Para encontrarmos a área da circunferência, realizaremos os cálculos abaixo:

\(\begin{align} & {{x}^{2}}+{{y}^{2}}+{{z}^{2}}={{a}^{2}} \\ & z=\sqrt{{{a}^{2}}-{{x}^{2}}-{{y}^{2}}} \\ & \frac{dz}{dx}=\frac{-x}{\sqrt{{{a}^{2}}-{{x}^{2}}-{{y}^{2}}}} \\ & \frac{dz}{dy}=\frac{-y}{\sqrt{{{a}^{2}}-{{x}^{2}}-{{y}^{2}}}} \\ & \\ & A=2\int_{{}}^{{}}{\int_{D}^{{}}{\sqrt{1+\frac{{{x}^{2}}}{{{a}^{2}}-{{x}^{2}}-{{y}^{2}}}+\frac{{{y}^{2}}}{{{a}^{2}}-{{x}^{2}}-{{y}^{2}}}}}}dA \\ & A=2a\int_{{}}^{{}}{\int_{D}^{{}}{\frac{dA}{\sqrt{{{a}^{2}}-{{x}^{2}}-{{y}^{2}}}}}} \\ & A=2a\int_{0}^{\pi /2}{\int_{0}^{a\cos \theta }{\frac{rdr}{\sqrt{{{a}^{2}}-{{r}^{2}}}}d\theta }} \\ & A=2a\int_{0}^{\pi /2}{\left( -\sqrt{{{a}^{2}}-{{r}^{2}}} \right)_{r=0}^{a\cos \theta }d\theta } \\ & A=2a\int_{0}^{\pi /2}{\left( 1-\sin \theta \right)d\theta } \\ & A=2{{a}^{2}}\left( \theta +\cos \theta \right)_{0}^{\pi /2} \\ & A=2\pi \left( {{a}^{2}}-a \right) \\ \end{align} \)

Portanto, a área da circunferência será de \(\begin{align} & A=2\pi \left( {{a}^{2}}-a \right) \\ \end{align} \).