Find the area of the cone $z=\sqrt{2x^2+2y^2}$ inscribed in the sphere $x^2+y^2+z^2=12^2$. 
Problem
Find the area of the cone $z=\sqrt{2x^2+2y^2}$ inscribed in the sphere $x^2+y^2+z^2=12^2$.

I think I have to solve this via the surface integral
$$\iint_S dS=\iint_T \|\Sigma_u\times \Sigma_v\| \, du \, dv$$
Where $\Sigma$ is a parametrization of the cone and $T=\operatorname {dom}\Sigma$.
Now
$$\Sigma(u,v):=(u,v,\sqrt{2u^2+2v^2})$$
Should work, but I have to figure out the domain, as $z\geq 0$ (and $0$ is achieved by $z$), we get that the domain has the form $[0,a]\times [0,b]$ (as $u=v=0$ is the only way to get $z=0$).
But I'm having problems getting the $a,b$: If I look for the intersection of the cone and the sphere I get a circle: $x^2+y^2=4\cdot 12$ and I'm not sure what to do next.
Could someone give a thorough explanation of how to solve this problem from the start? I'm pretty lost.

Also, on this site
I found the following formula, if $z=f(x,y)$
$$S=\iint_D\sqrt{f_x^2+f_y^2+1} \, dA$$
Then $S$ is the area of the surface $R=\operatorname{Im}_D(f)$.
Where does this formula come from? (Does it relate to surface integrals?)
 A: At the intersection 
$$
{z^2\over2}+z^2=12^2\implies z=4\sqrt6
$$
So we have to calculate the surface area of the cone from $z=0$ to $4\sqrt6$. We can use cylindrical polar to parametrize the cone :
$$
\vec r={z\over\sqrt2}\cos\phi\hat i+{z\over\sqrt2}\sin\phi\hat j+z\hat k,\qquad (z,\phi)\in(0,4\sqrt6)\times[0,2\pi)\\
\implies{\partial\vec r\over\partial z}=\left[{\cos\phi\over\sqrt2}\;{\sin\phi\over\sqrt2}\;1\right]^\top,\quad{\partial\vec r\over\partial \phi}=\left[-{z\sin\phi\over\sqrt2}\;{z\cos\phi\over\sqrt2}\;0\right]^\top\\
\implies\Big{|}{\partial\vec r\over\partial z}\times{\partial\vec r\over\partial \phi}\Big{|}={\sqrt3z\over2}
$$
So the required area is 
$$
\int_0^{4\sqrt6}\int_0^{2\pi}{\sqrt3z\over2}\,d\phi \,dz=48\sqrt3\pi
$$
A way to check that this is indeed the correct answer is to note that the right circular cone inside the sphere has height $4\sqrt6$ units and base-radius $4\sqrt3$ units and therefore has area
$$
\pi4\sqrt3\sqrt{4^2\times3+4^2\times6}=48\sqrt3\pi\text{ sq units}
$$

Suppose your surface is $z=f(x,y)$. So the surface is already parametrized as
$$
\vec r=x\hat i+y\hat j+f\hat k\implies \Big{|}{\partial \vec r\over\partial x}\times{\partial \vec r\over\partial y}\Big{|}=\Big{|}[1\,0\,f_x]^\top\times[0\,1\,f_y]^\top\Big{|}=\sqrt{f_x^2+f_y^2+1}
$$
So using your regular formula the area becomes
$$
\iint_D\sqrt{f_x^2+f_y^2+1}dxdy=\iint_D\sqrt{f_x^2+f_y^2+1}dA
$$
A: You could try to project the (double) cone and the sphere onto the $xz$-plane via substituting $y=0$, leaving you with the question of simply rotating the line $z=\sqrt{2}x$ around the $z$ axis to generate the (double) cone (where would this intersect with the circle the sphere would project on the $xz$-plane?). You can then proceed with the typical surface area of solid of rotation formulas. 
