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Question-61303




Question Number 61303 by mr W last updated on 31/May/19
Commented by mr W last updated on 31/May/19
A small sphere (say the moon) moves  around a big sphere (say the earth) in  a circular track. A point source (say  the sun) is at a distance e from the  center of the big sphere.  Find the area of shadow of the small  sphere on the surface of the big  sphere in terms of α. Find the  maximum shadow area.  We only need to treat the case that  the shadow is completely on the surface  of the big sphere.
$${A}\:{small}\:{sphere}\:\left({say}\:{the}\:{moon}\right)\:{moves} \\ $$$${around}\:{a}\:{big}\:{sphere}\:\left({say}\:{the}\:{earth}\right)\:{in} \\ $$$${a}\:{circular}\:{track}.\:{A}\:{point}\:{source}\:\left({say}\right. \\ $$$$\left.{the}\:{sun}\right)\:{is}\:{at}\:{a}\:{distance}\:{e}\:{from}\:{the} \\ $$$${center}\:{of}\:{the}\:{big}\:{sphere}. \\ $$$${Find}\:{the}\:{area}\:{of}\:{shadow}\:{of}\:{the}\:{small} \\ $$$${sphere}\:{on}\:{the}\:{surface}\:{of}\:{the}\:{big} \\ $$$${sphere}\:{in}\:{terms}\:{of}\:\alpha.\:{Find}\:{the} \\ $$$${maximum}\:{shadow}\:{area}. \\ $$$${We}\:{only}\:{need}\:{to}\:{treat}\:{the}\:{case}\:{that} \\ $$$${the}\:{shadow}\:{is}\:{completely}\:{on}\:{the}\:{surface} \\ $$$${of}\:{the}\:{big}\:{sphere}. \\ $$
Commented by ajfour last updated on 02/Jun/19
Plan Only (difficult to obtain   an answer this way)  __________________________  F(e,0,0)  T(dcos α+rsin φcos ψ,rsin φsin ψ,dsin α+rcos φ)  let  ∠POG=δ  P (Rcos δcos θ,Rcos δsin θ,Rsin δ)  __________________________  FT^2 =CF^2 +CT^2   (e−dcos α−rsin φcos ψ)^2 +r^2 sin^2 φsin^2 ψ       +(dsin α+rcos φ)^2 +r^2 =   (e−dcos α)^2 +d^2 sin^2 α   ...(i)  __________________________  As FTP is straight:  ((Rcos δcos θ−e)/(dcos α+rsin φcos ψ−e))=((Rcos δsin θ)/(rsin φsin ψ))          = ((Rsin δ)/(dsin α+rcos φ))    ..(ii), (iii)  from (i), (ii), (iii)  we find θ=f(δ) eliminating  φ and ψ from them.  A=2∫_δ_(min)  ^(  δ_(max) ) (Rcos δ)θ(Rdδ)  for obtaining δ_(min) , δ_(max)   we take  θ=0, ψ=0  ⇒ ((Rcos δ−e)/(dcos α+rsin φ−e))=((Rsin δ)/(dsin α+rcos φ))  for  φ_(min)  and φ_(max)   (e−dcos α−rsin φ)^2 +(dsin α+rcos φ)^2      +r^2  =(e−dcos α)^2 +d^2 sin^2 α  ⇒ (e−dcos α)sin φ−dsin αcos φ=r     sin (φ+tan^(−1) ((dsin α)/(e−dcos α)))=(r/( (√((e−dcos α)^2 +d^2 sin^2 α))))  ⇒ φ_1 =sin^(−1) ((r/( (√((e−dcos α)^2 +d^2 sin^2 α)))))                −tan^(−1) ((dsin α)/(e−dcos α))  &  φ_2 = π−sin^(−1) ((r/( (√((e−dcos α)^2 +d^2 sin^2 α)))))                   −tan^(−1) ((dsin α)/(e−dcos α))  We then obtain δ_(min)  and δ_(max)   from upper eq.  .....
