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Question Number 73673 by ajfour last updated on 14/Nov/19

Commented by ajfour last updated on 14/Nov/19

$Q 73503 (a n o t h e r w a y a t r y . .)$
Commented by ajfour last updated on 14/Nov/19

Commented by mind is power last updated on 15/Nov/19
$we can use parametric surface if f:IR^2 →IR^3 a surface paramtric f(u,v)=(a(u,v),b(u,v),c(u,v)) parametric of surface we will ginde parametrisation of ilimunated light A(u)=∫∫_A ∣(∂f/∂u)(u,v)∧(∂f/∂v)(u,v)∣dudv i show steps 1 put repert 2,equation of semis disc ∴ D ∵wich light pass throw 3,∀a∈D→a′∈A∩C since light horizontal C is Cone D=(0,pcos(t),psin(t)),t∈[(π/2),((3π)/2)],p∈[0,(a/(√3))] ⇒aa′^→ =s.i^→ i^→ =(1,0,0) use C={(x,y,z)∈R^3 }∣x^2 +y^2 =z^2 tg^2 (θ) we find s ther s (t,p) a′(t,p) is our paramtrisation of A our f(u,v) below i will poste it this methode use somm topological object surfaces$
$w e c a n u s e p a r a m e t r i c s u r f a c e$ $i f f : {I R}^{2} \to {I R}^{3} a s u r f a c e p a r a m t r i c$ $f (u, v) = (a (u, v), b (u, v), c (u, v)) p a r a m e t r i c o f s u r f a c e$ $w e w i l l g i n d e p a r a m e t r i s a t i o n o f i l i m u n a t e d l i g h t$ $A (u) = \int \int_{A} ∣ \frac{\partial f}{\partial u} (u, v) \land \frac{\partial f}{\partial v} (u, v) ∣ d u d v$ $i s h o w s t e p s$ $1 p u t r e p e r t$ $2, e q u a t i o n o f s e m i s d i s c ∴ D ∵ w i c h l i g h t p a s s t h r o w$ $3, \forall a \in D \to a^{'} \in A \cap C s i n c e l i g h t h o r i z o n t a l$ $C i s C o n e$ $D = (0, p c o s (t), p s i n (t)), t \in [\frac{π}{2}, \frac{3 π}{2}], p \in [0, \frac{a}{\sqrt{3}}]$ $\Rightarrow a a \vec{^{'}} = s . \vec{i}$ $\vec{i} = (1, 0, 0)$ $u s e C = {(x, y, z) \in R^{3}} ∣ x^{2} + y^{2} = z^{2} {t g}^{2} (θ)$ $w e f i n d s t h e r s (t, p)$ $a^{'} (t, p) i s o u r p a r a m t r i s a t i o n o f A o u r f (u, v) b e l o w$ $i w i l l p o s t e i t t h i s m e t h o d e u s e s o m m t o p o l o g i c a l o b j e c t$ $s u r f a c e s$
	Answered by ajfour last updated on 15/Nov/19
	$Let P be a point on circumference of circle y^2 +(z−r)^2 =r^2 P (0, rsin θ, r−rcos θ) Its shadow on slant surface of cone S(ssin αcos φ, ssin αsin φ, a(√3)−scos α) ⇒ ssin αsin φ=rsin θ (α=30°) a(√3)−scos α=r−rcos θ s=((a(√3)−r(1−cos θ))/(cos α)) sin φ=((rsin θcos α)/((a(√3)−r+rcos θ)sin α)) sin φ=((sin θ)/((√3)−1+cos θ)) ds=((−rsin θdθ)/(cos α)) (s/(2a))=(ρ/a) ⇒ ρ=(s/2) A=∫ρ(2φ)(−ds) = ∫(s)(−ds)φ =∫_0 ^( π) (((a(√3)−r(1−cos θ))/(cos α)))(((rsin θdθ)/(cos α)))(sin^(−1) {((rsin θcos α)/((a(√3)−r+rcos θ)sin α))}) =((4a^2 )/3)∫_0 ^( π) [1−(1/3)(1−cos θ)][sin^(−1) (((sin θ)/((√3)−(1/(√3))+((cos θ)/(√3)))))]sin θdθ =((4a^2 )/9)∫_(−1) ^( 1) (2+t)sin^(−1) ((((√3)(√(1−t^2 )))/(2+t)))dt A=((4a^2 )/9)((π/(√3))+(3/2))≈1.4728a^2 . A=((2a^2 (2π(√3)+9))/(27)) . (see Q. 73689 on how to solve the integral,special thanks to MjS Sir)$
	$L e t P b e a p o i n t o n c i r c u m f e r e n c e$ $o f c i r c l e y^{2} + {(z - r)}^{2} = r^{2}$ $P (0, r \sin θ, r - r \cos θ)$ $I t s s h a d o w o n s l a n t s u r f a c e o f c o n e$ $S (s \sin α \cos ϕ, s \sin α \sin ϕ, a \sqrt{3} - s \cos α)$ $\Rightarrow s \sin α \sin ϕ = r \sin θ (α = 30 °)$ $a \sqrt{3} - s \cos α = r - r \cos θ$ $s = \frac{a \sqrt{3} - r (1 - \cos θ)}{\cos α}$ $\sin ϕ = \frac{r \sin θ \cos α}{(a \sqrt{3} - r + r \cos θ) \sin α}$ $\sin ϕ = \frac{\sin θ}{\sqrt{3} - 1 + \cos θ}$ $d s = \frac{- r \sin θ d θ}{\cos α}$ $\frac{s}{2 a} = \frac{ρ}{a} \Rightarrow ρ = \frac{s}{2}$ $A = \int ρ (2 ϕ) (- d s) = \int (s) (- d s) ϕ$ $= \int_{0}^{π} (\frac{a \sqrt{3} - r (1 - \cos θ)}{\cos α}) (\frac{r \sin θ d θ}{\cos α}) (\sin^{- 1} {\frac{r \sin θ \cos α}{(a \sqrt{3} - r + r \cos θ) \sin α}})$ $= \frac{4 a^{2}}{3} \int_{0}^{π} [1 - \frac{1}{3} (1 - \cos θ)] [\sin^{- 1} (\frac{\sin θ}{\sqrt{3} - \frac{1}{\sqrt{3}} + \frac{\cos θ}{\sqrt{3}}})] \sin θ d θ$ $= \frac{4 a^{2}}{9} \int_{- 1}^{1} (2 + t) \sin^{- 1} (\frac{\sqrt{3} \sqrt{1 - t^{2}}}{2 + t}) d t$ $A = \frac{4 a^{2}}{9} (\frac{π}{\sqrt{3}} + \frac{3}{2}) \approx 1.4728 a^{2} .$ $A = \frac{2 a^{2} (2 π \sqrt{3} + 9)}{27} .$ $(s e e Q . 73689 o n h o w t o s o l v e$ $t h e i n t e g r a l, s p e c i a l t h a n k s t o$ $M j S S i r)$
	Commented by mr W last updated on 14/Nov/19

	$n i c e s o l u t i o n s i r!$
	Answered by liki last updated on 14/Nov/19

	Commented by ajfour last updated on 14/Nov/19

	$w r o n g p l a c e!$
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