Find the area of the part of the sphere $x^2+y^2+z^2=4$ cut off by the cone $z = \sqrt{x^2+y^2}$
First of all, I want to set up the integral only on the part of the sphere that lies inside the cone. In order to do that, I need to find the circle of intersection of these two objects.
I am using cylindrical coordinates.
sphere: $r^2 + z^2 = 4$
cone: $z = r$
Solving for the radius: $r = \sqrt2$
Angle: $0 \le \theta \le 2\pi$
And so the area of integration will be: $$A = \{(r, \theta): 0 \le r \le 2\pi \land 0 \le r \le \sqrt 2 \}$$ And the upper surface can be parametrized as $$T(r, \theta) = (r, \theta, \sqrt{4-r^2})$$ And now, we only need to set up the surface integral over $A$, remembering about the Jacobian $r$: $$\iint_A \|\frac{\partial z}{\partial r} \times \frac{\partial z}{\partial \theta}\|rdrd\theta$$
Is my method correct?