Section 15.2 : Iterated Integrals
6. Compute the following double integral over the indicated rectangle.
\[\iint\limits_{R}{{x{y^2}\,\sqrt {{x^2} + {y^3}} \,dA}}\hspace{0.5in}R = \left[ {0,3} \right] \times \left[ {0,2} \right]\]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. 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 the order still really doesn’t matter all that much as both will require a substitution. However, we might get a little more simplification of the integrand if we first do the \(y\) substitution since that will eliminate the \({y^2}\) in front of the root. Of course, the \(x\) substitution would have eliminated the \(x\) in front of the root so either way we’ll get some simplification, but the \(y\) integration seems to offer a little bit more simplification.
So, here is the integral set up to do the \(y\) integration first.
\[\iint\limits_{R}{{x{y^2}\,\sqrt {{x^2} + {y^3}} \,dA}} = \int_{0}^{3}{{\int_{0}^{2}{{x{y^2}\,\sqrt {{x^2} + {y^3}} \,dy}}\,dx}}\] Show Step 2Okay, let’s do the \(y\) integration now.
\[\begin{align*}\iint\limits_{R}{{x{y^2}\,\sqrt {{x^2} + {y^3}} \,dA}} & = \int_{0}^{3}{{\int_{0}^{2}{{x{y^2}\,{{\left( {{x^2} + {y^3}} \right)}^{\frac{1}{2}}}\,dy}}\,dx}}\hspace{0.5in}u = {x^2} + {y^3}\,\,\,\, \to \,\,\,\,\,\,du = 3{y^2}\,dy\\ & = \int_{0}^{3}{{\left. {\left( {\frac{2}{9}x{{\left( {{x^2} + {y^3}} \right)}^{\frac{3}{2}}}} \right)} \right|_0^2\,dx}} = \int_{0}^{3}{{\frac{2}{9}x{{\left( {{x^2} + 8} \right)}^{\frac{3}{2}}} - \frac{2}{9}{x^4}\,dx}}\end{align*}\]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.
Be careful with substitutions in the first integration. We showed the substitution we used above as well as the differential. When computing the differential we need to differentiate the right side of the substitution with respect to \(y\) (since we are doing a \(y\) integration). In other words, we need to do a partial derivative of the right side and so the \({x^2}\) will differentiate to zero when differentiating with respect to \(y\)!
One of the bigger mistakes that students make here is to leave the \({x^2}\) in the differential or to “differentiate” it to “2\(x\)” which in turn causes the substitution to not work because there is then no way to get rid of the \({y^2}\) in front of the root.
Mistakes with Calculus I substitutions at this stage is one of the biggest issues that students have when first doing these kinds of integrals so you need to be very careful and always pay attention to which variable you are integrating with respect to and then differentiate your substitution with respect to the same variable.
Show Step 3Now all we need to take care of is the \(x\) integration and that is just Calculus I integral with a simple substitution (we’ll leave the substitution work to you to verify). Here is that work.
\[\begin{align*}\iint\limits_{R}{{x{y^2}\,\sqrt {{x^2} + {y^3}} \,dA}} & = \int_{0}^{3}{{\frac{2}{9}x{{\left( {{x^2} + 8} \right)}^{\frac{3}{2}}} - \frac{2}{9}{x^4}\,dx}}\\ & = \left. {\left( {\frac{2}{{45}}{{\left( {{x^2} + 8} \right)}^{\frac{5}{2}}} - \frac{2}{{45}}{x^5}} \right)} \right|_0^3 = \require{bbox} \bbox[2pt,border:1px solid black]{{\frac{2}{{45}}\left( {{{17}^{\frac{5}{2}}} - 243 - 128\sqrt 2 } \right) = 34.1137}}\end{align*}\]Do not get excited about “messy” answers. They will happen fairly regularly with these kinds of problems. In those cases it might be easier to reduce everything down to a decimal as we’ve done here.