Section 15.2 : Iterated Integrals
1. Compute the following double integral over the indicated rectangle (a) by integrating with respect to \(x\) first and (b) by integrating with respect to \(y\) first.
\[\iint\limits_{R}{{12x - 18y\,dA}}\hspace{0.5in}R = \left[ { - 1,4} \right] \times \left[ {2,3} \right]\]Show All Steps Hide All Steps
a Evaluate by integrating with respect to \(x\) first. Show All Steps Hide All StepsStart Solution
Not too much to do with this problem other than to do the integral in the order given in the problem statement.
Let’s first get the integral set up with the proper order of integration.
\[\iint\limits_{R}{{12x - 18y\,dA}} = \int_{2}^{3}{{\int_{{ - 1}}^{4}{{12x - 18y\,dx}}\,dy}}\]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 outer 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.
\[\begin{align*}\iint\limits_{R}{{12x - 18y\,dA}} = \int_{2}^{3}{{\int_{{ - 1}}^{4}{{12x - 18y\,dx}}\,dy}}\\ & = \int_{2}^{3}{{\left. {\left( {6{x^2} - 18xy} \right)} \right|_{ - 1}^4\,dy}} = \int_{2}^{3}{{90 - 90y\,dy}}\end{align*}\]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 in the \(y\) integration and that is just a simple Calculus I integral. Here is that work.
\[\iint\limits_{R}{{12x - 18y\,dA}} = \int_{2}^{3}{{90 - 90y\,dy}} = \left. {\left( {90y - 45{y^2}} \right)} \right|_2^3 = \require{bbox} \bbox[2pt,border:1px solid black]{{ - 135}}\]b Evaluate by integrating with respect to \(y\) first. Show All Steps Hide All Steps
Start Solution
Not too much to do with this problem other than to do the integral in the order given in the problem statement.
Let’s first get the integral set up with the proper order of integration.
\[\iint\limits_{R}{{12x - 18y\,dA}} = \int_{{ - 1}}^{4}{{\int_{2}^{3}{{12x - 18y\,dy}}\,dx}}\]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 fist 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 \(y\) integration is first and so the \(y\) limits need to go on the inner integral and the \(x\) limits need to go on the outer integral. It is easy to get in a hurry and put them on the wrong integral.
Show Step 2Okay, let’s do the \(y\) integration now.
\[\begin{align*}\iint\limits_{R}{{12x - 18y\,dA}} = \int_{{ - 1}}^{4}{{\int_{2}^{3}{{12x - 18y\,dy}}\,dx}}\\ & = \int_{{ - 1}}^{4}{{\left. {\left( {12xy - 9{y^2}} \right)} \right|_2^3\,dx}} = \int_{{ - 1}}^{4}{{12x - 45\,dx}}\end{align*}\]Just remember that because we are integrating with respect to \(y\) in this step we treat all \(x\)’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 in the \(x\) integration and that is just a simple Calculus I integral. Here is that work.
\[\iint\limits_{R}{{12x - 18y\,dA}} = \int_{{ - 1}}^{4}{{12x - 45\,dx}} = \left. {\left( {6{x^2} - 45x} \right)} \right|_{ - 1}^4 = \require{bbox} \bbox[2pt,border:1px solid black]{{ - 135}}\]The same answer as from (a) which is what we should expect of course. The order of integration will not change the answer. One order may be easier/simpler than the other but the final answer will always be the same regardless of the order.