$${Plan}\:{Only}\:\left({difficult}\:{to}\:{obtain}\right. \\ $$$$\left.\:{an}\:{answer}\:{this}\:{way}\right) \\ $$$$\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_ \\ $$$${F}\left({e},\mathrm{0},\mathrm{0}\right) \\ $$$${T}\left({d}\mathrm{cos}\:\alpha+{r}\mathrm{sin}\:\phi\mathrm{cos}\:\psi,{r}\mathrm{sin}\:\phi\mathrm{sin}\:\psi,{d}\mathrm{sin}\:\alpha+{r}\mathrm{cos}\:\phi\right) \\ $$$${let}\:\:\angle{POG}=\delta \\ $$$${P}\:\left({R}\mathrm{cos}\:\delta\mathrm{cos}\:\theta,{R}\mathrm{cos}\:\delta\mathrm{sin}\:\theta,{R}\mathrm{sin}\:\delta\right) \\ $$$$\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_ \\ $$$${FT}^{\mathrm{2}} ={CF}^{\mathrm{2}} +{CT}^{\mathrm{2}} \\ $$$$\left({e}−{d}\mathrm{cos}\:\alpha−{r}\mathrm{sin}\:\phi\mathrm{cos}\:\psi\right)^{\mathrm{2}} +{r}^{\mathrm{2}} \mathrm{sin}\:^{\mathrm{2}} \phi\mathrm{sin}\:^{\mathrm{2}} \psi \\ $$$$\:\:\:\:\:+\left({d}\mathrm{sin}\:\alpha+{r}\mathrm{cos}\:\phi\right)^{\mathrm{2}} +\boldsymbol{{r}}^{\mathrm{2}} = \\ $$$$\:\left({e}−{d}\mathrm{cos}\:\alpha\right)^{\mathrm{2}} +{d}^{\mathrm{2}} \mathrm{sin}\:^{\mathrm{2}} \alpha\:\:\:…\left({i}\right) \\ $$$$\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_ \\ $$$${As}\:{FTP}\:{is}\:{straight}: \\ $$$$\frac{{R}\mathrm{cos}\:\delta\mathrm{cos}\:\theta−{e}}{{d}\mathrm{cos}\:\alpha+{r}\mathrm{sin}\:\phi\mathrm{cos}\:\psi−{e}}=\frac{{R}\mathrm{cos}\:\delta\mathrm{sin}\:\theta}{{r}\mathrm{sin}\:\phi\mathrm{sin}\:\psi} \\ $$$$\:\:\:\:\:\:\:\:=\:\frac{{R}\mathrm{sin}\:\delta}{{d}\mathrm{sin}\:\alpha+{r}\mathrm{cos}\:\phi}\:\:\:\:..\left({ii}\right),\:\left({iii}\right) \\ $$$${from}\:\left({i}\right),\:\left({ii}\right),\:\left({iii}\right) \\ $$$${we}\:{find}\:\theta={f}\left(\delta\right)\:{eliminating} \\ $$$$\phi\:{and}\:\psi\:{from}\:{them}. \\ $$$${A}=\mathrm{2}\int_{\delta_{{min}} } ^{\:\:\delta_{{max}} } \left({R}\mathrm{cos}\:\delta\right)\theta\left({Rd}\delta\right) \\ $$$${for}\:{obtaining}\:\delta_{{min}} ,\:\delta_{{max}} \:\:{we}\:{take} \\ $$$$\theta=\mathrm{0},\:\psi=\mathrm{0} \\ $$$$\Rightarrow\:\frac{{R}\mathrm{cos}\:\delta−{e}}{{d}\mathrm{cos}\:\alpha+{r}\mathrm{sin}\:\phi−{e}}=\frac{{R}\mathrm{sin}\:\delta}{{d}\mathrm{sin}\:\alpha+{r}\mathrm{cos}\:\phi} \\ $$$${for}\:\:\phi_{{min}} \:{and}\:\phi_{{max}} \\ $$$$\left({e}−{d}\mathrm{cos}\:\alpha−{r}\mathrm{sin}\:\phi\right)^{\mathrm{2}} +\left({d}\mathrm{sin}\:\alpha+{r}\mathrm{cos}\:\phi\right)^{\mathrm{2}} \\ $$$$\:\:\:+{r}^{\mathrm{2}} \:=\left({e}−{d}\mathrm{cos}\:\alpha\right)^{\mathrm{2}} +{d}^{\mathrm{2}} \mathrm{sin}\:^{\mathrm{2}} \alpha \\ $$$$\Rightarrow\:\left({e}−{d}\mathrm{cos}\:\alpha\right)\mathrm{sin}\:\phi−{d}\mathrm{sin}\:\alpha\mathrm{cos}\:\phi={r}\: \\ $$$$\:\:\mathrm{sin}\:\left(\phi+\mathrm{tan}^{−\mathrm{1}} \frac{{d}\mathrm{sin}\:\alpha}{{e}−{d}\mathrm{cos}\:\alpha}\right)=\frac{{r}}{\:\sqrt{\left({e}−{d}\mathrm{cos}\:\alpha\right)^{\mathrm{2}} +{d}^{\mathrm{2}} \mathrm{sin}\:^{\mathrm{2}} \alpha}} \\ $$$$\Rightarrow\:\phi_{\mathrm{1}} =\mathrm{sin}^{−\mathrm{1}} \left(\frac{{r}}{\:\sqrt{\left({e}−{d}\mathrm{cos}\:\alpha\right)^{\mathrm{2}} +{d}^{\mathrm{2}} \mathrm{sin}\:^{\mathrm{2}} \alpha}}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:−\mathrm{tan}^{−\mathrm{1}} \frac{{d}\mathrm{sin}\:\alpha}{{e}−{d}\mathrm{cos}\:\alpha} \\ $$$$\&\:\:\phi_{\mathrm{2}} =\:\pi−\mathrm{sin}^{−\mathrm{1}} \left(\frac{{r}}{\:\sqrt{\left({e}−{d}\mathrm{cos}\:\alpha\right)^{\mathrm{2}} +{d}^{\mathrm{2}} \mathrm{sin}\:^{\mathrm{2}} \alpha}}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:−\mathrm{tan}^{−\mathrm{1}} \frac{{d}\mathrm{sin}\:\alpha}{{e}−{d}\mathrm{cos}\:\alpha} \\ $$$${We}\:{then}\:{obtain}\:\delta_{{min}} \:{and}\:\delta_{{max}} \\ $$$${from}\:{upper}\:{eq}. \\ $$$$….. \\ $$
Commented by mr W last updated on 01/Jun/19
thanks for trying sir! i have not got  a solution. but your method would be  that what i had also used. but on the  other side i think there should be an  other maybe easier way. i′ll try it  later.
$${thanks}\:{for}\:{trying}\:{sir}!\:{i}\:{have}\:{not}\:{got} \\ $$$${a}\:{solution}.\:{but}\:{your}\:{method}\:{would}\:{be} \\ $$$${that}\:{what}\:{i}\:{had}\:{also}\:{used}.\:{but}\:{on}\:{the} \\ $$$${other}\:{side}\:{i}\:{think}\:{there}\:{should}\:{be}\:{an} \\ $$$${other}\:{maybe}\:{easier}\:{way}.\:{i}'{ll}\:{try}\:{it} \\ $$$${later}. \\ $$
Answered by ajfour last updated on 02/Jun/19
Answered by mr W last updated on 02/Jun/19
Commented by mr W last updated on 02/Jun/19
