 2:
y=2x or x=y/2
1
Let f(x, y) = e . The double integral may be set in
two ways over D.
2
D
x2

61. General Double Integrals
= dy
∫
y=0
2
dx
x=1
∫x=y/2
Its possible that a
domain is both type 1
and type 2:
y=2x or x=y/2
1
Let f(x, y) = e . The double integral may be set in
two ways over D. If we set the iterated integral as
∫ D
dA
∫
2
D
x2
x2
e x2
e

62. General Double Integrals
= dy
∫
y=0
2
dx
x=1
∫x=y/2
Its possible that a
domain is both type 1
and type 2:
y=2x or x=y/2
1
Let f(x, y) = e . The double integral may be set in
two ways over D. If we set the iterated integral as
∫D
dA
∫
2
D
it won't be not computable.
x2
x2
e x2
e

63. General Double Integrals
= dy
∫
y=0
2
dx
x=1
∫x=y/2
Its possible that a
domain is both type 1
and type 2:
y=2x or x=y/2
1
Let f(x, y) = e . The double integral may be set in
two ways over D. If we set the iterated integral as
∫D
dA
∫
2
D
it won't be not computable. But if we switch the order:
x2
x2
e x2
e
=
∫D
dA
∫ x2
e

64. General Double Integrals
= dy
∫
y=0
2
dx
x=1
∫x=y/2
Its possible that a
domain is both type 1
and type 2:
y=2x or x=y/2
1
Let f(x, y) = e . The double integral may be set in
two ways over D. If we set the iterated integral as
∫D
dA
∫
2
D
it won't be not computable. But if we switch the order:
x2
x2
e x2
e
= dx
∫
x=0
1
dy
y=2x
∫y=0
∫D
dA
∫ x2
e x2
e

65. General Double Integrals
= dy
∫
y=0
2
dx
x=1
∫x=y/2
Its possible that a
domain is both type 1
and type 2:
y=2x or x=y/2
1
Let f(x, y) = e . The double integral may be set in
two ways over D. If we set the iterated integral as
∫D
dA
∫
2
D
it won't be not computable. But if we switch the order:
x2
x2
e x2
e
= dx
∫
x=0
1
dy
y=2x
∫y=0
∫D
dA
∫ x2
e x2
e = dx
∫
x=0
1 y=2x
y=0
x2
e y|

66. General Double Integrals
= dy
∫
y=0
2
dx
x=1
∫x=y/2
Its possible that a
domain is both type 1
and type 2:
y=2x or x=y/2
1
Let f(x, y) = e . The double integral may be set in
two ways over D. If we set the iterated integral as
∫D
dA
∫
2
D
it won't be not computable. But if we switch the order:
x2
x2
e x2
e
= dx
∫
x=0
1
dy
y=2x
∫y=0
∫D
dA
∫ x2
e x2
e = dx
∫
x=0
1 y=2x
y=0
x2
e y| = dx
∫
x=0
1
x2
2xe

67. General Double Integrals
= dy
∫
y=0
2
dx
x=1
∫x=y/2
Its possible that a
domain is both type 1
and type 2:
y=2x or x=y/2
1
Let f(x, y) = e . The double integral may be set in
two ways over D. If we set the iterated integral as
∫D
dA
∫
2
D
it won't be not computable. But if we switch the order:
x2
x2
e x2
e
= dx
∫
x=0
1
dy
y=2x
∫y=0
∫D
dA
∫ x2
e x2
e = dx
∫
x=0
1 y=2x
y=0
x2
e y| = dx
∫
x=0
1
x2
2xe
x=0
1
x2
e |
= = e – 1

68. General Double Integrals
= dy
∫
y=0
2
dx
x=1
∫x=y/2
Its possible that a
domain is both type 1
and type 2:
y=2x or x=y/2
1
Let f(x, y) = e . The double integral may be set in
two ways over D. If we set the iterated integral as
∫D
dA
∫
2
D
it won't be not computable. But if we switch the order:
x2
x2
e x2
e
= dx
∫
x=0
1
dy
y=2x
∫y=0
∫D
dA
∫ x2
e x2
e = dx
∫
x=0
1 y=2x
y=0
x2
e y| = dx
∫
x=0
1
x2
2xe
x=0
1
x2
e |
= = e – 1
We say we reverse the order of integration in this case

69. General Double Integrals
The area of a 2D shape D is the same numerically
as the volume of the cylinder with height 1 over D.

