Section 15.2 : Iterated Integrals
2. Compute the following double integral over the indicated rectangle.
\[\iint\limits_{R}{{6y\sqrt x - 2{y^3}\,dA}}\hspace{0.5in}R = \left[ {1,4} \right] \times \left[ {0,3} \right]\]Show All Steps Hide All Steps
Start SolutionThe order of integration was not specified in the problem statement so we get to choose the order of integration. We know that the order will not affect the final answer so in that sense it doesn’t matter which order we decide to use.
Often one of the orders of integration will be easier than the other and so we should keep that in mind when choosing an order. Note as well that often each order will be just as easy/hard as the other order and so order won’t really matter all that much in those cases. Of course, there will also be integrals in which one of the orders of integration will be all but impossible, if not impossible, to compute and so you won’t really have a choice of orders in cases such as that.
Given all the possibilities discussed above it can seem quite daunting when you need to decide on the order of integration. It generally isn’t as bad as it seems however. When choosing an order of integration just take a look at the integral and think about what would need to be done for each order and see if there is one order that seems like it might be easier to take care of first or if maybe the resulting answer will make the second integration somewhat easier.
Sometimes it will not be readily apparent which order will be the easiest until you get into the problem. In these cases the only thing you can really do is just start with one order and see how it goes. If it starts getting too difficult you can always go back and give the other order a try to see if it is any easier.
The better you are at integration the easier/quicker it will be to choose an order of integration. If you are really rusty at integration and/or you didn’t learn it all that great back in Calculus I then you should probably head back into the Calculus I material and practice your integration skills a little bit. Multiple integrals will be much easier to deal with if you have good Calculus I skills.
As a final note about choosing an order of integration remember that for the vast majority of the integrals there is not a correct choice of order. There are a handful of integrals in which it will be impossible or very difficult to do one order first. In most cases however, either order can be done first and which order is easiest is often a matter of interpretation so don’t worry about it if you chose to do a different order here than we do. You will still get the same answer regardless of the order provided you do all the work correctly.
In this case neither orders seem to be particularly difficult so let’s integrate with respect to \(x\) first for no other reason that it will allow us to get rid of the root right away.
So, here is the integral set up to do the \(x\) integration first.
\[\iint\limits_{R}{{6y\sqrt x - 2{y^3}\,dA}} = \int_{0}^{3}{{\int_{1}^{4}{{6y{x^{\,\frac{1}{2}}} - 2{y^3}\,dx}}\,dy}}\]Remember that the first integration is always the “inner” integral and the second integration is always the “outer” integral.
When writing the integral down do not forget the differentials! Many students come out of a Calculus I course with the bad habit of not putting them in. At this point however, that will get you in trouble. You need to be able to recall which variable we are integrating with respect to with each integral and the differentials will tell us that so don’t forget about them.
Also, do not forget about the limits and make sure that they get attached to the correct integral. In this case the \(x\) integration is first and so the \(x\) limits need to go on the inner integral and the \(y\) limits need to go on the out integral. It is easy to get in a hurry and put them on the wrong integral.
Show Step 2Okay, let’s do the \(x\) integration now.
\[\begin{align*}\iint\limits_{R}{{6y\sqrt x - 2{y^3}\,dA}} & = \int_{0}^{3}{{\int_{1}^{4}{{6y{x^{\,\frac{1}{2}}} - 2{y^3}\,dx}}\,dy}}\\ & = \int_{0}^{3}{{\left. {\left( {4y{x^{\,\frac{3}{2}}} - 2x{y^3}} \right)} \right|_1^4\,dy}} = \int_{0}^{3}{{28y - 6{y^3}\,dy}}\end{align*}\]Just remember that because we are integrating with respect to \(x\) in this step we treat all \(y\)’s as if they were a constant and we know how to deal with constants in integrals.
Note that we are assuming that you are capable of doing the evaluation and so did not show the work in this problem and will rarely show it in any of the problems here unless there is a point that needs to be made.
Show Step 3Now all we need to take care of is the \(y\) integration and that is just a simple Calculus I integral. Here is that work.
\[\iint\limits_{R}{{6y\sqrt x - 2{y^3}\,dA}} = \int_{0}^{3}{{28y - 6{y^3}\,dy}} = \left. {\left( {14{y^2} - \frac{3}{2}{y^4}} \right)} \right|_0^3 = \require{bbox} \bbox[2pt,border:1px solid black]{{\frac{9}{2}}}\]