Answer
$$\frac{7}{3}\left( {e - 1} \right)$$
Work Step by Step
$$\eqalign{
& \iint\limits_R {{e^{x/{y^2}}}}dxdy;\,\,\,\,\,\,\,\,\,\,\,\,\,\,1 \leqslant y \leqslant 2,\,\,\,\,\,\,0 \leqslant x \leqslant {y^2} \cr
& {\text{Replacing the limits for the region }}R \cr
& \int_1^2 {\int_0^{{y^2}} {{e^{x/{y^2}}}dx} dy} \cr
& = \int_1^2 {\left[ {\int_0^{{y^2}} {{e^{x/{y^2}}}dx} } \right]dy} \cr
& {\text{solve the inner integral treat }}y{\text{ as a constant}} \cr
& = {y^2}\int_0^{{y^2}} {{e^{x/{y^2}}}\left( {\frac{1}{{{y^2}}}} \right)dx} \cr
& {\text{integrate by using }}\int {{e^u}} du = {e^u} + C \cr
& = {y^2}\left( {{e^{x/{y^2}}}} \right)_0^{{y^2}} \cr
& {\text{evaluating the limits in the variable }}x \cr
& = {y^2}\left( {{e^{{y^2}/{y^2}}} - {e^{0/{y^2}}}} \right) \cr
& = {y^2}\left( {e - 1} \right) \cr
& {\text{then}} \cr
& \cr
& \int_1^2 {\left[ {\int_0^{{y^2}} {{e^{x/{y^2}}}dx} } \right]dy} = \int_1^2 {{y^2}\left( {e - 1} \right)dy} \cr
& = \left( {e - 1} \right)\int_1^2 {{y^2}dy} \cr
& {\text{integrating}} \cr
& = \left( {e - 1} \right)\left[ {\frac{{{y^3}}}{3}} \right]_1^2 \cr
& {\text{evaluating}} \cr
& = \left( {e - 1} \right)\left[ {\frac{{{2^3}}}{3} - \frac{{{1^3}}}{3}} \right] \cr
& {\text{simplifying}} \cr
& = \left( {e - 1} \right)\left( {\frac{8}{3} - \frac{1}{3}} \right) \cr
& = \frac{7}{3}\left( {e - 1} \right) \cr} $$
