Multiple Integrals (MAT455)

Double Integrals

Before starting on double integrals let’s do a quick review of the definition of a definite integrals for functions of single variables.  First, when working with the integral,

we think of x’s as coming from the interval .  For these integrals we can say that we are integrating over the interval .  Note that this does assume that , however, if we have  then we can just use the interval .

Now, when we derived the definition of the definite integral we first thought of this as an area problem.  We first asked what the area under the curve was and to do this we broke up the interval  into n subintervals of width  and choose a point, , from each interval as shown below,

Each of the rectangles has height of  and we could then use the area of each of these rectangles to approximate the area as follows.

To get the exact area we then took the limit as n goes to infinity and this was also the definition of the definite integral.

In this section we want to integrate a function of two variables, .  With functions of one variable we integrated over an interval (i.e. a one-dimensional space) and so it makes some sense then that when integrating a function of two variables we will integrate over a region of  (two-dimensional space).

We will start out by assuming that the region in  is a rectangle which we will denote as follows,

This means that the ranges for x and y are  and .

Also, we will initially assume that  although this doesn’t really have to be the case.  Let’s start out with the graph of the surface S given by graphing  over the rectangle R.

Now, just like with functions of one variable let’s not worry about integrals quite yet.  Let’s first ask what the volume of the region under S (and above thexy-plane of course) is. 

We will first approximate the volume much as we approximated the area above.  We will first divide up  into n subintervals and divide up  into m subintervals.  This will divide up R into a series of smaller rectangles and from each of these we will choose a point .  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 .  Here is a sketch of that.

Each of the rectangles has a base area of  and a height of  so the volume of each of these boxes is .  The volume under the surfaceS is then approximately,

We will have a double sum since we will need to add up volumes in both the x and y directions.

To get a better estimation of the volume we will take n and m larger and larger and to get the exact volume we will need to take the limit as both n and m go to infinity.  In other words,

Now, this should look familiar.  This looks a lot like the definition of the integral of a function of single variable.  In fact this is also the definition of a double integral, or more exactly an integral of a function of two variables over a rectangle.

Here is the official definition of a double integral of a function of two variables over a rectangular region R as well as the notation that we’ll use for it.

Note the similarities and differences in the notation to single integrals.  We have two integrals to denote the fact that we are dealing with a two dimensional region and we have a differential here as well.  Note that the differential is dA instead of the dx and dy that we’re used to seeing.  Note as well that we don’t have limits on the integrals in this notation.  Instead we have the R written below the two integrals to denote the region that we are integrating over.

Note that one interpretation of the double integral of  over the rectangle R is the volume under the function  (and above the xy-plane).  Or,

We can use this double sum in the definition to estimate the value of a double integral if we need to.  We can do this by choosing  to be the midpoint of each rectangle.  When we do this we usually denote the point as .  This leads to the Midpoint Rule,

In the next section we start looking at how to actually compute double integrals.

Iterated Integrals

In the previous section we gave the definition of the double integral.  However, just like with the definition of a single integral the definition is very difficult to use in practice and so we need to start looking into how we actually compute double integrals.  We will continue to assume that we are integrating over the rectangle

We will look at more general regions in the next section.

The following theorem tells us how to compute a double integral over a rectangle.

Fubini’s Theorem

If  is continuous on  then,                             These integrals are called iterated integrals.

Note that there are in fact two ways of computing a double integral and also notice that the inner differential matches up with the limits on the inner integral and similarly for the outer differential and limits.  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.

Now, on some level this is just notation and doesn’t really tell us how to compute the double integral.  Let’s just take the first possibility above and change the notation a little.

We will compute the double integral by first computing

and we compute this by holding x constant and integrating with respect to y as if this were a single integral.  This will give a function involving only x’s which we can in turn integrate.

We’ve done a similar process with partial derivatives.  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. 

Double integrals work in the same manner.  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.

Let’s take a look at some examples.

