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Question Number 73663 by ajfour last updated on 14/Nov/19

	Answered by ajfour last updated on 15/Nov/19
	$See Q.73673 V=2∫((((ρ^2 (2φ))/2)×(−ds)) ρ=ssin α=ssin (π/6) = (s/2) s=((a(√3)−r(1−cos θ))/(cos α))=((2a)/3)(2+cos θ) −ds= ((rsin θ)/(cos α)) = (((2a)/3))sin θdθ φ=sin^(−1) {((rsin θcos α)/((a(√3)−r+rcos θ)sin α))} V=−(1/2)∫s^2 φds =(1/2)∫_0 ^( π) [((2a(2+cos θ))/3)]^2 sin^(−1) {((rsin θcos α)/((a(√3)−r+rcos θ)sin α))}(((2a)/3))sin θdθ V =((4a^3 )/(27))∫_(−1) ^( 1) (2+t)^2 sin^(−1) ((((√3)(√(1−t^2 )))/(2+t)))dt =((4a^3 )/(27))×((((√3)π)/2)+((15)/4))≈((4a^3 )/(27))×6.4707 ( Q.73689′s comment by MjS Sir on how to evaluate the integral). V_(cone) =((πa^3 )/(√3)) (V/V_(cone) ) = ((4(√3))/(27π))×6.4707 ≈ 0.5285 =((4(√3))/(27π))((((√3)π)/2)+((15)/4)) = ((2π+5(√3))/(9π)) . ⇒ 52.85% .$
	$S e e Q . 73673$ $V = 2 \int ((\frac{ρ^{2} (2 ϕ)}{2} \times (- d s))$ $ρ = s \sin α = s \sin \frac{π}{6} = \frac{s}{2}$ $s = \frac{a \sqrt{3} - r (1 - \cos θ)}{\cos α} = \frac{2 a}{3} (2 + \cos θ)$ $- d s = \frac{r \sin θ}{\cos α} = (\frac{2 a}{3}) \sin θ d θ$ $ϕ = \sin^{- 1} {\frac{r \sin θ \cos α}{(a \sqrt{3} - r + r \cos θ) \sin α}}$ $V = - \frac{1}{2} \int s^{2} ϕ d s$ $= \frac{1}{2} \int_{0}^{π} {[\frac{2 a (2 + \cos θ)}{3}]}^{2} \sin^{- 1} {\frac{r \sin θ \cos α}{(a \sqrt{3} - r + r \cos θ) \sin α}} (\frac{2 a}{3}) \sin θ d θ$ $V = \frac{4 a^{3}}{27} \int_{- 1}^{1} {(2 + t)}^{2} \sin^{- 1} (\frac{\sqrt{3} \sqrt{1 - t^{2}}}{2 + t}) d t$ $= \frac{4 a^{3}}{27} \times (\frac{\sqrt{3} π}{2} + \frac{15}{4}) \approx \frac{4 a^{3}}{27} \times 6.4707$ $(Q . 73689^{'} s c o m m e n t b y M j S S i r$ $o n h o w t o e v a l u a t e t h e i n t e g r a l) .$ $V_{c o n e} = \frac{π a^{3}}{\sqrt{3}}$ $\frac{V}{V_{c o n e}} = \frac{4 \sqrt{3}}{27 π} \times 6.4707 \approx 0.5285$ $= \frac{4 \sqrt{3}}{27 π} (\frac{\sqrt{3} π}{2} + \frac{15}{4})$ $= \frac{2 π + 5 \sqrt{3}}{9 π} .$ $\Rightarrow 52 . 85 % .$
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