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MULTIPLE INTEGRALS
Double integrals
We will start out by assuming that the region in
2 is a rectangle which we will denote
as follows,
R a , b c , d
This means that the ranges for x and y are a x b and c^ y d.
Also, we will initially assume that f ( x , y ) 0 although this doesn’t really have to be the
case. Let’s start out with the graph of the surface (^) S given by graphing f ( x , y ) over the
rectangle R.
We ask ourselves what the volume of the region under S (and above the xy - plane of
course) is.
We will approximate the volume much as we approximated the area above. We will first
divide up (^) a x b into n subintervals and divide up c y d into m subintervals. This
will divide up R into a series of smaller rectangles and from each of these we will choose
a point( , )
xi y j. Here is a sketch of this set up.
Now, over each of these smaller rectangles we will construct a box whose height is given
by ( , )
f xi y j. Here is a sketch of that.
Each of the rectangles has a base area of A and a height of ( , )
f xi y j so the volume of
each of these boxes is f^ (^ xi , yj ) A
. The volume under the surface S is then
approximately,
n
i
m
j
V f xi yj A 1 1
( , )
The integral can be worked in steps as follows;
f x ydA f x ydydx f x ydxdy
d
c
b
a
b
a
d
R c
Example
Compute each of the following double integrals over the indicated rectangles.
1. 6 , [ 2 , 4 ] [ 1 , 2 ]
2
xydA R
R
2. 2 4 , [ 5 , 4 ] [ 0 , 3 ]
3
x ydA R
R
- cos( ) sin( ) , [ 2 , 1 ] [ 0 , 1 ]
2 2
x y x ydA R
R
4. , [ 0 , 1 ] [ 1 , 2 ]
dA R
R x y
5. xe dA , R [ 1 , 2 ][ 0 , 1 ]
R
xy
Solutions
1. 6 , [ 2 , 4 ] [ 1 , 2 ]
2
xydA R
R
4
2
2
4
2
4
2
2
1
3
4
2
2
1
2 2
xydA xydydx xy dx xdx x R
2. 2 4 , [ 5 , 4 ] [ 0 , 3 ]
3
x ydA R
R
4
5
2
4
5
4
5
3 0
4
4
5
3
0
3 3
x ydA x ydydx xy y dx x dx x x R
- cos( ) sin( ) , [^2 ,^1 ] [^0 ,^1 ]
2 2
x y x ydA R R
cos( )
sin( ) 3
sin( ) sin( )
cos( ) sin( ) cos( ) sin( )
1
0
(^13)
0
2
1
2
1
0
(^132)
0
1
2
2 2 2 2
y
y ydy
y
x x y dy
x y x y x ydA x y x ydxdy R
4. , [ 0 , 1 ] [ 1 , 2 ]
dA R R x y
- xe dA , R [ 1 , 2 ][ 0 , 1 ]
R
xy
As we saw in the previous set of examples we can do the integral in either direction.
However, sometimes one direction of integration is significantly easier than the other so
make sure that you think about which one you should do first before actually doing the
integral.
A double integral can be evaluated easily when the integrand is a product of functions.
Let f ( x , y ) g ( x ) h ( y )where g ( x )is a function of x only and h ( y )is a function of y.
There are two types of regions that we need to look at. Here is a sketch of the regions
These regions are described as;
Case 1: D ( x , y ) a x b , g 1 ( x ) y g 2 ( x )
Case 2: D ( x , y ) h 1 ( x ) x h 2 ( x ), c y d
we are going to use all the points, ( x , y ), in which both of the coordinates satisfy the two
given inequalities. The double integral for both of these cases are defined in terms of
iterated integrals as follows.
Case 1: D ( x , y ) a x b , g 1 ( x ) y g 2 ( x )the integral is defined to be;
Case 2: D^ ^ (^ x , y ) h 1 ( x ) x h 2 ( x ), c y ^ d the integral is defined to be;
Some properties of double integrals are:
Examples
Evaluate each of the following integrals over the given region D.
Solutions
In this case we need to determine the two inequalities for x and y that we need to do the
integral. The best way to do this is the graph the two curves. Here is a sketch.
Now, there are two ways to describe this region. 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. In this case the region would be
given by D 1 (^) D 2 where,
If we do this, we’ll need to do two separate integrals, one for each of the regions.