Example 1  Compute each of the following double integrals over the indicated rectangles. (a) ,     [Solution] (b) ,     [Solution] (c) ,     [Solution] (d) ,     [Solution] (e) ,     [Solution] Solution (a) ,   It doesn’t matter which variable we integrate with respect to first, we will get the same answer regardless of the order of integration.  To prove that let’s work this one with each order to make sure that we do get the same answer. Solution 1 In this case we will integrate with respect to y first.  So, the iterated integral that we need to compute is,                                                   When setting these up make sure the limits match up to the differentials.  Since the dy is the inner differential (i.e. we are integrating with respect to yfirst) the inner integral needs to have y limits. To compute this we will do the inner integral first and we typically keep the outer integral around as follows,                                                     Remember that we treat the x as a constant when doing the first integral and we don’t do any integration with it yet.  Now, we have a normal single integral so let’s finish the integral by computing this.                                                        Solution 2 In this case we’ll integrate with respect to x first and then y.  Here is the work for this solution.                                                   Sure enough the same answer as the first solution. So, remember that we can do the integration in any order. [Return to Problems] (b) ,   For this integral we’ll integrate with respect to y first.                                             Remember that when integrating with respect to y all x’s are treated as constants and so as far as the inner integral is concerned the 2x is a constant and we know that when we integrate constants with respect to y we just tack on a y and so we get 2xy from the first term. [Return to Problems] (c) ,   In this case we’ll integrate with respect to x first.              Don’t forget your basic Calculus I substitutions! [Return to Problems] (d) ,   In this case because the limits for x are kind of nice (i.e. they are zero and one which are often nice for evaluation) let’s integrate with respect to x first.   We’ll also rewrite the integrand to help with the first integration.                                     [Return to Problems] (e) ,   Now, while we can technically integrate with respect to either variable first sometimes one way is significantly easier than the other way.  In this case it will be significantly easier to integrate with respect to y first as we will see.                                                     The y integration can be done with the quick substitution,                                                       which gives                                                    So, not too bad of an integral there provided you get the substitution.  Now let’s see what would happen if we had integrated with respect to x first.                                                     In order to do this we would have to use integration by parts as follows,                                                     The integral is then,                              We’re not even going to continue here as these are very difficult integrals to do. [Return to Problems]

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.

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.

Fact

If  and we are integrating over the rectangle  then,

So, if we can break up the function into a function only of x times a function of y then we can do the two integrals individually and multiply them together.

Let’s do a quick example using this integral.

Example 2  Evaluate , . Solution Since the integrand is a function of x times a function of y we can use the fact.

We have one more topic to discuss in this section.  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.

What we want to do is discuss single indefinite integrals of a function of two variables.  In other words we want to look at integrals like the following.

From Calculus I we know that these integrals are asking what function that we differentiated to get the integrand.  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. 

In the case of the first integral we are asking what function we differentiated with respect to y to get the integrand while in the second integral we’re asking what function differentiated with respect to x to get the integrand.  For the most part answering these questions isn’t that difficult.  The important issue is how we deal with the constant of integration.

Here are the integrals.

Notice that the “constants” of integration are now functions of the opposite variable.  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.

Likewise, in the second integral, the “constant” of integration must be a function of y since we are integrating with respect to x.  Again, remember if we differentiate the answer with respect to x then any function of only y’s will differentiate to zero.

Double Integrals Over General Regions

In the previous section we looked at double integrals over rectangular regions.  The problem with this is that most of the regions are not rectangular so we need to now look at the following double integral,

where D is any region.

There are two types of regions that we need to look at.  Here is a sketch of both of them.

      

We will often use set builder notation to describe these regions.  Here is the definition for the region in Case 1

and here is the definition for the region in Case 2.

This notation is really just a fancy way of saying we are going to use all the points, , 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.

In Case 1 where  the integral is defined to be,

In Case 2 where  the integral is defined to be,

Here are some properties of the double integral that we should go over before we actually do some examples.  Note that all three of these properties are really just extensions of properties of single integrals that have been extended to double integrals.

