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Section 15.2 : Iterated Integrals

7. Compute the following double integral over the indicated rectangle.

\[\iint\limits_{R}{{xy\cos \left( {y\,{x^2}} \right)\,dA}}\hspace{0.5in}R = \left[ { - 2,3} \right] \times \left[ { - 1,1} \right]\]

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Start Solution

The order of integration was not specified in the problem statement so we get to choose the order of integration. As we discussed in the first few problems of this section this can be daunting task and in those problems the order really did not matter. The order chosen for those problems was mostly a cosmetic choice in the sense that both orders were had pretty much the same level of difficulty.

With this problem however, we have a real difference in the orders. If we do the \(y\) integration first we will have to do integration by parts. On the other hand, if we do \(x\) first we only need to do a Calculus I substitution, which is almost always easier/quicker than integration by parts. On top of all that, if you think about how the substitution will go it looks like we’ll also lose the integration by parts for the \(y\) after the substitution is done (if you don’t see that don’t worry you will eventually with enough practice).

So, it looks like integrating with respect to \(x\) first is the way to go. Here is the integral set up to do the \(x\) integration first.

\[\iint\limits_{R}{{xy\cos \left( {y\,{x^2}} \right)\,dA}} = \int_{{ - 1}}^{1}{{\int_{{ - 2}}^{3}{{xy\cos \left( {y\,{x^2}} \right)\,dx}}\,dy}}\] Show Step 2

Okay, let’s do the \(x\) integration now.

\[\begin{align*}\iint\limits_{R}{{xy\cos \left( {y\,{x^2}} \right)\,dA}} & = \int_{{ - 1}}^{1}{{\int_{{ - 2}}^{3}{{xy\cos \left( {y\,{x^2}} \right)\,dx}}\,dy}}\hspace{0.5in}u = y{x^2}\,\,\,\, \to \,\,\,\,\,\,du = 2yx\,dx\\ & = \int_{{ - 1}}^{1}{{\left. {\left( {\frac{1}{2}\sin \left( {y\,{x^2}} \right)} \right)} \right|_{ - 2}^3\,dy}} = \int_{{ - 1}}^{1}{{\frac{1}{2}\left( {\sin \left( {9y} \right) - \sin \left( {4y} \right)} \right)\,dy}}\end{align*}\]

So, as noted above, upon doing the substitution we not only lost the \(x\) in front of the cosine but we also lost the \(y\) and that in turn means we won’t have to do integration by parts for the \(y\) integral.

So, in this case doing the \(x\) integration first completely eliminated the integration by parts from the \(y\) integration from the problem. This won’t always happen but when it does we’ll take advantage of it and when it doesn’t we’ll be doing integration by parts whether we want to or not.

Show Step 3

Now all we need to take care of is the \(y\) integration and that is just Calculus I integral with a couple of really simple substitutions (we’ll leave the substitution work to you to verify). Here is that work.

\[\begin{align*}\iint\limits_{R}{{xy\cos \left( {y\,{x^2}} \right)\,dA}} & = \int_{{ - 1}}^{1}{{\frac{1}{2}\left( {\sin \left( {9y} \right) - \sin \left( {4y} \right)} \right)\,dy}}\\ & = \frac{1}{2}\left. {\left( {\frac{1}{4}\cos \left( {4y} \right) - \frac{1}{9}\cos \left( {9y} \right)} \right)} \right|_{ - 1}^1\\ & = \require{bbox} \bbox[2pt,border:1px solid black]{{\frac{1}{2}\left[ {\frac{1}{4}\left( {\cos \left( 4 \right) - \cos \left( { - 4} \right)} \right) - \frac{1}{9}\left( {\cos \left( 9 \right) - \cos \left( { - 9} \right)} \right)} \right] = 0}}\end{align*}\]

Recall that cosine is an even function and so \(\cos \left( { - \theta } \right) = \cos \left( \theta \right)\)!