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Question Number 218813 by Spillover last updated on 15/Apr/25

Answered by nikif99 last updated on 15/Apr/25

Commented by Spillover last updated on 16/Apr/25

great work.thanks

$${great}\:{work}.{thanks}\: \\ $$

Answered by mr W last updated on 16/Apr/25

Commented by mr W last updated on 16/Apr/25

Commented by mr W last updated on 16/Apr/25

h_1 =R−13 V_W =((2πR^3 )/3)−πh_1 ^2 (R−(h_1 /3)) =((2πR^3 )/3)−π(R−13)^2 (R−((R−13)/3)) =((2πR^3 )/3)−((π(R−13)^2 (2R+13))/3) V_W =17^2 π(R−((17)/3))=((17^2 π(3R−17))/3) ((2πR^3 )/3)−((π(R−13)^2 (2R+13))/3)=((17^2 π(3R−17))/3) 39R^2 −867R+2716=0 ⇒R=((867+(√(867^2 −4×39×2716)))/(2×39))≈18.458 >17 ✓ 2V_W =πh^2 (R−(h/3)) πh^2 (R−(h/3))=2×17^2 π(R−((17)/3)) h^2 (3R−h)=2×17^2 (3R−17) ⇒h≈28.9997≈29 cm

$${h}_{\mathrm{1}} ={R}−\mathrm{13} \\ $$$${V}_{{W}} =\frac{\mathrm{2}\pi{R}^{\mathrm{3}} }{\mathrm{3}}−\pi{h}_{\mathrm{1}} ^{\mathrm{2}} \left({R}−\frac{{h}_{\mathrm{1}} }{\mathrm{3}}\right) \\ $$$$\:=\frac{\mathrm{2}\pi{R}^{\mathrm{3}} }{\mathrm{3}}−\pi\left({R}−\mathrm{13}\right)^{\mathrm{2}} \left({R}−\frac{{R}−\mathrm{13}}{\mathrm{3}}\right) \\ $$$$\:=\frac{\mathrm{2}\pi{R}^{\mathrm{3}} }{\mathrm{3}}−\frac{\pi\left({R}−\mathrm{13}\right)^{\mathrm{2}} \left(\mathrm{2}{R}+\mathrm{13}\right)}{\mathrm{3}} \\ $$$${V}_{{W}} =\mathrm{17}^{\mathrm{2}} \pi\left({R}−\frac{\mathrm{17}}{\mathrm{3}}\right)=\frac{\mathrm{17}^{\mathrm{2}} \pi\left(\mathrm{3}{R}−\mathrm{17}\right)}{\mathrm{3}} \\ $$$$\:\frac{\mathrm{2}\pi{R}^{\mathrm{3}} }{\mathrm{3}}−\frac{\pi\left({R}−\mathrm{13}\right)^{\mathrm{2}} \left(\mathrm{2}{R}+\mathrm{13}\right)}{\mathrm{3}}=\frac{\mathrm{17}^{\mathrm{2}} \pi\left(\mathrm{3}{R}−\mathrm{17}\right)}{\mathrm{3}} \\ $$$$\:\mathrm{39}{R}^{\mathrm{2}} −\mathrm{867}{R}+\mathrm{2716}=\mathrm{0} \\ $$$$\Rightarrow{R}=\frac{\mathrm{867}+\sqrt{\mathrm{867}^{\mathrm{2}} −\mathrm{4}×\mathrm{39}×\mathrm{2716}}}{\mathrm{2}×\mathrm{39}}\approx\mathrm{18}.\mathrm{458}\:>\mathrm{17}\:\checkmark \\ $$$$ \\ $$$$\mathrm{2}{V}_{{W}} =\pi{h}^{\mathrm{2}} \left({R}−\frac{{h}}{\mathrm{3}}\right) \\ $$$$\pi{h}^{\mathrm{2}} \left({R}−\frac{{h}}{\mathrm{3}}\right)=\mathrm{2}×\mathrm{17}^{\mathrm{2}} \pi\left({R}−\frac{\mathrm{17}}{\mathrm{3}}\right) \\ $$$${h}^{\mathrm{2}} \left(\mathrm{3}{R}−{h}\right)=\mathrm{2}×\mathrm{17}^{\mathrm{2}} \left(\mathrm{3}{R}−\mathrm{17}\right) \\ $$$$\Rightarrow{h}\approx\mathrm{28}.\mathrm{9997}\approx\mathrm{29}\:{cm} \\ $$

Commented by Spillover last updated on 16/Apr/25

great work.thanks

$${great}\:{work}.{thanks}\: \\ $$

Answered by Spillover last updated on 17/Apr/25

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