the light rays form a right circular  cone. our task is to determine  the intersection from  this cone and the big sphere.  to simplify the calcultion i select the  source S as origin and SM as x−axis.  the shape of cone is described by the  angle β.  the position of the big sphere is  described by h and k.  SO=e, OM=d  SM=(√((e−d cos α)^2 +(d sin α)^2 ))  =(√(e^2 +d^2 −2ed cos α))  sin β=(r/( (√(e^2 +d^2 −2ed cos α))))  ⇒β=sin^(−1) (r/( (√(e^2 +d^2 −2ed cos α))))  tan β=(r/( (√(e^2 +d^2 −r^2 −2ed cos α))))=m (say)  (k/(d sin α))=(e/(SM))  ⇒k=((ed sin α)/( (√(e^2 +d^2 −2ed cos α))))  h=(√(e^2 −k^2 ))=(√(e^2 −((e^2 d^2  sin^2  α)/(e^2 +d^2 −2ed cos α))))  ⇒h=((e(e−d cos α))/( (√(e^2 +d^2 −2ed cos α))))    condition such that the shadow is  completely on the big sphere is  sin β=((R−(k/(cos β)))/(h−k tan β))=((R cos β−k)/(h cos β−k sin β))
$${the}\:{light}\:{rays}\:{form}\:{a}\:{right}\:{circular} \\ $$$${cone}.\:{our}\:{task}\:{is}\:{to}\:{determine} \\ $$$${the}\:{intersection}\:{from} \\ $$$${this}\:{cone}\:{and}\:{the}\:{big}\:{sphere}. \\ $$$${to}\:{simplify}\:{the}\:{calcultion}\:{i}\:{select}\:{the} \\ $$$${source}\:{S}\:{as}\:{origin}\:{and}\:{SM}\:{as}\:{x}−{axis}. \\ $$$${the}\:{shape}\:{of}\:{cone}\:{is}\:{described}\:{by}\:{the} \\ $$$${angle}\:\beta. \\ $$$${the}\:{position}\:{of}\:{the}\:{big}\:{sphere}\:{is} \\ $$$${described}\:{by}\:{h}\:{and}\:{k}. \\ $$$${SO}={e},\:{OM}={d} \\ $$$${SM}=\sqrt{\left({e}−{d}\:\mathrm{cos}\:\alpha\right)^{\mathrm{2}} +\left({d}\:\mathrm{sin}\:\alpha\right)^{\mathrm{2}} } \\ $$$$=\sqrt{{e}^{\mathrm{2}} +{d}^{\mathrm{2}} −\mathrm{2}{ed}\:\mathrm{cos}\:\alpha} \\ $$$$\mathrm{sin}\:\beta=\frac{{r}}{\:\sqrt{{e}^{\mathrm{2}} +{d}^{\mathrm{2}} −\mathrm{2}{ed}\:\mathrm{cos}\:\alpha}} \\ $$$$\Rightarrow\beta=\mathrm{sin}^{−\mathrm{1}} \frac{{r}}{\:\sqrt{{e}^{\mathrm{2}} +{d}^{\mathrm{2}} −\mathrm{2}{ed}\:\mathrm{cos}\:\alpha}} \\ $$$$\mathrm{tan}\:\beta=\frac{{r}}{\:\sqrt{{e}^{\mathrm{2}} +{d}^{\mathrm{2}} −{r}^{\mathrm{2}} −\mathrm{2}{ed}\:\mathrm{cos}\:\alpha}}={m}\:\left({say}\right) \\ $$$$\frac{{k}}{{d}\:\mathrm{sin}\:\alpha}=\frac{{e}}{{SM}} \\ $$$$\Rightarrow{k}=\frac{{ed}\:\mathrm{sin}\:\alpha}{\:\sqrt{{e}^{\mathrm{2}} +{d}^{\mathrm{2}} −\mathrm{2}{ed}\:\mathrm{cos}\:\alpha}} \\ $$$${h}=\sqrt{{e}^{\mathrm{2}} −{k}^{\mathrm{2}} }=\sqrt{{e}^{\mathrm{2}} −\frac{{e}^{\mathrm{2}} {d}^{\mathrm{2}} \:\mathrm{sin}^{\mathrm{2}} \:\alpha}{{e}^{\mathrm{2}} +{d}^{\mathrm{2}} −\mathrm{2}{ed}\:\mathrm{cos}\:\alpha}} \\ $$$$\Rightarrow{h}=\frac{{e}\left({e}−{d}\:\mathrm{cos}\:\alpha\right)}{\:\sqrt{{e}^{\mathrm{2}} +{d}^{\mathrm{2}} −\mathrm{2}{ed}\:\mathrm{cos}\:\alpha}} \\ $$$$ \\ $$$${condition}\:{such}\:{that}\:{the}\:{shadow}\:{is} \\ $$$${completely}\:{on}\:{the}\:{big}\:{sphere}\:{is} \\ $$$$\mathrm{sin}\:\beta=\frac{{R}−\frac{{k}}{\mathrm{cos}\:\beta}}{{h}−{k}\:\mathrm{tan}\:\beta}=\frac{{R}\:\mathrm{cos}\:\beta−{k}}{{h}\:\mathrm{cos}\:\beta−{k}\:\mathrm{sin}\:\beta} \\ $$
