Calculus (3rd Edition)

$\dfrac{1}{16}$
Given: $f(x, y, z)=xyz$ The iterated triple integral can be calculated as: $\iiint_{\mathcal{w}} xyz \ d V = \iiint_{\mathcal{D}} (\int_0^1 xyz) d A \\ =\iint_{\mathcal{D}} (\dfrac{ xyz^2}{2})_0^1 d A \\ = \int_{0}^{1} (\int_{0}^{\sqrt {1-x^2}} \dfrac{xy}{2} \ dy) dx \\=\int_{0}^{1} \dfrac{x(1-x^2)}{4} \ dx \\=\int_0^1 (\dfrac{x-x^3}{4}) \ dx \\=[ \dfrac{x^2}{8}-\dfrac{x^4}{16}]_0^1 \ dx \\=\dfrac{1}{16}$