## Calculus (3rd Edition)

Please see the figure attached. The iterated integral in the opposite order is $\mathop \smallint \limits_1^{\rm{e}} \mathop \smallint \limits_0^{\ln y} f\left( {x,y} \right){\rm{d}}x{\rm{d}}y$ We have $\mathop \smallint \limits_0^1 \mathop \smallint \limits_{{{\rm{e}}^x}}^{\rm{e}} f\left( {x,y} \right){\rm{d}}y{\rm{d}}x$. From the limits of the integrals we obtain the domain description: ${\cal D} = \left\{ {\left( {x,y} \right)|0 \le x \le 1,{{\rm{e}}^x} \le y \le {\rm{e}}} \right\}$ Notice that this is a vertically simple region. Using this description we sketch the the domain of integration. Please see the figure attached. We change the order of integration such that the domain becomes a horizontally simple region. In this case, the lower and upper boundaries are $y=1$ and $y = {\rm{e}}$, respectively. Whereas, the left boundary is $x=0$. To find the right boundary, we use $y = {{\rm{e}}^x}$ and get $x = \ln y$. Thus, the new domain description: ${\cal D} = \left\{ {\left( {x,y} \right)|1 \le y \le {\rm{e}},0 \le x \le \ln y} \right\}$ So, the iterated integral in the opposite order is $\mathop \smallint \limits_1^{\rm{e}} \mathop \smallint \limits_0^{\ln y} f\left( {x,y} \right){\rm{d}}x{\rm{d}}y$