Question Number 221057 by fantastic last updated on 23/May/25
Commented by fantastic last updated on 23/May/25
$${Please}\:{find}\:{the}\:{area}\:{of}\:{shaded}\:{region}\:{using}\:{calculus} \\ $$
Answered by mr W last updated on 24/May/25
Commented by mr W last updated on 24/May/25
$${blue}\:{hatched}\:{area}\:=\frac{\pi{R}^{\mathrm{2}} }{\mathrm{4}}−\frac{{R}^{\mathrm{2}} }{\mathrm{2}} \\ $$$${red}\:{shaded}\:{area}\:=\mathrm{2}×\:{blue}\:{hatched}\:{area}\: \\ $$$$\:\:\:\:\:\:\:=\mathrm{2}\left(\frac{\pi{R}^{\mathrm{2}} }{\mathrm{4}}−\frac{{R}^{\mathrm{2}} }{\mathrm{2}}\right)=\frac{\left(\pi−\mathrm{2}\right){R}^{\mathrm{2}} }{\mathrm{2}} \\ $$$$\:\:\:\:\:\:=\frac{\left(\pi−\mathrm{2}\right)×\mathrm{25}}{\mathrm{2}} \\ $$
Answered by Frix last updated on 24/May/25
$$\mathrm{We}\:\mathrm{need}\:\mathrm{only}\:\mathrm{one}\:\mathrm{circle}: \\ $$$${x}^{\mathrm{2}} +\left({y}+\frac{\mathrm{5}\sqrt{\mathrm{2}}}{\mathrm{2}}\right)^{\mathrm{2}} =\mathrm{5}^{\mathrm{2}} \\ $$$$\mathrm{Upper}\:\mathrm{half}: \\ $$$${f}\left({x}\right)=−\frac{\mathrm{5}\sqrt{\mathrm{2}}}{\mathrm{2}}+\sqrt{\mathrm{25}−{x}^{\mathrm{2}} } \\ $$$$\mathrm{Red}\:\mathrm{area}\:=\:\mathrm{4}\underset{\mathrm{0}} {\overset{\frac{\mathrm{5}\sqrt{\mathrm{2}}}{\mathrm{2}}} {\int}}{f}\left({x}\right)=\frac{\mathrm{25}\left(\pi−\mathrm{2}\right)}{\mathrm{2}} \\ $$
Commented by fantastic last updated on 24/May/25
$$\:{Thanks}\:{sir}. \\ $$