With this problem the region can be dealt with either order of integration. The integral on the other hand can’t. We simply cannot integrate \(y\) first as there is no \({y^2}\) in front of the cosine we’d need to do the substitution. So, we’ll have no choice but to do the \(x\) integration first.

So, here are the limits for this integral.

\[\begin{array}{c}0 \le y \le 2\\ 0 \le x \le 2\sqrt y \end{array}\] Show Step 3

Here is the integral set up for \(x\) integration first.

\[\iint\limits_{D}{{5{x^3}\cos \left( {{y^3}} \right)\,dA}} = \int_{0}^{2}{{\int_{0}^{{2\sqrt y }}{{5{x^3}\cos \left( {{y^3}} \right)\,dx}}\,dy}}\] Show Step 4

Here is the \(x\) integration.

\[\begin{align*}\iint\limits_{D}{{5{x^3}\cos \left( {{y^3}} \right)\,dA}} & = \int_{0}^{2}{{\int_{0}^{{2\sqrt y }}{{5{x^3}\cos \left( {{y^3}} \right)\,dx}}\,dy}}\\ & = \int_{0}^{2}{{\left. {\left( {\frac{5}{4}{x^4}\cos \left( {{y^3}} \right)} \right)} \right|_0^{2\sqrt y }\,dy}}\\ & = \int_{0}^{2}{{20{y^2}\cos \left( {{y^3}} \right)\,dy}}\end{align*}\] Show Step 5

Note that while we couldn’t do the \(y\) integration first we can do it now. The \({y^2}\) we need for the substitution is now there after doing the \(x\) integration.

Here is the \(y\) integration.

\[\begin{align*}\iint\limits_{D}{{5{x^3}\cos \left( {{y^3}} \right)\,dA}} & = \int_{0}^{2}{{20{y^2}\cos \left( {{y^3}} \right)\,dy}}\\ & = \left. {\left( {\frac{{20}}{3}\sin \left( {{y^3}} \right)} \right)} \right|_0^2 = \require{bbox} \bbox[2pt,border:1px solid black]{{\frac{{20}}{3}\sin \left( 8 \right) = 6.5957}}\end{align*}\]

Remember to have your calculator set to radians when computing to decimals.