Commented by mr W last updated on 02/Jun/19
Commented by mr W last updated on 03/Jun/19
eqn. of cone:  x=((√(y^2 +z^2 ))/(tan β))=((√(y^2 +z^2 ))/m)  or y^2 =m^2 x^2 −z^2     eqn. of sphere:  (x−h)^2 +y^2 +(z+k)^2 =R^2     intersection of cone and sphere:  (x−h)^2 +m^2 x^2 −z^2 +(z+k)^2 =R^2   ⇒(1+m^2 )x^2 −2hx+2kz+(h^2 +k^2 −R^2 )=0   ...(i)  (this is a parabola in xz−plane)    at θ:  let ρ=R−Δh  x=h−ρ cos θ  z=−k+ρ sin θ  put this into (i):  (1+m^2 )[h−ρ cos θ]^2 −2h[h−ρ cos θ]+2k[−k+ρ sin θ]+(h^2 +k^2 −R^2 )=0  (1+m^2 )cos^2  θ ρ^2 −2(m^2 h cos θ−k sin θ)ρ+(m^2 h^2 −k^2 −R^2 )=0  ⇒ρ=(((m^2 h cos θ−k sin θ)+(√(m^4 h^2  cos^2  θ+k^2  sin^2  θ−2m^2 hk sin θ cos θ−(1+m^2 )cos^2  θ(m^2 h^2 −k^2 −R^2 ))))/((1+m^2 )cos^2  θ))  (two intersections, we take + for the left one)  ⇒ρ=R cos ϕ=(((m^2 h cos θ−k sin θ)+(√(k^2 +[(1+m^2 )R^2 −m^2 (h^2 −k^2 )]cos^2  θ−m^2 hk sin 2θ)))/((1+m^2 )cos^2  θ))  ⇒ϕ=cos^(−1) {(((m^2 h cos θ−k sin θ)+(√(k^2 +[(1+m^2 )R^2 −m^2 (h^2 −k^2 )]cos^2  θ−m^2 hk sin 2θ)))/((1+m^2 )R cos^2  θ))}  for ρ=(((m^2 h cos θ−k sin θ)+(√(k^2 +[(1+m^2 )R^2 −m^2 (h^2 −k^2 )]cos^2  θ−m^2 hk sin 2θ)))/((1+m^2 )cos^2  θ))=R  we get θ_1  and θ_2 .    Area of shadow=A  A=∫_θ_1  ^θ_2  2RϕRdθ  ⇒A=2R^2 ∫_θ_1  ^θ_2  cos^(−1) {(((m^2 h cos θ−k sin θ)+(√(k^2 +[(1+m^2 )R^2 −m^2 (h^2 −k^2 )]cos^2  θ−m^2 hk sin 2θ)))/((1+m^2 )R cos^2  θ))}dθ  with μ=(k/R),λ=(h/R)  ⇒A=2R^2 ∫_θ_1  ^θ_2  cos^(−1) {((m^2 λ cos θ−μ sin θ+(√(μ^2 +[1+m^2 (1+μ^2 −λ^2 )]cos^2  θ−m^2 λμ sin 2θ)))/((1+m^2 )cos^2  θ))}dθ
$${eqn}.\:{of}\:{cone}: \\ $$$${x}=\frac{\sqrt{{y}^{\mathrm{2}} +{z}^{\mathrm{2}} }}{\mathrm{tan}\:\beta}=\frac{\sqrt{{y}^{\mathrm{2}} +{z}^{\mathrm{2}} }}{{m}} \\ $$$${or}\:{y}^{\mathrm{2}} ={m}^{\mathrm{2}} {x}^{\mathrm{2}} −{z}^{\mathrm{2}} \\ $$$$ \\ $$$${eqn}.