Question Number 73663 by ajfour last updated on 14/Nov/19
Answered by ajfour last updated on 15/Nov/19
![See Q.73673 V=2∫((((ρ^2 (2φ))/2)×(−ds)) ρ=ssin α=ssin (π/6) = (s/2) s=((a(√3)−r(1−cos θ))/(cos α))=((2a)/3)(2+cos θ) −ds= ((rsin θ)/(cos α)) = (((2a)/3))sin θdθ φ=sin^(−1) {((rsin θcos α)/((a(√3)−r+rcos θ)sin α))} V=−(1/2)∫s^2 φds =(1/2)∫_0 ^( π) [((2a(2+cos θ))/3)]^2 sin^(−1) {((rsin θcos α)/((a(√3)−r+rcos θ)sin α))}(((2a)/3))sin θdθ V =((4a^3 )/(27))∫_(−1) ^( 1) (2+t)^2 sin^(−1) ((((√3)(√(1−t^2 )))/(2+t)))dt =((4a^3 )/(27))×((((√3)π)/2)+((15)/4))≈((4a^3 )/(27))×6.4707 ( Q.73689′s comment by MjS Sir on how to evaluate the integral). V_(cone) =((πa^3 )/(√3)) (V/V_(cone) ) = ((4(√3))/(27π))×6.4707 ≈ 0.5285 =((4(√3))/(27π))((((√3)π)/2)+((15)/4)) = ((2π+5(√3))/(9π)) . ⇒ 52.85% .](Q73719.png)
$${See}\:{Q}.\mathrm{73673} \\ $$$${V}=\mathrm{2}\int\left(\left(\frac{\rho^{\mathrm{2}} \left(\mathrm{2}\phi\right)}{\mathrm{2}}×\left(−{ds}\right)\right)\right. \\ $$$$\:\rho={s}\mathrm{sin}\:\alpha={s}\mathrm{sin}\:\frac{\pi}{\mathrm{6}}\:=\:\frac{{s}}{\mathrm{2}} \\ $$$$\:{s}=\frac{{a}\sqrt{\mathrm{3}}−{r}\left(\mathrm{1}−\mathrm{cos}\:\theta\right)}{\mathrm{cos}\:\alpha}=\frac{\mathrm{2}{a}}{\mathrm{3}}\left(\mathrm{2}+\mathrm{cos}\:\theta\right) \\ $$$$\:−{ds}=\:\frac{{r}\mathrm{sin}\:\theta}{\mathrm{cos}\:\alpha}\:=\:\left(\frac{\mathrm{2}{a}}{\mathrm{3}}\right)\mathrm{sin}\:\theta{d}\theta \\ $$$$\:\phi=\mathrm{sin}^{−\mathrm{1}} \left\{\frac{{r}\mathrm{sin}\:\theta\mathrm{cos}\:\alpha}{\left({a}\sqrt{\mathrm{3}}−{r}+{r}\mathrm{cos}\:\theta\right)\mathrm{sin}\:\alpha}\right\} \\ $$$$\:{V}=−\frac{\mathrm{1}}{\mathrm{2}}\int{s}^{\mathrm{2}} \phi{ds} \\ $$$$\:\:\:\:=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\:\:\pi} \left[\frac{\mathrm{2}{a}\left(\mathrm{2}+\mathrm{cos}\:\theta\right)}{\mathrm{3}}\right]^{\mathrm{2}} \mathrm{sin}^{−\mathrm{1}} \left\{\frac{{r}\mathrm{sin}\:\theta\mathrm{cos}\:\alpha}{\left({a}\sqrt{\mathrm{3}}−{r}+{r}\mathrm{cos}\:\theta\right)\mathrm{sin}\:\alpha}\right\}\left(\frac{\mathrm{2}{a}}{\mathrm{3}}\right)\mathrm{sin}\:\theta{d}\theta \\ $$$${V}\:=\frac{\mathrm{4}{a}^{\mathrm{3}} }{\mathrm{27}}\int_{−\mathrm{1}} ^{\:\:\mathrm{1}} \left(\mathrm{2}+{t}\right)^{\mathrm{2}} \mathrm{sin}^{−\mathrm{1}} \left(\frac{\sqrt{\mathrm{3}}\sqrt{\mathrm{1}−{t}^{\mathrm{2}} }}{\mathrm{2}+{t}}\right){dt} \\ $$$$\:\:=\frac{\mathrm{4}{a}^{\mathrm{3}} }{\mathrm{27}}×\left(\frac{\sqrt{\mathrm{3}}\pi}{\mathrm{2}}+\frac{\mathrm{15}}{\mathrm{4}}\right)\approx\frac{\mathrm{4}{a}^{\mathrm{3}} }{\mathrm{27}}×\mathrm{6}.\mathrm{4707} \\ $$$$\:\left(\:{Q}.\mathrm{73689}'{s}\:\:{comment}\:{by}\:{MjS}\:{Sir}\right. \\ $$$$\left.\:\:\:{on}\:{how}\:{to}\:{evaluate}\:{the}\:{integral}\right). \\ $$$${V}_{{cone}} =\frac{\pi{a}^{\mathrm{3}} }{\sqrt{\mathrm{3}}} \\ $$$$\:\:\frac{{V}}{{V}_{{cone}} }\:=\:\frac{\mathrm{4}\sqrt{\mathrm{3}}}{\mathrm{27}\pi}×\mathrm{6}.\mathrm{4707}\:\approx\:\mathrm{0}.\mathrm{5285} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{4}\sqrt{\mathrm{3}}}{\mathrm{27}\pi}\left(\frac{\sqrt{\mathrm{3}}\pi}{\mathrm{2}}+\frac{\mathrm{15}}{\mathrm{4}}\right) \\ $$$$\:\:\:\:=\:\frac{\mathrm{2}\pi+\mathrm{5}\sqrt{\mathrm{3}}}{\mathrm{9}\pi}\:. \\ $$$$\:\:\Rightarrow\:\:\mathrm{52}.\mathrm{85\%}\:. \\ $$