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Double integrals over
general regions

Double Integrals Over General Regions
In the previous section we looked at double integrals over rectangular regions
...

There are two types of regions that we need to look at
...


We will often use set builder notation to describe these regions
...


D = {( x, y ) | h1 ( y ) ≤ x ≤ h2 ( y ) , c ≤ y ≤ d }

This notation is really just a fancy way of saying we are going to use all the points, ( x, y ) , in
which both of the coordinates satisfy the two given inequalities
...

In Case 1 where D =

{( x, y ) | a ≤ x ≤ b, g ( x ) ≤ y ≤ g ( x )} the integral is defined to be,
1

∫∫
D

In Case 2 where D =

2

g 2 ( x)
⌠
f ( x, y ) dA = ⎮ ∫
f ( x, y ) dy dx
⌡ a g 1( x)
b

{( x, y ) | h ( y ) ≤ x ≤ h ( y ) , c ≤ y ≤ d } the integral is defined to be,
1

∫∫
D

2

d

f ( x, y ) dA = ⌠
⎮
⌡c

h2 ( y)

∫h ( y ) f ( x, y ) dx dy
1

Here are some properties of the double integral that we should go over before we actually do
some examples
...

Properties

1
...


D

D

∫∫ cf ( x, y ) dA = c ∫∫ f ( x, y ) dA , where c is any constant
...
If the region D can be split into two separate regions D1 and D2 then the integral can be written
as

∫∫ f ( x, y ) dA = ∫∫ f ( x, y ) dA + ∫∫ f ( x, y ) dA
D

D1

D2

Let’s take a look at some examples of double integrals over general regions
...

x
y

(a) ⌠⌠ e dA , D =

⌡⌡

{( x, y ) |1 ≤ y ≤ 2,

y ≤ x ≤ y 3 } [Solution]

D

(b)

3
3
xy
−
y
dA
,
D
is
the
region
bounded
by
y
=
x
and
y
=
x

...


D

[Solution]

Solution
x
y

(a) ⌠⌠ e dA , D =

⌡⌡

{( x, y ) |1 ≤ y ≤ 2,

y ≤ x ≤ y3}

D

Okay, this first one is set up to just use the formula above so let’s do that
...

4
∫∫
D

In this case we need to determine the two inequalities for x and y that we need to do the integral
...
Here is a sketch
...

D

We got even less information about the region this time
...


Since we have two points on each edge it is easy to get the equations for each edge and so we’ll
leave it to you to verify the equations
...
If we use functions of x, as shown in the image
we will have to break the region up into two different pieces since the lower function is different
depending upon the value of x
...
If we do this then we’ll need to do two separate integrals, one for each of the regions
...

Either way should give the same answer and so we can get an example in the notes of splitting a
region up let’s do both integrals
...
Notice however, that after we did the first substitution that we didn’t
multiply everything out
...
We’ll do that on occasion to make
some of these integrals a little easier
...

3

⌠
6
x
−
40
y
dA
=
⎮
∫∫D
⌡1
2

∫

2 y −1

2
6
x
− 40 y dx dy
3

1
− y+
2 2

3

2 y −1
⌠
3
= ⎮ ( 2 x − 40 xy ) 1 3 dy
− y+
⌡1
2 2

= ∫ 100 y − 100 y + 2 ( 2 y − 1) − 2 ( − y +
3

3

2

1

(

1
2

3
1
1
y
y
= 50 y 2 − 100
+
2
−
1
+
−
(
)
(
3
4
2 y+
4

)

3 4
2

)

3 3
2

)

dy

3
1

935
=−
3
So, the numbers were a little messier, but other than that there was much less work for the same
result
...

[Return to Problems]

As the last part of the previous example has shown us we can integrate these integrals in either
order (i
...
x followed by y or y followed by x), although often one order will be easier than the
other
...

Let’s see a couple of examples of these kinds of integrals
...

3

9
3 y3
⌠
x e dy dx [Solution]
(a)
2
∫
⌡0 x

⎮
(b) ⌠

8

2

⌡ 0 ∫3 y

x 4 + 1 dx dy [Solution]

Solution
(a) ⌠

3

⌡0

9

∫x

2

3 y3

x e dy dx

First, notice that if we try to integrate with respect to y we can’t do the integral because we would
need a y2 in front of the exponential in order to do the y integration
...

Now, when we say that we’re going to reverse the order of integration this means that we want to
integrate with respect to x first and then y
...
This would not fix our original
problem and in order to integrate with respect to x we can’t have x’s in the limits of the integrals
...

