Answer
$1$
Work Step by Step
We have the triple integral:
$\int^1_0 \int^1_0 \int^1_0 (x^2+y^2+z^2) dz dy dx=\int^1_0 \int^1_0 (x^2+y^2+\dfrac{1}{3}) dy dx $
or, $=\int^1_0 (x^2+\dfrac{2}{3})$
Thus, we have
$\int^1_0 \int^1_0 \int^1_0 (x^2+y^2+z^2) dz \space dy \space dx =[\dfrac{x^3}{3}]_0^1+\dfrac{2}{3}=1$
