I have to evaluate the are of the paraboloid $$z=7.2^2-x^2-y^2$$ which is above the plane $z=2.17^2$.
Since the area is $$S=\iint\sqrt{1+z_x^2+z_y^2}\;dA$$ then $$S=\iint\sqrt{1+4x^2+4y^2}\;dA=\iint\sqrt{1+4(x^2+y^2)}\;dA$$
If $z=2.17^2$, I get the circumference $$x^2+y^2=7.2^2-2.17^2$$ Therefore, in polar coordinates, the area of the surface will be $$S=\int_{0}^{2\pi}\int_{0}^{\sqrt{7.2^2-2.17^2}}\sqrt{1+4r^2}\cdot r dr d\theta$$ Is it correct?