Question

Explain why it would be difficult to evaluate the double integrals in Exercises 18 and 19 as iterated integrals.

$$\begin{gathered} \int {\int }_{\mathfrak{R}}\cos \left(xy\right)dA \\ where\mathfrak{R}=\left\{\left(x,y\right)I\frac{\mathrm{\pi }}{4}\le x\le \frac{\mathrm{\pi }}{2}and\frac{\mathrm{\pi }}{2}\le y\le \mathrm{\pi }\right\} \end{gathered}$$

Short answer

The double integral is difficult to evaluate using iterated integral.

Step-by-step solution

Step 1. Given Information.

The iterated integral is,

$$\begin{gathered} \int {\int }_{\mathfrak{R}}\cos \left(xy\right)dA \\ where\mathfrak{R}=\left\{\left(x,y\right)I\frac{\mathrm{\pi }}{4}\le x\le \frac{\mathrm{\pi }}{2}and\frac{\mathrm{\pi }}{2}\le y\le \mathrm{\pi }\right\} \end{gathered}$$

Step 2. Explanation.

Evaluating the first integral with respect to $y$and treating $x$as a constant.

$$\begin{gathered} \int {\int }_{\mathfrak{R}}\cos \left(xy\right)dA \\ ={\int }_{\raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$4$}\right.}^{\raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}{\int }_{\raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}^{\mathrm{\pi }}\cos \left(xy\right)dydx \\ ={\int }_{\raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$4$}\right.}^{\raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\left({\int }_{\raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}^{\mathrm{\pi }}\cos \left(xy\right)dy\right)dx \\ ={\int }_{\raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$4$}\right.}^{\raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}{\left[\frac{1}{x}\sin \left(xy\right)\right]}_{y=\raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}^{y=\mathrm{\pi }}dx \\ ={\int }_{\raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$4$}\right.}^{\raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\left[\frac{1}{x}\sin \left(\mathrm{\pi x}\right)-\frac{1}{x}\sin \left(\frac{\mathrm{\pi }}{2}x\right)\right]dx \end{gathered}$$

Now the resultant function $\frac{1}{x}\sin \left(\mathrm{\pi x}\right)-\frac{1}{x}\sin \left(\frac{\mathrm{\pi }}{2}x\right)$does not have a simple anti-derivative to evaluate the next step which is integrating with respect to variable$x$. The double integral is difficult to evaluate using iterated integral.