Question Number 124651 by benjo_mathlover last updated on 05/Dec/20
Answered by liberty last updated on 05/Dec/20
$${for}\:−\mathrm{3}\leqslant{x}\leqslant\mathrm{3}\:\rightarrow\:{x}^{\mathrm{2}} +{y}^{\mathrm{2}} \:=\:\mathrm{9}\:;\:{g}\left({x}\right)=\sqrt{\mathrm{9}−{x}^{\mathrm{2}} } \\ $$$${for}\:\mathrm{3}\leqslant{x}\leqslant\mathrm{6}\rightarrow\left({x}−\frac{\mathrm{9}}{\mathrm{2}}\right)^{\mathrm{2}} +{y}^{\mathrm{2}} =\:\frac{\mathrm{9}}{\mathrm{4}} \\ $$$$;\:{g}\left({x}\right)=−\sqrt{\frac{\mathrm{9}}{\mathrm{4}}−\left({x}−\frac{\mathrm{9}}{\mathrm{2}}\right)^{\mathrm{2}} } \\ $$$${therefore}\:{g}\left({x}\right)=\begin{cases}{\sqrt{\mathrm{9}−{x}^{\mathrm{2}} }\:;\:−\mathrm{3}\leqslant{x}\leqslant\mathrm{3}}\\{−\sqrt{\frac{\mathrm{9}}{\mathrm{4}}−\left({x}−\frac{\mathrm{9}}{\mathrm{2}}\right)^{\mathrm{2}} }\:;\:\mathrm{3}\leqslant{x}\leqslant\mathrm{6}}\end{cases} \\ $$$${thus}\:{f}\left(\mathrm{6}\right)=\int_{−\mathrm{3}} ^{\:\mathrm{3}} {g}\left({x}\right){dx}\:+\:\int_{\mathrm{3}} ^{\:\mathrm{6}} {g}\left({x}\right)\:{dx}\: \\ $$