\:{of}\:{sphere}: \\ $$$$\left({x}−{h}\right)^{\mathrm{2}} +{y}^{\mathrm{2}} +\left({z}+{k}\right)^{\mathrm{2}} ={R}^{\mathrm{2}} \\ $$$$ \\ $$$${intersection}\:{of}\:{cone}\:{and}\:{sphere}: \\ $$$$\left({x}−{h}\right)^{\mathrm{2}} +{m}^{\mathrm{2}} {x}^{\mathrm{2}} −{z}^{\mathrm{2}} +\left({z}+{k}\right)^{\mathrm{2}} ={R}^{\mathrm{2}} \\ $$$$\Rightarrow\left(\mathrm{1}+{m}^{\mathrm{2}} \right){x}^{\mathrm{2}} −\mathrm{2}{hx}+\mathrm{2}{kz}+\left({h}^{\mathrm{2}} +{k}^{\mathrm{2}} −{R}^{\mathrm{2}} \right)=\mathrm{0}\:\:\:…\left({i}\right) \\ $$$$\left({this}\:{is}\:{a}\:{parabola}\:{in}\:{xz}−{plane}\right) \\ $$$$ \\ $$$${at}\:\theta: \\ $$$${let}\:\rho={R}−\Delta{h} \\ $$$${x}={h}−\rho\:\mathrm{cos}\:\theta \\ $$$${z}=−{k}+\rho\:\mathrm{sin}\:\theta \\ $$$${put}\:{this}\:{into}\:\left({i}\right): \\ $$$$\left(\mathrm{1}+{m}^{\mathrm{2}} \right)\left[{h}−\rho\:\mathrm{cos}\:\theta\right]^{\mathrm{2}} −\mathrm{2}{h}\left[{h}−\rho\:\mathrm{cos}\:\theta\right]+\mathrm{2}{k}\left[−{k}+\rho\:\mathrm{sin}\:\theta\right]+\left({h}^{\mathrm{2}} +{k}^{\mathrm{2}} −{R}^{\mathrm{2}} \right)=\mathrm{0} \\ $$$$\left(\mathrm{1}+{m}^{\mathrm{2}} \right)\mathrm{cos}^{\mathrm{2}} \:\theta\:\rho^{\mathrm{2}} −\mathrm{2}\left({m}^{\mathrm{2}} {h}\:\mathrm{cos}\:\theta−{k}\:\mathrm{sin}\:\theta\right)\rho+\left({m}^{\mathrm{2}} {h}^{\mathrm{2}} −{k}^{\mathrm{2}} −{R}^{\mathrm{2}} \right)=\mathrm{0} \\ $$$$\Rightarrow\rho=\frac{\left({m}^{\mathrm{2}} {h}\:\mathrm{cos}\:\theta−{k}\:\mathrm{sin}\:\theta\right)+\sqrt{{m}^{\mathrm{4}} {h}^{\mathrm{2}} \:\mathrm{cos}^{\mathrm{2}} \:\theta+{k}^{\mathrm{2}} \:\mathrm{sin}^{\mathrm{2}} \:\theta−\mathrm{2}{m}^{\mathrm{2}} {hk}\:\mathrm{sin}\:\theta\:\mathrm{cos}\:\theta−\left(\mathrm{1}+{m}^{\mathrm{2}} \right)\mathrm{cos}^{\mathrm{2}} \:\theta\left({m}^{\mathrm{2}} {h}^{\mathrm{2}} −{k}^{\mathrm{2}} −{R}^{\mathrm{2}} \right)}}{\left(\mathrm{1}+{m}^{\mathrm{2}} \right)\mathrm{cos}^{\mathrm{2}} \:\theta} \\ $$$$\left({two}\:{intersections},\:{we}\:{take}\:+\:{for}\:{the}\:{left}\:{one}\right) \\ $$$$\Rightarrow\rho={R}\:\mathrm{cos}\:\varphi=\frac{\left({m}^{\mathrm{2}} {h}\:\mathrm{cos}\:\theta−{k}\:\mathrm{sin}\:\theta\right)+\sqrt{{k}^{\mathrm{2}} +\left[\left(\mathrm{1}+{m}^{\mathrm{2}} \right){R}^{\mathrm{2}} −{m}^{\mathrm{2}} \left({h}^{\mathrm{2}} −{k}^{\mathrm{2}} \right)\right]\mathrm{cos}^{\mathrm{2}} \:\theta−{m}^{\mathrm{2}} {hk}\:\mathrm{sin}\:\mathrm{2}\theta}}{\left(\mathrm{1}+{m}^{\mathrm{2}} \right)\mathrm{cos}^{\mathrm{2}} \:\theta} \\ $$$$\Rightarrow\varphi=\mathrm{cos}^{−\mathrm{1}} \left\{\frac{\left({m}^{\mathrm{2}} {h}\:\mathrm{cos}\:\theta−{k}\:\mathrm{sin}\:\theta\right)+\sqrt{{k}^{\mathrm{2}} +\left[\left(\mathrm{1}+{m}^{\mathrm{2}} \right){R}^{\mathrm{2}} −{m}^{\mathrm{2}} \left({h}^{\mathrm{2}} −{k}^{\mathrm{2}} \right)\right]\mathrm{cos}^{\mathrm{2}} \:\theta−{m}^{\mathrm{2}} {hk}\:\mathrm{sin}\:\mathrm{2}\theta}}{\left(\mathrm{1}+{m}^{\mathrm{2}} \right){R}\:\mathrm{cos}^{\mathrm{2}} \:\theta}\right\} \\ $$$${for}\:\rho=\frac{\left({m}^{\mathrm{2}} {h}\:\mathrm{cos}\:\theta−{k}\:\mathrm{sin}\:\theta\right)+\sqrt{{k}^{\mathrm{2}} +\left[\left(\mathrm{1}+{m}^{\mathrm{2}} \right){R}^{\mathrm{2}} −{m}^{\mathrm{2}} \left({h}^{\mathrm{2}} −{k}^{\mathrm{2}} \right)\right]\mathrm{cos}^{\mathrm{2}} \:\theta−{m}^{\mathrm{2}} {hk}\:\mathrm{sin}\:\mathrm{2}\theta}}{\left(\mathrm{1}+{m}^{\mathrm{2}} \right)\mathrm{cos}^{\mathrm{2}} \:\theta}={R} \\ $$$${we}\:{get}\:\theta_{\mathrm{1}} \:{and}\:\theta_{\mathrm{2}} . \\ $$$$ \\ $$$${Area}\:{of}\:{shadow}={A} \\ $$$${A}=\int_{\theta_{\mathrm{1}} } ^{\theta_{\mathrm{2}} } \mathrm{2}{R}\varphi{Rd}\theta \\ $$$$\Rightarrow{A}=\mathrm{2}{R}^{\mathrm{2}} \int_{\theta_{\mathrm{1}} } ^{\theta_{\mathrm{2}} } \mathrm{cos}^{−\mathrm{1}} \left\{\frac{\left({m}^{\mathrm{2}} {h}\:\mathrm{cos}\:\theta−{k}\:\mathrm{sin}\:\theta\right)+\sqrt{{k}^{\mathrm{2}} +\left[\left(\mathrm{1}+{m}^{\mathrm{2}} \right){R}^{\mathrm{2}} −{m}^{\mathrm{2}} \left({h}^{\mathrm{2}} −{k}^{\mathrm{2}} \right)\right]\mathrm{cos}^{\mathrm{2}} \:\theta−{m}^{\mathrm{2}} {hk}\:\mathrm{sin}\:\mathrm{2}\theta}}{\left(\mathrm{1}+{m}^{\mathrm{2}} \right){R}\:\mathrm{cos}^{\mathrm{2}} \:\theta}\right\}{d}\theta \\ $$$${with}\:\mu=\frac{{k}}{{R}},\lambda=\frac{{h}}{{R}} \\ $$$$\Rightarrow{A}=\mathrm{2}{R}^{\mathrm{2}} \int_{\theta_{\mathrm{1}} } ^{\theta_{\mathrm{2}} } \mathrm{cos}^{−\mathrm{1}} \left\{\frac{{m}^{\mathrm{2}} \lambda\:\mathrm{cos}\:\theta−\mu\:\mathrm{sin}\:\theta+\sqrt{\mu^{\mathrm{2}} +\left[\mathrm{1}+{m}^{\mathrm{2}} \left(\mathrm{1}+\mu^{\mathrm{2}} −\lambda^{\mathrm{2}} \right)\right]\mathrm{cos}^{\mathrm{2}} \:\theta−{m}^{\mathrm{2}} \lambda\mu\:\mathrm{sin}\:\mathrm{2}\theta}}{\left(\mathrm{1}+{m}^{\mathrm{2}} \right)\mathrm{cos}^{\mathrm{2}} \:\theta}\right\}{d}\theta \\ $$
Commented by mr W last updated on 02/Jun/19
Commented by mr W last updated on 02/Jun/19
Commented by mr W last updated on 02/Jun/19
Commented by mr W last updated on 02/Jun/19
Commented by ajfour last updated on 02/Jun/19
Great Sir, but you′ll have to  guide me to the idea used, i dont  think i′ll follow it otherwise..
