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**Opalg** As **Danny** says, you should write the equation of the sphere as $\displaystyle z = \sqrt{4-x^2-y^2}$. To find the region over which to integrate, you need to know where the sphere meets the cone, namely where $\displaystyle \sqrt{x^2+y^2} = \sqrt{4-x^2-y^2}$ or in other words $\displaystyle x^2+y^2=2$. So the integral will be over the region $\displaystyle x^2+y^2\leqslant2$, and the surface area is given by the integral $\displaystyle \iint_{x^2+y^2\leqslant2}\sqrt{1+z_x^2+z_y^2}dxdy = \iint_{x^2+y^2\leqslant2}\frac2{\sqrt{4-x^2-y^2}}dxdy $. This cries out for conversion to polar coordinates, and it then becomes $\displaystyle \int_0^{2\pi}\!\!\int_0^{\sqrt2}\!\!\frac2{\sqrt{4-r^2}}\,rdrd\theta$. From there on, it's easy.