Properties

1.  2. , where c is any constant. 3. If the region D can be split into two separate regions D1 and D2 then the integral can be written as

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

Example 1  Evaluate each of the following integrals over the given region D. (a) ,     [Solution] (b) ,  D is the region bounded by  and .   [Solution] (c) ,  D is the triangle with vertices , , and .         [Solution] Solution (a)  ,   Okay, this first one is set up to just use the formula above so let’s do that.                                         [Return to Problems] (b) ,  D is the region bounded by  and . 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. So, from the sketch we can see that that two inequalities are,                                                     We can now do the integral,                                     [Return to Problems] (c) ,  D is the triangle with vertices , , and . We got even less information about the region this time.  Let’s start this off by sketching the triangle. 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. 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  where,                                           Note the  is the “union” symbol and just means that D is the region we get by combing the two regions.  If we do this then 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,                                  If we do this we can 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 we’d have to do otherwise. 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. Solution 1     That was a lot of work.  Notice however, that after we did the first substitution that we didn’t multiply everything out.  The two quadratic terms can be easily integrated with a basic Calc I substitution and so we didn’t bother to multiply them out.  We’ll do that on occasion to make some of these integrals a little easier. Solution 2 This solution will be a lot less work since we are only going to do a single integral.                        So, the numbers were a little messier, but other than that there was much less work for the same result.  Also notice that again we didn’t cube out the two terms as they are easier to deal with using a Calc I substitution. [Return to Problems]

As the last part of the previous example has shown us 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.

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

Example 2  Evaluate the following integrals by first reversing the order of integration. (a)    [Solution] (b)    [Solution] Solution (a)   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 they 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.  Note as well that we can’t just interchange the integrals, keeping the original limits, and be done with it.  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.  Even if we ignored that the answer would not be a constant as it should be. So, let’s see how we reverse the order of integration.  The best way to reverse the order of integration is to first sketch the region given by the original limits of integration.  From the integral we see that the inequalities that define this region are,                                                                  These inequalities tell us that we want the region with  on the lower boundary and  on the upper boundary that lies between  and .  Here is a sketch of that region. 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.  Here they are for this region.                                                                  Any horizontal line drawn in this region will start at  and end at  and so these are the limits on the x’s and the range of y’s for the regions is 0 to 9. The integral, with the order reversed, is now,                                            and notice that we can do the first integration with this order.  We’ll also hope that this will give us a second integral that we can do.  Here is the work for this integral.                                            [Return to Problems] (b)   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.  Here are the limits for the variables that we get from this integral.                                                                 and here is a sketch of this region. So, if we reverse the order of integration we get the following limits.                                                                  The integral is then,                                  [Return to Problems]

The final topic of this section is two geometric interpretations of a double integral.  The first interpretation is an extension of the idea that we used to develop the idea of a double integral in the first section of this chapter.  We did this by looking at the volume of the solid that was below the surface of the function  and over the rectangle R in the xy-plane.  This idea can be extended to more general regions.

The volume of the solid that lies below the surface given by  and above the region D in the xy-plane is given by,

Example 3  Find the volume of the solid that lies below the surface given by  and lies above the region in the xy-plane bounded by  and . Solution Here is the graph of the surface and we’ve tried to show the region in the xy-plane below the surface.     Here is a sketch of the region in the xy-plane by itself. By setting the two bounding equations equal we can see that they will intersect at  and .  So, the inequalities that will define the region D in the xy-plane are,                                                               The volume is then given by,

Example 4  Find the volume of the solid enclosed by the planes , , , . Solution This example is a little different from the previous one.  Here the region D is not explicitly given so we’re going to have to find it.  First, notice that the last two planes are really telling us that we won’t go past the xy-plane and the yz-plane when we reach them. The first plane, , is the top of the volume and so we are really looking for the volume under, and above the region D in the xy-plane.  The second plane,  (yes that is a plane), gives one of the sides of the volume as shown below.       The region D will be the region in the xy-plane (i.e.  ) that is bounded by , , and the line where  intersects the xy-plane.  We can determine where  intersects the xy-plane by plugging  into it.                      So, here is a sketch the region D. The region D is really where this solid will sit on the xy-plane and here are the inequalities that define the region.                                                             Here is the volume of this solid.

The second geometric interpretation of a double integral is the following.

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.

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

Or in terms of a double integral we have,

This is exactly the same formula we had in Calculus I.

Demonstration for Solving Double Integral

Example: Introduction to double integral and volume

Example: Find the area by using double integral

Example: Reversing the order of integration

Example: Reversing the order of integration 2