Double integral surface area with polar coordinates

Problem: Find the surface area of the paraboloid $\displaystyle z=4-x^2-y^2$ that lies above the xy-plane. Finding where the function crosses the xy plane- $\displaystyle 4=x^2+y^2$ taking the gradient- $\displaystyle \{2x,2y\}$ Useing the surface area formula- <br>$\displaystyle <br>\int\int\sqrt{(4x^2+4y^2+1)}dxdy$<br><br>This is where I'm unsure of how to continue. It seems that I could just set my bounds to be [-2,2] and [0,2] but this gives and incorrect answer.

The solution book says to convert the function to polar coordinates as such -

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$\displaystyle \int_{0}^{2\pi}\int_{0}^{2}\sqrt{(4r^2+1)}drd\thet a$,<br><br>I'm unsure of the bounds for theta, wouldn't 0-2\pi give the surface area of the whole sphere and not the area above $\displaystyle xy$? How would you compute the total surface area if this equation just gives it to you above the xy-plane?<br>