To avoid this, we could turn things around and solve the two equations for x to get;
y 2 x 3 , x y
y x x y
we notice that the same function is always on the right and the same function is always
on the left and so the region is;
Writing the region in this form means doing a single integral instead of the two integrals.
NB: We can integrate these integrals in either order (i.e. x followed by y or y followed
by x ), although often one order will be easier than the other. In fact, there will be times
when it will not even be possible to do the integral in one order while it will be possible
to do the integral in the other order.
Example
Evaluate the following integrals by first reversing the order of integration.
Solutions:
First, notice that if we try to integrate with respect to y we can’t do the integral because
we would need a
2 y in front of the exponential in order to do the y integration. We are
going to hope that if we reverse the order of integration we will get an integral that we
can do.
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. We can’t just interchange the
integrals, keeping the original limits, and be done with it. This would not fix our original
So, as we hoped, we were able to do the integral once we interchanged the order of
integration.
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. Here are
the limits for the variables that we get from this integral.
and the sketch of this region is;
So, if we reverse the order of integration we get the following limits.
The integral is then,
We now look at 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. This is given by;
Example
Find the volume of the solid that lies below the surface given by z 16 xy 200 and lies
above the region in the xy - plane bounded by
2 y x and
2 y 8 x.
Solution
By setting the two bounding equations equal we can see that they will intersect at x 2
and (^) x 2. So, the inequalities that will define the region D in the xy - plane are
The volume is then given by;
The region D is really where this solid will sit on the xy - plane and here are the
inequalities that define the region.
The volume of this solid is;
In general,
gives the net volume between the graph of z f ( x , y ) and the region D in the xy - plane.
Regions that are below the xy - plane have a negative volume and regions that are above
the xy - plane have a positive volume.
gives the net area between the curve given by y f ( x ) and the x - axis on the interval
[ a , b ].
The second geometric interpretation of a double integral is
This is easy to see why this is true in general. Let’s suppose that we want to find the area
of the region shown below.
We know that this area can be found by the integral,
Or in terms of a double integral we have,
Double Integrals in Polar Coordinates
We have seen to this point the region D could be easily described in terms of simple
functions in Cartesian coordinates. In this section we want to look at some regions that
are much easier to describe in terms of polar coordinates. For instance, we might have a
region that is a disk, ring, or a portion of a disk or ring. In these cases, using Cartesian
coordinates could be somewhat cumbersome. For instance, let’s suppose we wanted to
do the following integral,
This general region will be defined by inequalities,
Now, to find dA let’s redo the figure above as follows,
As shown, we’ll break up the region into a mesh of radial lines and arcs. Now, if we pull
one of the pieces of the mesh out as shown we have something that is almost, but not
quite a rectangle. The area of this piece is A. The two sides of this piece both have length
r r 0 ri where r 0 is the radius of the outer arc and ri is the radius of the inner arc. Basic
geometry then tells us that the length of the inner edge is ri while the length of the out
edge is r 0 where is the angle between the two radial lines that form the sides of
this piece.
Now, let’s assume that we’ve taken the mesh so small that we can assume that ri r 0 r
and with this assumption we can also assume that our piece is close enough to a rectangle
that we can also then assume that,
Also, if we assume that the mesh is small enough then we can also assume that,
With these assumptions we then get,
In order to arrive at this, we had to make the assumption that the mesh was very small.
This is not an unreasonable assumption. Recall that the definition of a double integral is
in terms of two limits and as limits go to infinity the mesh size of the region will get
smaller and smaller. In fact, as the mesh size gets smaller and smaller the formula above
becomes more and more accurate and so we can say that,
Before moving on it is again important to note that dA drd . The actual formula for dA
has an (^) r in it. It will be easy to forget this (^) r on occasion, but as you’ll see without it some
integrals will not be possible to do.
Now, if we’re going to be converting an integral in Cartesian coordinates into an integral
in polar coordinates we are going to have to make sure that we’ve also converted all the
x ’s and y ’s into polar coordinates as well. To do this we’ll need to remember the
following conversion formulas,
x r cos , y^ r^ sin,
2 2 2 r x y
We are now able to write down a formula for the double integral in terms of polar
coordinates.
Example
Evaluate the following integrals by converting them into polar coordinates.
