Question Number 215782 by universe last updated on 18/Jan/25
Answered by mr W last updated on 18/Jan/25
Commented by mr W last updated on 18/Jan/25
Commented by mr W last updated on 18/Jan/25
$${r}={a} \\ $$$${l}={r}\:\mathrm{sin}\:\theta \\ $$$${dS}={lrd}\theta={r}^{\mathrm{2}} \mathrm{sin}\:\theta\:{d}\theta \\ $$$$\frac{{S}}{\mathrm{16}}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {r}^{\mathrm{2}} \:\mathrm{sin}\:\theta\:{d}\theta={r}^{\mathrm{2}} \\ $$$$\Rightarrow{S}=\mathrm{16}{r}^{\mathrm{2}} =\mathrm{16}{a}^{\mathrm{2}} \:\:\:\checkmark \\ $$
Commented by mr W last updated on 18/Jan/25
$${dV}={l}\:\left({r}\:\mathrm{cos}\:\theta\right)^{\mathrm{2}} {d}\theta={r}^{\mathrm{3}} \mathrm{cos}^{\mathrm{2}} \:\theta\:\mathrm{sin}\:\theta\:{d}\theta \\ $$$$\frac{{V}}{\mathrm{16}}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {r}^{\mathrm{3}} \mathrm{cos}^{\mathrm{2}} \:\theta\:\mathrm{sin}\:\theta\:{d}\theta=\frac{{r}^{\mathrm{3}} }{\mathrm{3}} \\ $$$$\Rightarrow{V}=\frac{\mathrm{16}{r}^{\mathrm{3}} }{\mathrm{3}}=\frac{\mathrm{16}{a}^{\mathrm{3}} }{\mathrm{3}}\:\checkmark \\ $$
Commented by universe last updated on 18/Jan/25
$${thank}\:{you}\:{sir} \\ $$
Commented by ajfour last updated on 19/Jan/25
https://youtu.be/LAUiOcJsf48?si=N2W3CCYzG3qqjfC-
Commented by mr W last updated on 19/Jan/25
$$\mathrm{sin}\:\mathrm{22}.\mathrm{5}°=\sqrt{\frac{\mathrm{1}−\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}}{\mathrm{2}}}=\frac{\sqrt{\mathrm{2}−\sqrt{\mathrm{2}}}}{\mathrm{2}} \\ $$$${r}=\mathrm{1}+\frac{\mathrm{1}}{\mathrm{sin}\:\mathrm{22}.\mathrm{5}°}=\mathrm{1}+\frac{\mathrm{2}}{\:\sqrt{\mathrm{2}−\sqrt{\mathrm{2}}}}=\mathrm{1}+\sqrt{\mathrm{4}+\mathrm{2}\sqrt{\mathrm{2}}} \\ $$