Answer
$\dfrac{3e-7}{6}$
Work Step by Step
Let us consider that $I=\iiint_E e^{z/y} dV$
$ I=\int_{0}^1 \int_{y}^{1} \int_{0}^{xy}(e^{z/y}) dz dx dy= \int_{0}^{1} \int_{y}^{1} [y(e^{z/y})]_{0}^{xy} dx dy=\int_{0}^1 [ye^{x}-xy]_y^1 dy$
Further, $\int_{0}^{3} yx+y^2-yx+y^2 dy dx= \int_0^{1}[(e-1)y-ye^y+y^2] dy$
and $[\dfrac{1}{2}(e-1)y^2-(ye^y-e^y) +(\dfrac{1}{3})y^3]_0^1=\dfrac{(e-1)}{2}-e+e+\dfrac{1}{3}-1=\dfrac{(e-1)}{2}+\dfrac{1}{3}-1$
or, $\iiint_E e^{z/y} dV=\dfrac{3e-7}{6}$
