Let S be an octant of sphere of radius $2$ centered at the origin. Precisely we take the surface $z=f(x,y)=\sqrt{4-x^2-y^2}$ and restrict to $x\geq 0$ and $y\geq 0$ we want to compute the surface integral over S of the function $z^2$.
To solve it we can use symmetry which shows that:
$$\int(x^2)=\int(y^2)=\int(z^2)$$
But I haven’t understood why? Please geometrical explanation would be very appreciated Thank you in advance for your help!
Note, with the symmetry and spherical coordinates, the surface integral is,
$$S=\int_Az^2dA=\frac13\int(x^2+y^2+z^2)dA=\frac13\int_0^{\pi/2}\int_0^{\pi/2}r^4\sin\theta d\theta d\phi=\frac{8\pi}3$$