Calculate the surface integral $f(x,y,z)=x^2+y^2+z^2$ in the upper hemisphere of the sphere $x^2+y^2+(z-1)^2=1$
I tried to compute the value of the surface integral $\iint_S{F.n} dS$ with the explicit representation z=$\sqrt{-x^2-y^2-2z}$ but the term $2z$ brings me conflict.
I don´t know if it should be done like this or it is convenient to change coordinates.
I would appreciate someone helping me. Since it is for a job that is delivered today and I am stuck.
Hoping that by "upper" hemisphere, you mean "with respect to $z$:
Let $$ Q = \int_S x^2+y^2+z^2 ~ dA $$, where $S$ is the upper hemisphere of the sphere $x^2+y^2+(z-1)^2=1$.
Substituting $u = z-1; u+1 = z; du = dz$, we get $$ Q = \int_T x^2+y^2+(u+1)^2 ~ dA $$, where $T$ is the upper hemisphere of the sphere $x^2+y^2+u^2=1$, which I'll rewrite as $$ Q = \int_U x^2+y^2+(z+1)^2 ~ dA, $$ where $U$ is the upper hemisphere of the sphere $x^2+y^2+z^2=1$.
Expanding the integrand, we get \begin{align} Q &= \int_U x^2+y^2+z^2 + 2z + 1 ~ dA \\ &= \int_U x^2+y^2+z^2 ~dA+ \int_U 2z~dA + \int_U 1 ~ dA \\ &= \int_U 1 ~dA+ \int_U 2z~dA + \int_U 1 ~ dA \\ &= \int_U 2 ~dA+ \int_U 2z~dA\\ &= 4\pi+ 2\int_U z~dA \end{align}
The latter integral is the average height of a point on the hemisphere characterized by $z \ge 0$, muktiplied by the area ($\2\pi$) of that hemisphere.
Because horizontal radial projection from the $z$-axis to a cylinder surrounding the sphere, i.e., the map $$ (x, y, z) \mapsto (\frac{x}{\sqrt{x^2 + y^2}}, \frac{y}{\sqrt{x^2 + y^2}}, z) $$ is area-preserving (which requires a little proof, but was mentioned by Boy in his dissertation somewhere around 1895-1905, so it's a pretty well-established result!), this is the same as the average-height-times-area of a point on a cylinder of radius $1$ and height $1$, i.e., $\frac{1}{2} \cdot 2\pi = \pi$.
So the overall value of your integral is $6\pi$.