Plotting surfaces, it's pretty much clear that we will have the following figure.
Stokes theorem states that the path integral of a vector field around those dotted line is equal to the the surface vector integral of the curl of the field bound by surface. i.e.
$$ \oint_\gamma \vec F \cdot dr = \iint_\Omega \nabla \times \vec F \cdot \hat n d\sigma $$
First, let's calculate the left part, the path integral around along the curve going by route $OA \to AB \to BO$.
Along $OA$, the curve is $z=x^2$ and $y=0$.
$$\int_{OA} ( y^2 \hat i + 2xy \hat j + xz^2 \hat k) \cdot dr = \int_{OA} xz^2 \hat k \cdot (dx \hat i + dz \hat k ) = \int_{OA} xz^2 dz $$
Putting $z=x^2$ and $x = 0 \to 1$, we get
$$
\int_0^1 2x^6dx = \frac 2 7 \hspace{1cm} (1)$$
Now along AB, $z=1$, constant.
$$\int_{AB} ( y^2 \hat i + 2xy \hat j + xz^2 \hat k) \cdot dr = \int_{AB} ( y^2 \hat i + 2xy \hat j + xz^2 \hat k) \cdot ( dx \hat i + dy \hat j)$$
The circular curve can be parametrize $x = \cos(\theta)$ and $y = \sin(\theta)$, and we know that in along the curve $\theta = 0 \to \frac \pi 2 $, and that gives $dx = -\sin\theta d\theta$ and $dy = \cos\theta d\theta$, and simplifying we get
$$\int_0^{\frac \pi 2 } \left( -\sin^3 \theta + 2 \sin\theta \cos^2 \theta \right) d\theta = 0 \hspace{1cm} (2)$$
Along $OB$ we proceed in the same as along $OA$ but here $x=0$ and we get
$$\int_{BO} \vec F \cdot dr = 0 \hspace{1cm} (3)$$
Adding $(1), (2)$ and $(3)$, we get $
\displaystyle \oint_\gamma \vec F \cdot dr = \frac 2 7 $.
The right part::
The curl of the field turns out to be $- z^2 \hat j $ and the integral is
$$ \iint_\Sigma (-z^2 \hat j)\cdot \hat n ds $$
The surface $\displaystyle z = x^2 + y^2$ can be parameterized as $S = (r \cos\theta, r\sin\theta, r^2)$ where $r =0\to 1$ and $\theta = 0 \to \frac \pi 2$ and the surface integral can be evaluated as
$$\int_0^1 \int_0^{\frac \pi 2 } -r^4 \hat j \cdot \left( \frac{\partial S}{\partial r } \times \frac{\partial S}{\partial \theta } \right ) dr \; d\theta $$
The cross product can be calculated as $
-2 r^2 \cos (\theta ) \hat i -2 r^2 \sin (\theta ) \hat j + r \sin ^2(\theta )+r \cos ^2(\theta ) \hat k$
And the integral turns out as
$$ \int_0^1\int_0^{\frac \pi 2 } 2r^6 \sin ( \theta) d\theta dr = \frac 2 7 $$
ADDED:: For plotting surfaces, just plot all surfaces and take the surface of the parabloid that is bounded by $x=0, y=0, z=1$ on first quadrant, that is the one on the first quadrant which looks like the strip above.
