Answer
$\dfrac{27}{2}$
Work Step by Step
$\iiint_E y dV=\int_{0}^3 \int_{0}^{x} \int_{x-y}^{x+y} dz dy dx= \int_{0}^3 \int_{0}^{x} [yz]_{x-y}^{x+y} dy dx \\
=\int_{0}^1 \int_{0}^{1} [xy[e^{2-x^2-y^2}]-xy(1)] dy dx\\
= \int_{0}^{3} [yx+y^2-yx+y^2] dy dx\\
=(2/3)[ \int_0^3 y^3]_0^x\\=\dfrac{2}{3} \int_0^3 x^3\\
=\dfrac{27}{2}$
