Answer
$\dfrac{1}{144}$
Work Step by Step
$\iiint_T xyz dV= \int_{0}^1 \int_{0}^{x} \int_{0}^{x-y} xyz dz dy dx$
$= \int_{0}^1 \int_{0}^{x}[\dfrac{xyz^2}{2}]_{0}^{x-y} dy dx$
$=\int_{0}^1 \int_{0}^{x}\dfrac{xy(x-y)^2}{2} dy dx$
$=\int_{0}^1 \int_{0}^{x}\dfrac{xy(x^2+y^2-2xy)}{2} dy dx$
$=\dfrac{1}{2}) \int_{0}^{1}\int_0^x [x^3y-2x^2y^2+xy^3] dy dx$
$=(1/2)\int^{0}_{1} [\dfrac{x^3y^2}{2}-\dfrac{2y^2x}{3}+\dfrac{xy^4}{4}]]_0^x dx$
$=(\dfrac{1}{2})[\dfrac{x^6}{(6)(12)}]_0^1$
$=(\dfrac{1}{2})[\dfrac{(1-0)^6}{72}]$
$=\dfrac{1}{144}$
