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Iterated Integrals
Iterated Integrals
In the previous section we gave the definition of the double integral
...
We will continue to assume that we
are integrating over the rectangle
R = [ a,b ] × [ c, d ]
We will look at more general regions in the next section
...
Fubini’s Theorem
If f ( x, y ) is continuous on R = [ a,b ] × [ c, d ] then,
∫∫
R
b
d
f ( x, y ) dA = ⌠ ∫ f ( x, y ) dy dx = ⌠
⌡a c
⌡c
d
∫ a f ( x, y ) dx dy
b
These integrals are called iterated integrals
...
In other words, if the inner differential is dy then the limits on the inner integral must
be y limits of integration and if the outer differential is dy then the limits on the outer integral
must be y limits of integration
...
Let’s just take the first possibility above and change the notation a little
...
This will give a function involving only x’s which we can in turn integrate
...
To take the derivative of a function with
respect to y we treated the x’s as constants and differentiated with respect to y as if it was a
function of a single variable
...
We think of all the x’s as constants and integrate with
respect to y or we think of all y’s as constants and integrate with respect to x
...
Example 1 Compute each of the following double integrals over the indicated rectangles
...
To prove that let’s work this one with each order to make
sure that we do get the same answer
...
So, the iterated integral that we need to
compute is,
4
⌠
6
xy
dA
=
∫∫R
⌡2
2
2
∫1
6 xy 2 dy dx
When setting these up make sure the limits match up to the differentials
...
e
...
To compute this we will do the inner integral first and we typically keep the outer integral around
as follows,
⌠ ( 2 xy
6
xy
dA
=
∫∫R
⌡2
4
2
3
)
2
1
dx
4
= ∫ 16 x − 2 x dx
2
4
= ∫ 14 x dx
2
Remember that we treat the x as a constant when doing the first integral and we don’t do any
integration with it yet
...
∫∫ 6 xy
R
2
dA = 7 x
2 4
2
= 84
Solution 2
In this case we’ll integrate with respect to x first and then y
...
2
⌠
6
xy
dA
=
∫∫R
⌡1
2
4
∫2
6 xy 2 dx dy
= ⌠ ( 3x y
⌡1
2
2
2
)
4
2
dy
2
= ∫ 36 y 2 dy
1
= 12 y
3 2
1
= 84
Sure enough the same answer as the first solution
...
[Return to Problems]
(b)
3
x
−
4
y
dA , R = [ −5, 4] × [ 0,3]
2
∫∫
R
For this integral we’ll integrate with respect to y first
...
[Return to Problems]
(c)
2 2
x
∫∫ y + cos (π x ) + sin (π y ) dA , R = [ −2, −1] × [0,1]
R
In this case we’ll integrate with respect to x first
...
e
...
We’ll also rewrite the integrand to help
with the first integration
...
In this case it will be significantly easier to integrate with
respect to y first as we will see
...
Now let’s see what would
happen if we had integrated with respect to x first
...
[Return to Problems]
As we saw in the previous set of examples we can do the integral in either direction
...
The next topic of this section is a quick fact that can be used to make some iterated integrals
somewhat easier to compute on occasion
...
Let’s do a quick example using this integral
...
⎣ 2⎦
R
2
Solution
Since the integrand is a function of x times a function of y we can use the fact
...
This topic really doesn’t have anything to do
with iterated integrals, but this is as good a place as any to put it and there are liable to be some
questions about it at this point as well so this is as good a place as any
...
In other
words we want to look at integrals like the following
...
However, in this case we need to pay attention to the differential (dy or dx) in
the integral, because that will change things a little
...
For the most part answering these questions isn’t that difficult
...
Here are the integrals
...
In the first
integral we are differentiating with respect to y and we know that any function involving only x’s
will differentiate to zero and so when integrating with respect to y we need to acknowledge that
there may have been a function of only x’s in the function and so the “constant” of integration is a
function of x
...
Again, remember if we differentiate the answer with respect to x
then any function of only y’s will differentiate to zero