I need to calculate the surface integral $\iint_S x \mathbf{i}+y \mathbf{j} \ dS$ over a hemisphere $z = \sqrt{1-x^2-y^2}$ with $z \geq 0.$
My attempt was the following: calculate $\frac{\partial}{\partial x} (\sqrt{1-x^2-y^2}) = \frac{-x}{\sqrt{1-x^2-y^2}}$ and $\frac{\partial}{\partial y} (\sqrt{1-x^2-y^2}) = \frac{-y}{\sqrt{1-x^2-y^2}}.$ Then I have the integral $$ \iint_S x \mathbf{i}+y \mathbf{j} \ dS = \iint_S \frac{-x^2-y^2}{\sqrt{1-x^2-y^2}} dx\ dy,$$ and that's it; I wanted to change the integral to polar coordinates, but if the radio of the hemisphere is $1$, the denominator of the integrand is $0,$ and I can't compute the integral. What am I doing wrong here?
It looks like you're getting spherical coordinates confused with polar coordinates. Polar coordinates are of the form $(r, \theta)$ and are used in two dimensions. They're handy when describing things like circles. In polar coordinates you do indeed have $r^2=x^2+y^2$. However, in your problem you're integrating over a sphere, not a circle. Since the radius of your sphere is 1, for a general point $(x, y, z)$ on the sphere you get $1=x^2+y^2+z^2$, not $1=x^2+y^2$. Spherical coordinates $(r, \theta, \phi)$ tend to be the most appropriate when integrating over the surface of a sphere. If you use the following spherical coordinate transformations I think you'll find your problem solved. $$x = r\sin\phi\cos\theta$$ $$y = r\sin\phi\sin\theta$$ $$z = r\cos\phi$$