By symmetry, I can calculate as $$2\int_0^{\pi/2} \int_0^{a\cos\theta} \frac a{\sqrt{a^2-r^2}} r\,dr\,d\theta= a^2(\pi-2)$$
but when I calculate as $$\int_{-\pi/2}^{\pi/2} \int_0^{a\cos\theta} \frac a{\sqrt{a^2-r^2}} r\,dr\,d\theta= a^2\pi$$
I thought the sphere $x^2+y^2+z^2=a^2$ consider as even function, both of them should get same result, thus, I don't know why the second solution is wrong.