$${Great}\:{Sir},\:{but}\:{you}'{ll}\:{have}\:{to} \\ $$$${guide}\:{me}\:{to}\:{the}\:{idea}\:{used},\:{i}\:{dont} \\ $$$${think}\:{i}'{ll}\:{follow}\:{it}\:{otherwise}.. \\ $$
Commented by mr W last updated on 02/Jun/19
the boundary of the shadow on the  surface of big sphere is the intersection  (a curve in the space) from the cone  and the big sphere. the shape of the  cone is given by the size of the small  sphere. with the coordinate system  i select the eqn. of the cone and the  big sphere can be described in simple  form. the eqn. of the intersection  line in the xz−plane can be obtained  quite easily, and it′s a parabola. with  this eqn. we can determine the  parameters of the shadow area on  the sphere surface. the parameters  are: θ and ρ (i.e. ϕ).
$${the}\:{boundary}\:{of}\:{the}\:{shadow}\:{on}\:{the} \\ $$$${surface}\:{of}\:{big}\:{sphere}\:{is}\:{the}\:{intersection} \\ $$$$\left({a}\:{curve}\:{in}\:{the}\:{space}\right)\:{from}\:{the}\:{cone} \\ $$$${and}\:{the}\:{big}\:{sphere}.\:{the}\:{shape}\:{of}\:{the} \\ $$$${cone}\:{is}\:{given}\:{by}\:{the}\:{size}\:{of}\:{the}\:{small} \\ $$$${sphere}.\:{with}\:{the}\:{coordinate}\:{system} \\ $$$${i}\:{select}\:{the}\:{eqn}.\:{of}\:{the}\:{cone}\:{and}\:{the} \\ $$$${big}\:{sphere}\:{can}\:{be}\:{described}\:{in}\:{simple} \\ $$$${form}.\:{the}\:{eqn}.\:{of}\:{the}\:{intersection} \\ $$$${line}\:{in}\:{the}\:{xz}−{plane}\:{can}\:{be}\:{obtained} \\ $$$${quite}\:{easily},\:{and}\:{it}'{s}\:{a}\:{parabola}.\:{with} \\ $$$${this}\:{eqn}.\:{we}\:{can}\:{determine}\:{the} \\ $$$${parameters}\:{of}\:{the}\:{shadow}\:{area}\:{on} \\ $$$${the}\:{sphere}\:{surface}.\:{the}\:{parameters} \\ $$$${are}:\:\theta\:{and}\:\rho\:\left({i}.{e}.\:\varphi\right). \\ $$
Commented by ajfour last updated on 06/Jun/19
I am very thankful but i lack  the required patience (these days)  Sir, i have saved your solution,  and shall analyse it soon.
$${I}\:{am}\:{very}\:{thankful}\:{but}\:{i}\:{lack} \\ $$$${the}\:{required}\:{patience}\:\left({these}\:{days}\right) \\ $$$${Sir},\:{i}\:{have}\:{saved}\:{your}\:{solution}, \\ $$$${and}\:{shall}\:{analyse}\:{it}\:{soon}. \\ $$
Commented by mr W last updated on 06/Jun/19
thank you sir!  i wasn′t able to provide a better and  complete solution.  but i′ll try to  continue to think about the question.  i hope one day you may give a perfect  solution sir.
$${thank}\:{you}\:{sir}! \\ $$$${i}\:{wasn}'{t}\:{able}\:{to}\:{provide}\:{a}\:{better}\:{and} \\ $$$${complete}\:{solution}.\:\:{but}\:{i}'{ll}\:{try}\:{to} \\ $$$${continue}\:{to}\:{think}\:{about}\:{the}\:{question}. \\ $$$${i}\:{hope}\:{one}\:{day}\:{you}\:{may}\:{give}\:{a}\:{perfect} \\ $$$${solution}\:{sir}. \\ $$

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