4.3 Iterated Integrals

4.3 Iterated Integrals

  1. The Iterated Integral Theorem
  2. Integrating Over Non-rectangles
  3. Examples
  4. Problems

    

The Iterated Integral Theorem

The main theorem of this section tells us how to compute multi-variables integrals in practice. The point is that it reduces integration over 2- and higher-dimensions regions to (repeated) 1-d integrations, which we in principle know how to evaluate, at least some of the time.

Suppose that R=[a,b]×[c,d] is a rectangle in R2. Let f be a function on R, and suppose that f is integrable on R and for every y[c,d], the function fy:[a,b]R defined by fy(x)=f(x,y) is integrable on [a,b].

Then the function g:[c,d]R defined by g(y)=abf(x,y)dx is integrable on [c,d], and (1)RfdA=cd(abf(x,y)dx)dy.

In the above equation, abf(x,y)dx means that we consider y to be a constant, and we integrate f(x,y)=fy(x). If this seems unclear, take a look at a few of the examples presented below.

Remark 1

There are three integrals in formula (1): a 2-dimensional integral dA on the left, and two 1-dimensional integrals on the right.

Remark 2

If the roles of x and y are switched in hypotheses, then the conclusion (1) also holds with the roles of x and y switched. Note that this means we require fx(y) to be integrable on [c,d] for every x[a,b], and we can conclude that g(x)=cdf(x,y)dy is integrable. This can be useful if both fx(y) and fy(x) are integrable, we can choose which order we integrate. As a result, ab(cdf(x,y)dy)dx = RfdA = cd(abf(x,y)dx)dy  whenever all the integrals appearing in the formula make sense (that is, all the integrands are integrable).

It follows from theory developed in Sections 4.2 that all the hypotheses of the above theorem will in practice be satisfied in almost every integral we encounter in this class (generally, integrals of continuous functions over compact sets whose boundaries have content zero). There is an example that things that can go wrong if the hypotheses are not all satisfied — see the advanced problem at the end of this section.

Remark 3

A generalization of Theorem 1 holds for integrals in n dimensions, for any n3. The hypotheses have the same character but are more complicated to state, and in practice are satisfied by more or less every integral we will meet in this class.

Example 1.

Let R=[0,1]×[0,1], and compute Rxcos(xy)dA.
According to Theorem 1, Rxcos(xy)dA = 01(01xcos(xy)dx)dy = 01(01xcos(xy)dy)dx. Which of these two iterated integrals is easier to evaluate? Let’s try both.
First, 01(01xcos(xy)dy)dx=01(sin(xy)|y=0y=1) dx =01sin(x) dx =1cos(1). Note that after we carry out the first integral 01xcos(xy)dy, the result, sin(x), is a function only of x; we have “integrated out” all the dependence on y.
If we do the integrals in the other order, we get 01(01xcos(xy)dx)dy=01(xysin(xy)+1y2cos(xy))x=0x=1 dy =011ysin(y)+1y2(cos(y)1) dy. One can check that it is integrable and that its integral equals 1cos(1), but neither of these claims is at all obvious. Since we have already found the answer by integrating in the other order, we are not obliged to worry about it.

But see here for more, if you are interested.

The key points are

  • Using l’Hospital’s rule, for example, one can check that limy01ysin(y)+1y2(cos(y)1) exists. From this one can deduce that this function is integrable on [0,1].
  • the antiderivative of $ 1 y (y) + 1{y^2}((y) -1)$ on (0,) is 1y(cosy1), and
  • One can laso check that limy01y(cosy1)=0. Using this one can finish evaluating the integral.
Note also that, again, after we integrate 01xcos(xy)dx, the result is a function of y alone; we have integrated out all of the dependence on x.

Integrating Over Non-rectangles

The main difficulty we will have arises when integrating over a set S that is not a rectangle. If S has the form (2)S={(x,y)R2:axb,ϕ(x)yψ(x)} for some continuous functions ϕ,ψ:[a,b]R , then SfdA=ab(ϕ(x)ψ(x)f(x,y)dy)dx Similarly it is straightforward to write down a suitable interated integral when the set S is described as S={(x,y)R2:cyd,ϕ(y)xψ(y)}.

The difficulty is that S is usally not described in a convenient form such as (2). So the challenge one often ecounters is, given a set S, translate its description into a set of inequalities, such as (x,y)S{axb ϕ(x)yψ(x) or maybe (x,y)S{cyd ϕ(y)xψ(y). It may be necessary to split a set S into mutiple pieces, each one of which is expressed by inequalities of the above form.

The same issues arise in higher dimensions where they are even more difficult. For example, given a set SR3, to write iterated integrals over S, it is necessary to translate description of S into inequalities (x,y,z)S{axb ϕ(x)yψ(x) Φ(x,y)zΨ(x,y) or similar inequalities with the roles of x,y,z permuted in some way.

The rest of this section consists of examples. Many of them will focus on how to set up integrals, rather than how to carry them out (which in some examples we will omit entirely), since setting them up is the new element here.

Examples

Example 2.

Let S be the subset of R2 that is bounded by the curves y=12x2 and y=x22, and let f:SR be continuous. Express SfdA as an interated integral in two different ways.