So, let’s see how we reverse the order of integration
...
From the
integral we see that the inequalities that define this region are,

0≤ x≤3

x2 ≤ y ≤ 9
These inequalities tell us that we want the region with y = x 2 on the lower boundary and y = 9
on the upper boundary that lies between x = 0 and x = 3
...


Since we want to integrate with respect to x first we will need to determine limits of x (probably
in terms of y) and then get the limits on the y’s
...


0≤ x≤

y

0≤ y≤9

Any horizontal line drawn in this region will start at x = 0 and end at x =
the limits on the x’s and the range of y’s for the regions is 0 to 9
...
We’ll also hope that this will give
us a second integral that we can do
...

3

9

3 y
⌠
⌠
e
dy
dx
x
=
⎮
2
∫
⌡0 x
⌡0
9

3

∫

y
0

3 y3

x e dx dy

9

⌠ 1 4 y3
=⎮ x e
⌡0 4

y

dy
0

9

1 2 y3
⌠
y e dy
=⎮
⌡0 4
1 y3
= e
12

(

9

0

)

1 729
e −1
=
12

[Return to Problems]
8

2
⌠
(b) ⎮ ∫ 3
x 4 + 1 dx dy
⌡0 y

As with the first integral we cannot do this integral by integrating with respect to x first so we’ll
hope that by reversing the order of integration we will get something that we can integrate
...

3

y ≤x≤2
0≤ y≤8

and here is a sketch of this region
...


0≤ x≤2
0 ≤ y ≤ x3
The integral is then,
8

2

⌠
⎮ ∫ 3 x 4 + 1 dx dy = ⌠
⎮
⌡0 y
⌡0
2

∫

x3

x 4 + 1 dy dx

0

2

x3

4
=⌠
+ 1 dx
y
x
⎮
0
⌡0
2

= ∫ x3
0

3
⎛
⎞
1
4
2
x + 1 dx = ⎜17 − 1⎟
6⎝
⎠

[Return to Problems]

The final topic of this section is two geometric interpretations of a double integral
...
We did this by looking at the volume of the solid that was below
the surface of the function z = f ( x, y ) and over the rectangle R in the xy-plane
...

The volume of the solid that lies below the surface given by z = f ( x, y ) and above the region D
in the xy-plane is given by,

V = ∫∫ f ( x, y ) dA
D

Example 3 Find the volume of the solid that lies below the surface given by z = 16xy + 200
and lies above the region in the xy-plane bounded by y = x 2 and y = 8 − x 2
...


Here is a sketch of the region in the xy-plane by itself
...
So, the inequalities that will define the region D in the xy-plane are,

−2 ≤ x ≤ 2

x2 ≤ y ≤ 8 − x2
The volume is then given by,

V = ∫∫ 16 xy + 200 dA
D

2

8− x 2

= ⌠ ∫ 2 16 xy + 200 dy dx
⌡−2 x
⎮ ( 8 xy 2 + 200 y ) 2 dx
=⌠
x
⌡−2
2

2

8− x2

= ∫ −128 x − 400 x + 512 x + 1600 dx
3

2

−2

2

400 3
12800
⎛
⎞
4
2
= ⎜ 32 x −
x + 256 x + 1600 x ⎟ =
3
3
⎝
⎠ −2
Example 4 Find the volume of the solid enclosed by the planes 4x + 2 y + z = 10 , y = 3x ,
z = 0, x = 0
...
Here the region D is not
explicitly given so we’re going to have to find it
...

The first plane, 4 x + 2 y + z = 10 , is the top of the volume and so we are really looking for the
volume under,

z = 10 − 4x − 2 y

and above the region D in the xy-plane
...


The region D will be the region in the xy-plane (i
...
z = 0 ) that is bounded by y = 3x , x = 0 ,
and the line where z + 4x + 2 y = 10 intersects the xy-plane
...


0 + 4x + 2 y = 10

⇒

2x + y = 5

⇒

y = −2x + 5

So, here is a sketch the region D
...


0 ≤ x ≤1
3 x ≤ y ≤ −2 x + 5

Here is the volume of this solid
...


Area of D = ∫∫ dA
D

This is easy to see why this is true in general
...


From Calculus I we know that this area can be found by the integral,

A = ∫ g 2 ( x ) − g1 ( x ) dx
b

a

Or in terms of a double integral we have,

Area of D = ∫∫ dA
D

g2 ( x )
⌠
=⎮ ∫
dy dx
⌡ a g1 ( x )

b

g2 ( x )

= ∫ y g ( x ) dx = ∫ g 2 ( x ) − g1 ( x ) dx
b

a

1

b

a

This is exactly the same formula we had in Calculus I

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