Given surface integral
$$ \iint_{S(V)} y^2dzdx + z^2dxdy $$
Where $ S(V) $ is
$$ x^2 + y^2 + z^2 = 4 $$ $$ x \geq 0,\quad y \geq 0, \quad z \geq 0$$
If I haven't made any mistake, using Gauss–Ostrogradsky formula outer surface area should be $ 4\pi + \frac{32}{3}$.
I'm having issue with using projection method; essentially I find it confusing how projection should be done and the respective double integrals obtained. I'm not looking for exact solution, just suggestions on how to approach this problem.