Answer
$1$
Work Step by Step
Consider $Volume=\int_{0}^{1} \int_{1-x}^{1} x^2 \ dy \ dx ....(1)$
The region is symmetric with respect to both the x and y axes. Therefore, we will multiply the volume of the one quadrant by $4$.
The bounds for $x$ are from 0 to 1 and the bounds for $y$ are from line $y=1-x$ to $1$.
Thus, the equation (1) becomes:
$Volume=4 \times \int_{0}^{1} \int_{1-x}^{1} x^2 \ dy \ dx $
or, $=4 \times \int_{0}^{1} (yx^2)_{1-x}^{1} x^2 \ dx $
or, $=4 \times \int_{0}^{1} (x^2-x^2+x^3) \ dx $
or, $=4 \times \int_{0}^{1} x^3 \ dx $
or, $=4 \times [\dfrac{x^4}{4}]_0^1 $
or, $=4 (\dfrac{1}{4}-0)$
or, $=1$
