Answer
$\dfrac{4}{3}$
Work Step by Step
Consider $Volume=\int_{0}^{1} \int_{x}^{2-x} (x^2+y^2) \ dy \ dx$
or, $=\int_{0}^{1} (2x^2+\dfrac{(2-x)^3}{3}-\dfrac{7x^3}{3}] \ dx$
or, $=\int_{0}^{1} (x^2y+\dfrac{y^3}{3}]_{x}^{2-x} \ dx$
or, $=[ \dfrac{2x^3}{3}+\dfrac{(2-x)^4}{12}-\dfrac{7x^4}{12}]_0^1$
or, $=\dfrac{2}{3}-\dfrac{1}{12}-\dfrac{1}{12}+\dfrac{(2)^4}{12}$
or, $=\dfrac{4}{3}$
