Answer
$\dfrac{11}{6} $
Work Step by Step
Given: $ f(x, y, z)=z$
The iterated triple integral can be calculated as:
$\iiint_{\mathcal{w}} z \ d V =\int_0^1 \int_{x^2}^{2} \int_{x-y}^{x+y} (z) \ dz dy dx \\=\int_0^1 \int_{x^2}^{2} [\dfrac{z^2}{2}] dy dx\\=\int_0^1 [\int_{x^2}^2 (2xy) \ dy ] dx \\=\int_0^1 [xy^2]_{x^2}^2\\=\int_0^1 (4x-x^5) \ dx\\=[2x^2-\dfrac{x^6}{6}]_0^1\\=\dfrac{11}{6} $
