The surface area of S can be calculated by $\iint_R \sqrt{f_x^2 + f_y^2 + 1} dydx.$
In reference to this question surface area of a sphere above a cylinder, I first computed the square root in terms of cartesian coordinates before converting it into spherical coordinates: $$x = 2 \cos\theta\sin\varphi \\ y = 2 \sin\theta\sin\varphi $$ where $0 \le \theta\le2\pi$ and $0 \le \varphi \le\pi/4 $
and then multiplying it by the absolute value of the Jacobian, $r^2\sin\varphi $.
However this did not give me $4\sin(\varphi)$.
Why does this method work for polar coordinates but not for spherical coordinates?
My working out: $$ z = \sqrt{4 - x^2 - y^2} \\ f_x = \frac{-x}{\sqrt{4 - x^2 - y^2}} \\ f_y = \frac{-y}{\sqrt{4 - x^2 - y^2}} \\\sqrt{f_x^2 + f_y^2 + 1} = \sqrt{(\frac{-x}{\sqrt{4 - x^2 - y^2}})^2 + (\frac{-y}{\sqrt{4 - x^2 - y^2}})^2 + 1} = \sqrt{\frac{4}{4 - x^2 - y^2}} $$
Substituting the spherical coordinates into the above equation: $$ \sqrt{f_x^2 + f_y^2 + 1} = \sqrt{\frac{1}{1 - \sin^2(\varphi)}} = \sqrt{\frac{1}{\cos^2(\varphi)}} = \frac{1}{\cos\varphi} $$
Hence, we get $$\int_0^{2\pi} \int_0^{\pi/4} \frac{1}{\cos\varphi}*4\sin\varphi d\varphi d\theta$$