I have a cone $x^2+y^2=z^2$. It is cut off by a sphere $\left(x-a\right)^2+y^2+z^2=a^2$. I have to find the area of that portion which lies above the $xy$ plane. I parametrized the cone as $\Phi\left(z,\,\theta\right)=\left(z\cos\theta,\,z\sin\theta,\,z\right)$. Then I calculated norm of $\dfrac{\delta\Phi}{\delta z}\times\dfrac{\delta\Phi}{\delta \theta}$, which is $\sqrt2 z$. Now, due to the sphere, the constraint will be $z=a\cos\theta$.
Now, if I take $z$ going from $0$ to $a$, and $\theta$ going from $-\cos^{-1}(z/a)$ to $\cos^{-1}(z/a)$, and calculate $\displaystyle\iint\sqrt2z\,d\theta\, dz$, I get the answer.
However, I don't understand why we took $\theta$ in those limits. Can we take $z$ going from $0$ to $a\cos\theta$ and $\theta$ going from $0$ to $2\pi$?
The projection of the intersection of the cone and the sphere in the $xy$ plane is the disc:
$$D:=\{(x,y)\;|\; (x-\frac{a}{2})^2+y^2\le(\frac{a}{2})^2\} $$
which can be written with polar coordinates as:
$$ D:=\{(r,\theta)\;|\; -\frac{\pi}{2}\le \theta \le \frac{\pi}{2}, 0\le r \le a\cos{\theta} \} $$
But in your parametrization, $z=r$, so it should work if you take $\theta$ from $-\frac{\pi}{2}$ to $\frac{\pi}{2}$.