## Calculus (3rd Edition)

$e -2$
The iterated integral can be calculated as: $\iint_{D} f(x,y) d A= \int_{1}^{e} \int_{(\ln y)^2}^{\ln y} (\ln y)^{-1} \ dx \ dy \\= \int_{1}^{e} [x (\ln y) ]_{(\ln y)^2}^{\ln y} \ dy\\= \int_{1}^{e} (1-\ln y) \ dy \\= [y]_1^e-[y \ln y-y]_1^e \\=(e-1)-(e \ln e-e -\ln (1)+1) \\=e -2$