Find the surface area of the sphere inside the cylinder

Question

The given equations are of a sphere and cylinder respectively $$x^2+y^2+z^2=400$$ $$x^2+y^2=256$$ Solving the sphere equation for $z$ yields $$z=\sqrt{400-x^2-y^2}$$ Now to find $ds$ we take the partial derivative of both $x$ and $y$, then square them to get $$f_x=\frac{x^2}{400-x^2-y^2}$$ $$f_y=\frac{y^2}{400-x^2-y^2}$$ Then take those and adding one, and getting a common denominator, we get $$\frac{20}{\sqrt{400-x^2-y^2}}$$ Then converting the above fraction to polar coordinates and taking the iterated integral $$20\int_0^{16}\int_0^{2\pi}\frac{r}{\sqrt{400-r^2}}\,d\theta\,dr$$ and I get $320\pi$ but I guess I'm incorrect because the website we are using says I'm wrong. So where did I go wrong?

Answer 1

The surface area inside the cylinder is

$$A=2\int_{S} \sqrt{1+ z_x^2+z_y^2}\> dx dy = 2\int_{S} \frac{20}{\sqrt{20^2 -(x^2+y^2)}}\> dx dy \\ =2\int_0^{2\pi}\int_0^{16} \frac{20}{\sqrt{20^2 -r^2}}\> rdrd\theta =640\pi $$