It is often helpful to draw a picture:
Ex4.3.2

S is the region between the two parabolas. First, we have to determine the points where the curves intersect. Both equations y=x22 and y=12x2 are satisfied at these points, so x22=y=12x2. We solve to find that (x,y)=(±2,2). (We can see this in the picture, but had we drawn by hand, we would not know where to put the intersection points without reasoning as above.)

For 2x2, we can see that x2212x2, so “between the two curves” means “above $ x^2-2$ and below 12x2”. We conclude that (x,y)S{2x2 x22y12x2

and thus Sf(x)dA=22x2212x2f(x,y)dydx. Note that, if you start by drawing a careful sketch, you can probably figure this out from looking at your picture.

Switching the roles of x and y Writing the SdA as an interated integal in the opposite order is more complicated. So if you are just asked to evaluate the integral you should probably solve it as above, unless there is a very good reason for doing otherwise.

First, we have express the curves that comprise S in the form x=f(y). This is straightforward:

y=x22x=±y+2(and y2), y=12x2x=±2y(and y0),

Next, it is convenient to split S into 3 sub-regions — the part below the x-axis, and the parts in the first and the second quadrants — since in each of these regions, the range of x (as a function of y) has a different description.

Below the x-axis:

(x,y)S and y0{2y0 y+2xy+2\.

First quadrant:

(x,y)Sfirst quadrant{0y2 2yxy+2 .

Second quadrant:

(x,y)Ssecond quadrant{0y2 y+2x2y .

Putting these together, we conclude that $ _S f(x), dA $ can be written as 20y+2y+2f(x,y)dxdy+022yy+2f(x,y)dxdy+02y+22yf(x,y)dxdy.

Example 3

Consider the integral 1428/yf(x,y)dxdy.

Rewrite this in the forms SfdA (that is, determine the region S) and abϕ(x)ψ(x)f(x,y)dydx.

From looking at the limits of integration, we can see that (x,y)S{1y4 2x8/y\.

In other words, the integral can be written as Sf dA, where S={(x,y):1y4, 2x8/y}. The easiest way to change the order of integration is to draw a picture. To understand the problem, you should do this. For this region, by rearranging the inequalities, it is also possible to figure out that (x,y)S{2x8 1y8/x\. Thus we can also write the integral as 2818/xf(x,y)dydx.

An interesting class of examples involves iterated integrals that look impossible as written, but that become possible once one exchanges the order of integration.

Example 4

Consider 01y1ex2dxdy.
The antiderivative of ex2 is not an elementary function. However, from looking at the limits of integration, we can see that we are asked to integrate over a region — call it S — defined by the inequalities (x,y)S{0y1 yx1 and this is the same as (x,y)S{0x1 0yx . This can be seen either by sketching S or else just by manipulating the inequalities, as in the previous example. It follows that 01y1ex2dxdy=010xex2dydx. Now the iterated integral on the right can be computed easily: 010xex2dydx=01xex2dx=1201eudu=12(e1).

For more examples of this sort, see the exercises.

Example 5

Find the volume of a sphere in R3. That is, compute the integral of f(x,y,z)=1 over the 3-dimensional ball of radius r, S=B(0;r)={(x,y,z)R3:x2+y2+z2r2}. It does not matter whether we consider the open ball or the closed ball, since the boundary is a set of zero content.

First we have to set up the integral. Since x,y,z appear appear in a symmetrical way in both f and S, the order of integration does not matter. So we can choose (arbitrarily) to write (x,y,z)S{azb ϕ(z)yψ(z) Φ(y,z)xΨ(y,z) First we consider the range of z. The smallest and largest possible values of z in the set S are z=r and z=r respectively. We can see this either from drawing a picture, or from noting that z2r2x2y2r2, and that points where x,y both equal 0 (hence z2=r2) belong to S. Thus (3)rzr About the range of y, we know that y2r2z2x2. If we know the value of z but not that of x, then the extreme case occurs when x=0, leading to z2=r2y2, hence (4)r2z2yr2z2 Finally, we see that (5)r2y2z2xr2y2z2 Thus S1dV=rrr2z2r2z2r2z2y2r2z2y21 dxdydz. We now carry out the integration. The innermost intergral leads to S1dV=2rrr2z2r2z2r2z2y2dydz. To evaluate this, we remember that in general $ _{-}^, dt $ can be evaluated by the substitution t=αsinθ, with π/2θπ/2, leading to

ααα2t2dt=α2π/2π/2cos2θdθ=π2α2. Substituting this into our above computation of S1dV (with α=r2z2 we can compute the integral and find that S1dV = 43πr3.

Problems

Basic

  1. For the following sets S, write down an iterated integral that corresponds to SfdA. For some of these, it may be necessary to divide S into two or more pieces, as in Example 2.

  2. For the following iterated integrals, determine the set SR2 corresponding to the region of integration, and rewrite the integral after changing the order of integration. For some of these, it may be necessary to divide S into two or more pieces.

  3. Evaluate the following integrals. It may be necessary to change the order of integration in order to do so.

  4. Find the volume of a sphere in R4.

Advanced

  1. For x,y0, define f(x,y)={y2 if y>x>0 x2 if x>y>0 0 if x=y or xy=0. Compute
    0101f(x,y)dxdy and 0101f(x,y)dydx. In the process of doing this, you will see that these iterated integrals are well-defined, in the sense that every function you have to integrate is in fact integrable. Are they equal?

  2. Prove Theorem 1. (You will do this in Problem Set 6, with the proof being divided into several smaller questions.)

    

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