Surface area of a sphere of radius a using double integral area formula

Question

I have $A=\int^{2\pi}_0\int_0^a(\sqrt{1+\frac{x^2}{a^2-x^2-y^2}+\frac{y^2}{a^2-x^2-y^2}})$. The next step in my notes was $A=\int^{2\pi}_0\int_0^a(\frac{a}{\sqrt{a^2-x^2-y^2}})$I understand what's there from that point on to the answer but going between these two steps I'm not sure. The 1 in the square root part is giving me issues trying to deal with it.

Answer 1

Just add all the fractions in there. You will have $$\sqrt{\frac{a^2-x^2-y^2+x^2+y^2}{a^2-x^2-y^2}}=\frac{a}{\sqrt{a^2-x^2-y^2}}$$