Answer
$ \dfrac{1}{2} (e^{16}-17) \approx 4.44 \times 10^{6}$
Work Step by Step
We are given that the domain $D$ is bounded by $y=x; y=4 $ and $x=0$
We can define the area of surface as: $\iint_{D} y^2 e^{xy} dA=\int_{0}^{4} \int_{0}^{y} y^2 e^{xy} dx \ dy \\= \int_{0}^{4} [ye^{xy}]_{0}^{y} \ dy \\ = \int_{0}^{4} [ye^{y^2}-y] \ dy \\= [\dfrac{1}{2}e^{y^2} -\dfrac{y^2}{2}]_0^4 \\ = \dfrac{1}{2} (e^{4^2} -(4)^2-(1-0)] \\= \dfrac{1}{2} (e^{16}-17) \approx 4.44 \times 10^{6}$
