Given region consist of $\triangle OAB$ as shown in fig.

Line AB equation can be given as

$$\dfrac {x-x_1}{y-y_1}=\dfrac {x_1-x_2}{y_1-y_2}\\ i.e. \dfrac {x-1}{y-2}=\dfrac {1-1}{2-0}\\ \therefore (x-1)2=0\times (y-2)\\ \therefore 2x=2\\ \therefore x=1$$

Thus the vertical strip will slide from $x=0 \space to\space x=1 $

1) Outer limit $x=0 \space to\space x=1 $

2) Inner limit y

a) Upper limit is equation of line OA

$$i.e.\dfrac {x-x_1}{y-y_1}=\dfrac {x_1-x_2}{y_1-y_2}\\ \therefore \dfrac {x-1}{y-2}=\dfrac {1-0}{2-0}\\ \therefore 2x-2=y-2\\ \therefore y=2x$$

b) Lower limit $y=0$

$$\therefore I=\int\limits_0^1\int\limits_0^{2x}(x^2+y^2)dydx\\ =\int\limits_0^1\Bigg[x^2y+\dfrac {y^3}3\Bigg]^{2x}_0dx\\ =\int\limits_0^1\Bigg[2x^3+\dfrac {8x^3}3\Bigg]dx\\ \therefore I=\Bigg[\dfrac {2x^4}4+\dfrac {8x^4}{12}\Bigg]_0^1\\ =\dfrac 12+\dfrac 8{12}\\ I=7/6$$