# Surface area integral

How do you find the surface area of the part of a circular paraboloid $z=x^2+y^2$ that lies inside the cylinder $x^2+y^2=4$
$z_{x}=2x, \;\ z_{y}=2y, \;\ z_{x}^{2}+z_{y}^{2}+1=4x^{2}+4y^{2}+1$
$\int_{0}^{2\pi}\int_{0}^{2}r\sqrt{4r^{2}+1}drd{\th eta}$