Answer
$\dfrac{1}{6}$
Work Step by Step
Given: $ f(x, y, z)=(x+y)$
The iterated triple integral can be calculated as:
$\iiint_{\mathcal{B}} f(x,y,z)d V = \iiint_{\mathcal{B}} (x+y) d V \\
=\int_{0}^{1} \int_{0}^{x} \int_{y}^{x} (x+y)\ dz \ dy \ dx \\
= \int_{0}^{1} \int_{0}^{x} [(x+y+z)]_y^x \ dy dx \\=\int_{0}^{1} \int_0^x [x^2-y^2] dy \ dx \\=\int_0^1 [x^2(x)-\dfrac{x^3}{3}] \ dx \\=\int_0^1 \dfrac{2x^3}{3} \ dx \\=\dfrac{1}{6}$
