Now, with this problem, the region will allow either order of integration without any real change of difficulty in the integration.

So, here are the limits for each order of integration that we could use.

\[\begin{array}{*{20}{c}}\begin{array}{c}0 \le x \le 3\\ - {x^3} \le y \le 0\end{array}&{\hspace{0.25in}{\mbox{OR}}\hspace{0.25in}}&\begin{array}{c} - 27 \le y \le 0\\ - {y^{\frac{1}{3}}} \le x \le 3\end{array}\end{array}\] Show Step 3

As noted above either order could be done without much real change in difficulty. However, it looks like the limits are a little nicer if we integrate with respect to \(y\) first. One of the \(y\) limits is zero and the \(x\) limits are smaller so it looks like this might be a little easier to deal with.

So, for this problem let’s integrate with respect to \(y\) first. Here is the integral set up for \(y\) integration first.

\[\iint\limits_{D}{{\frac{1}{{{y^{\frac{1}{3}}}\left( {{x^3} + 1} \right)}}\,dA}} = \int_{0}^{3}{{\int_{{ - {x^3}}}^{0}{{\frac{1}{{{y^{\frac{1}{3}}}\left( {{x^3} + 1} \right)}}\,dy}}\,dx}}\] Show Step 4

Here is the \(y\) integration.

\[\begin{align*}\iint\limits_{D}{{\frac{1}{{{y^{\frac{1}{3}}}\left( {{x^3} + 1} \right)}}\,dA}} & = \int_{0}^{3}{{\int_{{ - {x^3}}}^{0}{{\frac{1}{{{y^{\frac{1}{3}}}\left( {{x^3} + 1} \right)}}\,dy}}\,dx}}\\ & = \int_{0}^{3}{{\left. {\left( {\frac{{\frac{3}{2}{y^{\frac{2}{3}}}}}{{\left( {{x^3} + 1} \right)}}} \right)} \right|_{ - {x^3}}^0\,dx}}\\ & = \int_{0}^{3}{{ - \frac{{\frac{3}{2}{{\left( { - {x^3}} \right)}^{\frac{2}{3}}}}}{{\left( {{x^3} + 1} \right)}}\,\,dx}} = \int_{0}^{3}{{ - \frac{{\frac{3}{2}{x^2}}}{{\left( {{x^3} + 1} \right)}}\,\,dx}}\end{align*}\]

Be careful with the minus signs in this integration. The minus sign in front of the integrand comes from the evaluation process (recall we first evaluate at zero!). The minus sign in the numerator will be eliminated when we deal with the exponent (the numerator of the exponent is a two so we need to square things!).

Show Step 5

Now the \(x\) integration is a simple Calculus I substitution. We’ll leave the substitution work to you to verify. Here is the \(x\) integration.

\[\iint\limits_{D}{{\frac{1}{{{y^{\frac{1}{3}}}\left( {{x^3} + 1} \right)}}\,dA}} = \int_{0}^{3}{{ - \frac{{\frac{3}{2}{x^2}}}{{\left( {{x^3} + 1} \right)}}\,\,dx}} = \left. {\left( { - \frac{1}{2}\ln \left( {{x^3} + 1} \right)} \right)} \right|_0^3 = \require{bbox} \bbox[2pt,border:1px solid black]{{ - \frac{1}{2}\ln \left( {28} \right) = - 1.6661}}\]

Remember that \(\ln \left( 1 \right) = 0\)!