Integrate[x*x + y, {x, y} ∈ Disk[{0, 0}, Sqrt [2]]]
which returns $\pi$.
However, if it is school work, the "manual" way to do that is a change of variable like this one: $(x,y)=(r\cos \theta,r\sin \theta)$
After Jacobian calculation you get:
Integrate[r*(r*r*Cos[θ]^2 + r*Sin[θ]), {r, 0, Sqrt[2]}, {θ, 0, 2 π}]
Which (fortunately) also gives $\pi$
Update: Now if you want area of the surface, you can use this formula:
$$
\int_\Omega \| \left( \begin{array}{c} 1 \\ 0 \\ \partial_x f \end{array} \right)\times \left( \begin{array}{c} 0 \\ 1 \\ \partial_y f \end{array} \right) \| dxdy
$$
with $\Omega$ your centered disk of radius $\sqrt{2}$ and $f:(x,y)\rightarrow x^2+y$. You get, after change of variable
$$
\int_0^\sqrt{2}\int_0^{2\pi} r \sqrt{4 r^2 \cos ^2(\theta )+2}\ drd\theta
$$
which unfortunately involves Elliptic functions...
Under Mathematica:
f[x_, y_] := x*x + y
vx = D[{x, y, f[x, y]}, x];
vy = D[{x, y, f[x, y]}, y];
dareaCartesian =
Simplify[Norm[Cross[vx, vy], 2], Assumptions -> {x ∈ Reals}]
dareaPolar =
r*Simplify[dareaCartesian /. x -> r*Cos[θ] /. y -> r*Sin[θ] ]
area = Integrate[dareaPolar, {r, 0, Sqrt[2]}, {θ, 0, 2*Pi}]
N[area]
which prints:
$$
\sqrt{4 x^2+2}
$$
$$
r \sqrt{4 r^2 \cos ^2(\theta )+2}
$$
$$
\frac{2}{3} \sqrt{2} (5 EllipticK(-4)+3 EllipticE(-4))
$$
$$
12.212
$$