Calculate the area of the part of the cone $x^2+y^2=z^2$ with $z \geq 0$ that is inside of the sphere $x^2+y^2+z^2=2Rz$ where $R>0$
Attempt Notice that we should apply the formula $$\int_{S}f \dot dS=\int_{D} ||T_{u} \times T_{v}|| dudv$$ where $T_{u}$ and $T_{v}$ are the tangent vectors to our surface when $S$ is a parametric surface.\ We need find where is the intersection of the cone with the sphere, to can get superior límit and integrate the area of the cone from $0$ to the intersection point namely $P$.\ $$x^2+y^2+z^2=2Rz$$ completing the square $$x^2+y^2+(z-R)^2=R^2$$ This meaning is our sphere have center in $(0,0,R)$ with radius $R$.\ Now using the fact that $x^2+y^2=z^2$ and substitute it in $x^2+y^2+(z-R)^2=R^2$ $$z^2+(z-R)^2=R^2$$ $$(z^2-R^2)+(z-R)^2=0$$ $$(z-R)[(z+R)+(z-R)]=0$$ from here the intersection are $z=R$ and $z=0$ but since $z \geq 0$ and get that the intersection are in the plane $z=R$ that is $p=(0,0,R)$, and $Q=(0,0,0)$
Now parametrizing the surface $S$ given by the intersection we get $$x=rcos \theta$$ $$y=rsin \theta$$ $$z=r$$ where $r\in[0,R]$ and $\theta \in [0, 2\pi]$ and calculatin the tangent vectors $$T_{r}=(cos\theta)i+(\sin \theta)j+k$$ $$T_{\theta}=(-r \sin \theta )i+ (r \cos \theta )k+0k$$ $$T_{r}\times T_{\theta}=(-r \cos \theta)i+(-r\sin \theta )k+(r)k$$ $$||T_{r}\times T_{\theta}||=\sqrt{2}|r|$$ finally getting the integral given by the formula we get $$ \int_{0}^{2\pi} \int_{0}^{R}\sqrt{2}r dr d\theta=\sqrt{2}\pi R^2$$ Is my answer right or i make a mistake in the problem interpretation.
Your answer is correct but your intersection points are not, due to a typo you made I think.
$z^2+(z-R)^2=R^2 \implies (z^2-R^2) + (z-R)^2 = R^2$
$(z^2-R^2) + (z-R)^2 = R^2 \implies (z-R)(z + R + z - R) = 0$
$2z(z-R) = 0 \,$ gives you intersection points $z = 0, R$.
This makes sense as the cone is an inverted cone with vertex at $(0, 0, 0)$ and the sphere is of radius $R$ with center at $(0, 0, R)$ and so it touches the point $(0, 0, 0)$. It never goes to $z = - R$.
The cone is inside the sphere starting at $z = 0$ and comes out of the sphere at $z = R$ which is exactly at the hemisphere.
So, $0 \leq z \leq R$. Since you discarded $z = - R$ and took the integral from $z = 0$, everything eventually worked out.