70. General Double Integrals
The area of a 2D shape D is the same numerically
as the volume of the cylinder with height 1 over D.
=
Area Volume (numerically)
1

71. General Double Integrals
The area of a 2D shape D is the same numerically
as the volume of the cylinder with height 1 over D.
Therefore we may compute area of D by computing
the volume of the cylinder with ht=1 over D.
=
Area Volume (numerically)
1

72. General Double Integrals
The area of a 2D shape D is the same numerically
as the volume of the cylinder with height 1 over D.
Therefore we may compute area of D by computing
the volume of the cylinder with ht=1 over D.
Example:
Find the area
enclosed by y = x2
and y = 2x + 3.
=
Area Volume (numerically)
1

73. General Double Integrals
The area of a 2D shape D is the same numerically
as the volume of the cylinder with height 1 over D.
Therefore we may compute area of D by computing
the volume of the cylinder with ht=1 over D.
Example:
Find the area
enclosed by y = x2
and y = 2x + 3.
=
Area Volume (numerically)
-1 3
1
1

74. General Double Integrals
The area of a 2D shape D is the same numerically
as the volume of the cylinder with height 1 over D.
Therefore we may compute area of D by computing
the volume of the cylinder with ht=1 over D.
Example:
Find the area
enclosed by y = x2
and y = 2x + 3.
=
Area Volume (numerically)
Set x2 = 2x + 3, we get x = -1, and 3. Set the integral
-1 3
1

75. General Double Integrals
Let z = f(x, y) = 1.
-1 3
1

76. General Double Integrals
Let z = f(x, y) = 1. It defines the cylinder whose
volume is the integral:
dx
∫
x=-1
3
dy
y=2x+3
∫y=x2
1
-1 3
1

77. General Double Integrals
Let z = f(x, y) = 1. It defines the cylinder whose
volume is the integral:
dx
∫
x=-1
3
dy
y=2x+3
∫y=x2
1 = dx
∫
x=-1
3
y |
y=2x+3
y=x2
=
-1 3
1

78. General Double Integrals
Let z = f(x, y) = 1. It defines the cylinder whose
volume is the integral:
dx
∫
x=-1
3
dy
y=2x+3
∫y=x2
1 = dx
∫
x=-1
3
y |
y=2x+3
y=x2
= dx
∫
x=-1
3
2x + 3 – x2
-1 3
1

79. General Double Integrals
Let z = f(x, y) = 1. It defines the cylinder whose
volume is the integral:
dx
∫
x=-1
3
dy
y=2x+3
∫y=x2
1 = dx
∫
x=-1
3
y |
y=2x+3
y=x2
= dx
∫
x=-1
3
2x + 3 – x2
= x2 + 3x – x3/3 |
3
=
x=-1
-1 3
1

80. General Double Integrals
Let z = f(x, y) = 1. It defines the cylinder whose
volume is the integral:
dx
∫
x=-1
3
dy
y=2x+3
∫y=x2
1 = dx
∫
x=-1
3
y |
y=2x+3
y=x2
= dx
∫
x=-1
3
2x + 3 – x2
= x2 + 3x – x3/3 |
3
=
28
3
x=-1
which is the measurement of the area.
-1 3
1

81. General Double Integrals
Let z = f(x, y) = 1. It defines the cylinder whose
volume is the integral:
dx
∫
x=-1
3
dy
y=2x+3
∫y=x2
1 = dx
∫
x=-1
3
y |
y=2x+3
y=x2
= dx
∫
x=-1
3
2x + 3 – x2
= x2 + 3x – x3/3 |
3
=
28
3
x=-1
which is the measurement of the area.
To summarize: ∫ dA = area of D.
∫ 1
D
-1 3
1