Integrate $x^2+y^2$ over the upper hemisphere of radius $a>0$ centered at $(0,0,0)$.
$\textbf{Edit}$
Consider the parametrization of the upper hemisphere given by
$$X(\phi, \theta) = (a \sin \phi \cos \theta, a \sin \phi \sin \theta, a \cos \phi)$$
where $ 0 \leq \theta \leq 2 \pi, 0 \leq \phi \leq \pi/2$. Then $f(X( \phi, \theta)) = a^2 \sin^2 \phi $ and $N(X(\phi,\theta)) = a^2 \sin^2 \phi$ hence we have,
$$\iint\limits_{S} \psi \ d\sigma = \int_{0}^{2\pi} \int_{0}^{\frac{\pi}{2}} a^4\sin^3 \phi \ d\phi\ d\theta = \frac{4\pi a^2}{3}$$
No. You can integrate scalar fields over surfaces, by doing: $$\int_S f\,{\rm d}S = \int_D f(X(u,v)) \|X_u \times X_v\|\,{\rm d}u\,{\rm d}v,$$where $X: D \subset \Bbb R^2 \to S$ is a convenient parametrization.
You were supposed to compute: $$\int_0^{2\pi}\int_{0}^{\pi/2} a^2 \sin^2\phi \,(a \sin\phi)\,{\rm d}\phi\,{\rm d}\theta,$$which gives the expected result.