bcb12e ppt 8 7

Chapter 8

Multivariable

Calculus

Section 7

Learning Objectives for Section 8.7 Double

Integrals over General Regions

 The student will be able to identify what is meant by a regular region.

 The student will be able to evaluate double integrals over regular regions.

 The student will be able to

reverse the order of integration.  The student will be able to

Regular x Regions

In this section we extend the concept of double integration discussed previously to non-rectangular regions. We begin with an example.

f (x) = 6x - x2

g(x) = x

Let R be the region shaded. We can describe R with the inequalities R = { (x,y) | x  y  6x - x2_{, 0}  _x  _5}

The region R can be viewed as a union of vertical line

Definition of Regular x Region

A region R in the xy plane is a regular x region if there exist functions f (x) and g(x) and numbers a and b so that

Regular y Regions

Consider the shaded region R in the figure. It can be described with the following inequalities:

R = { (x,y) | y2  _x  _{y + 2, -1}  _y  _{2 }}

x = y2

x = y+2

The region R can be viewed as a union of horizontal line segments going from the graph of h(y) = x2 _{to the graph of k(y) = y + 2 on} the interval [–1, 2]. Regions that can be described in this manner are called regular y regions.

Definition: A region R is a regular y region if there exist functions

Examples

The shaded region at the top is both an x

and a y region, because it satisfies the requirements for both.

The shaded region to the right in the

lower graph is neither an x region nor a y

region. A region is not a regular y region if some horizontal line has a nonempty intersection with the region that is

neither a closed interval nor a point.

Similarly, a region is not a regular x region if some vertical line has a nonempty intersection with the regions that is neither a closed

Describing A Regular Region

R Problem: The region R is bounded by the

graphs of x + y2 _{= 9 and x + 3y = 9. Graph R} and describe it as a regular x region, a regular

Region R can be covered by vertical line segments that go from the graph of x + 3y = 9 to the graph of x + y2 _{= 9. Thus R is a} regular x region. If we solve each of the equations for y, we get and , so

Region R is also a regular y region and could be described by solving each equation for x.

Describing A Regular Region

R Problem: The region R is bounded by the

graphs of x + y2 _{= 9 and x + 3y = 9. Graph R} and describe it as a regular x region, a regular

y region, both or neither. Represent R using set notation and double inequalities.

y = 3−1

3 x y = 9 −x R = {(x, y) | 3− 1

Double Integrals over

Regular Regions

Now we want to extend the definition of double integration to include regular x regions and regular y regions. Notice in the following slide that the order of integration now depends on the nature of the region R. If R is a regular x region, we

integrate with respect to y first, while if R is a regular y region, we integrate with respect to x first.

It is also important to note that the variable limits of

Double Integration Over Regular

Regions (continued)

R a g x

F x y dA = F x y dy dx

∫∫

∫ ∫

y = g(x) y = f (x)

a b

Regular x region

If R = { (x, y) | g(x)  y  f (x), a  x  b}, then

Regular y region

If R = { (x,y) | h(y)  x  k(y), c  y  d },

then ( )

( )

( , ) [ ( , ) ]

k y d

R c h y

F x y dA = F x y dx dy

∫∫

∫ ∫

R

c d

Example of Evaluating a

Double Integral

Evaluate

where R is the region bounded by the graphs of y = –x, y = x2, and x = 1.

2

R

xydA

Example of Evaluating a

Double Integral

Evaluate

where R is the region bounded by the graphs of y = –x, y = x2, and x = 1.

From the graph we can see that R is a regular x region described by

R = {(x,y) | –x  y  x2, 0  x  1}, so

2 R xydA

∫∫

∫∫

−x x2

∫

∫

= xy2 0

1

∫

y=x2 _dx

= [x(x2₎2 0

1

∫

= (x5 0 1

∫

= (x6

6 −

x4

4 ) |x=0

x=1

Reversing the Order of Integration

Problem: Reverse the order of integration in:

3 1

1 0

[ ( , ) ]

x

f x y dy dx

−

Reversing the Order of Integration

Problem: Reverse the order of integration in:

3 1

1 0

[ ( , ) ]

x

f x y dy dx

−

∫ ∫

Solution: The limits of integration indicate that the region of

integration is a regular x region: R = {(x,y)| 0  y  x – 1, 1  x  3}.

The graph shown below is a regular x region, and also a regular y

region. Thus,

3 1

1 0

[

( , ) ]

x

f x y dy dx

−

∫ ∫

[

( , ) ]

f x y dx dy

+

=

_{∫ ∫}

y = x – 1 or

Volume and Double Integrals

In general, if a solid can be described by the graph of a positive function f (x,y) over a regular region R (not necessarily a rectangle), then the double integral of the

Example of Volume

Problem: The region R is bounded by the graphs of x + y = 1,

y = 0, and x = 0. Find the volume of the solid under the graph of

z = 1 – x – y over the region R.

The graph of R indicates that R is both a regular x region and a regular y region. We choose to use the regular x region:

Example of Volume

(continued)

Thus, the volume of the solid is V =

1 1 0 0 1 2 1 0 0 1 2 0 1 2 0

2 3 1 0

(1 ) [ (1 ) ]

1

[( ) | ] 2

1

[(1 ) (1 ) (1 ) ] 2

1 1

( )

2 2

1 1 1

( ) |

2 2 6

1 1 1 1

( ) 0

x R y x y x x

x y dA x y dy dx

y xy y dx

x x x x dx

x x dx

x x x