In a scalar field I need to calculate the surface integral of this:
$$\iint_{\Sigma}\frac{d \sigma}{\sqrt{x^2+y^2+(z+R)^2}}$$ with $\Sigma$ the upper half of the sphere $x^2+y^2+z^2=R^2$
The formula for surface integrals we got is this: $$\iint_{\Sigma}f(x,y,z)d\sigma=\int du\int f(\phi_{1}(u,v), \phi_{2}(u,v), \phi_{3}(u,v))\cdot \left\|\frac{\partial \phi}{\partial u}\times \frac{\partial \phi}{\partial v}\right\|dv$$
I didn't do it with polar coordinates because I think it's possible with just carthesian coordinates(?), so for my paremeterisation I had $\phi = (\sqrt{R^2-y^2-z^2}, y, z)$. With this I could calculate the norm in that formula, which I got $$\left\|\frac{\partial \phi}{\partial y}\times \frac{\partial \phi}{\partial z}\right\|= \sqrt{1+\frac{y^2+z^2}{R^2-y^2-z^2}}$$ for.
Now for the integral itself I got: $$\int_{-R}^{R}dy\int_{0}^{R}\frac{dz}{\sqrt{2zR}}$$ but working this out confuses me, I think I have the wrong borders or so but the solution should be: $2\pi R(2-\sqrt{2})$. How do I get this pi? Should I have pi in my integration borders
Substituting $\phi(y,z)=\left\langle \pm\sqrt{R^2-y^2-z^2}, y, z \right\rangle$ into $f$ yields
$$f(\phi(y,z)) = \frac1{\sqrt{(R^2-y^2-z^2) + y^2 + (z+R)^2}} = \frac1{\sqrt{2R}} \frac1{\sqrt{R+z}}$$
The area element would be
$$d\sigma = \left\|\frac{\partial\phi}{\partial y} \times \frac{\partial\phi}{\partial z}\right\| \, dy\, dz = \frac R{\sqrt{R^2-y^2-z^2}} \, dy\, dz$$
Then the overall integral would be
$$2 \iint_\Sigma f(x,y,z)\,d\sigma = \sqrt{2R} \int_{-R}^R \int_0^R \frac{dz\,dy}{\sqrt{(R+z)(R^2-y^2-z^2)}} $$
where we multiply by $2$ to account for both halves of the hemisphere to either side of the plane $x=0$.