Question Number 197027 by sonukgindia last updated on 06/Sep/23
Answered by MM42 last updated on 06/Sep/23
$${c}_{\mathrm{1}} :\:{x}^{\mathrm{2}} +{y}^{\mathrm{2}} =\mathrm{36}\:\:\:\:\&\:\:{c}_{\mathrm{2}} :\:\left({x}−\mathrm{10}\right)^{\mathrm{2}} +{y}^{\mathrm{2}} =\mathrm{36} \\ $$$${c}_{\mathrm{1}} ,{c}_{\mathrm{2}} \Rightarrow\:{A}=\mathrm{5}\:\:\:,\:\:{B}=\mathrm{6} \\ $$$${v}_{\mathrm{1}} =\frac{\mathrm{1}}{\mathrm{2}}×\frac{\mathrm{4}}{\mathrm{3}}\pi×\mathrm{6}^{\mathrm{3}} =\mathrm{144}\pi \\ $$$${v}_{\mathrm{2}} =\pi\int_{\mathrm{0}} ^{\mathrm{5}} \left(\mathrm{36}−{x}^{\mathrm{2}} \right){dx}=\pi\left(\mathrm{36}{x}−\frac{{x}^{\mathrm{3}} }{\mathrm{3}}\right)\mid_{\mathrm{0}} ^{\mathrm{5}} =\frac{\mathrm{415}\pi}{\mathrm{3}} \\ $$$${v}_{\mathrm{3}} =\pi\int_{\mathrm{5}} ^{\:\mathrm{6}} \left(\mathrm{36}−\left({x}^{\mathrm{2}} −\mathrm{10}\right)^{\mathrm{2}} \right){dx}=\pi\left(\mathrm{36}{x}−\frac{\left({x}−\mathrm{10}\right)^{\mathrm{3}} }{\mathrm{3}}\right)\mid_{\mathrm{5}} ^{\mathrm{6}} =\frac{\mathrm{47}\pi}{\mathrm{3}} \\ $$$${v}=\mathrm{307}\pi\:\checkmark \\ $$$$ \\ $$
Answered by MM42 last updated on 06/Sep/23
