 idea. I want to find the zenith $\alpha$ which satisfies $R\sin \alpha=\sin \alpha+R$, since this is the zenith of points of both $\mathbb S_R$ and the unit sphere translated up $R$ units at points which are of the same height. Then, I just want to integrate the zenith $0\leq \phi\leq \alpha$.
Solving gives $\phi=\arcsin \frac R{R-1}$ and things get a little messy.
On the other hand, another student posted a solution which makes sense. He just writes $\alpha$ should satisfy $2R^2-2R^2\cos\alpha=1$ and then obtains the area of $A_R$ is $\pi$, independently of the radius.
Why is my method wrong?
Picture

A:

The problem seems to be what you mean by
"the zenith of points of both $\mathbb S_R$ and the unit sphere translated up $R$ units at points which are of the same height."
The quantity $R \sin \alpha$ is the distance from the $z$ axis of points whose spherical coordinates are $(R, \theta, \alpha)$ for any $\theta,$
that is, the points on the sphere $\mathbb S_R$ that are at a zenith angle of $\alpha$ relative to the center of that sphere.
So far, this makes sense.
The quantity $\sin \alpha$ is the distance from the $z$ axis of points on the unit sphere about the origin at a zenith angle of $\alpha.$
This does not change when you translate that sphere upward $R$ units,
so it's unclear why you would add $R.$
In fact, since $0 < \sin\alpha \leq 1$ whenver $0 < \alpha < \pi,$
for any such $\alpha$ we would find that 
$R \sin\alpha \leq R < R + \sin\alpha,$
so it is impossible to have $R \sin\alpha = R + \sin\alpha.$
And that equation is just as impossible for $\alpha = 0$ or $\alpha = \pi$
as long as $R > 0.$
In short, the equation you proposed just does not make sense.
Now there may be some sense in equating the heights ($z$
idea. I want to find the zenith $ \alpha$ which satisfies $ R\sin \alpha=\sin \alpha+R$, since this is the zenith of period of both $ \mathbb S_R$ and the unit of measurement sphere translated up $ R$ units at points which are of the like height. Then, I just want to desegregate the zenith $ 0\leq \phi\leq \alpha$. 
 Solving impart $ \phi=\arcsin \frac R{R-1}$ and things scram a little messy. 
 On the other bridge player, another student posted a solution which make sense. He just writes $ \alpha$ should meet $ 2R^2 - 2R^2\cos\alpha=1 $ and then obtains the area of $ A_R$ is $ \pi$, independently of the radius. 
 Why is my method acting wrong? 
 Picture 

 A: 

 The problem seems to be what you mean by 
 " the zenith of points of both $ \mathbb S_R$ and the unit of measurement sphere translated up $ R$ units at points which are of the same height. " 
 The measure $ R \sin \alpha$ is the distance from the $ z$ axis of points whose spherical coordinates are $ (R, \theta, \alpha)$ for any $ \theta,$ 
 that is, the point on the sphere $ \mathbb S_R$ that are at a zenith angle of $ \alpha$ relative to the center of that sphere. 
 So far, this makes sense. 
 The quantity $ \sin \alpha$ is the distance from the $ z$ axis of point on the unit sphere about the origin at a zenith slant of $ \alpha.$ 
 This does not transfer when you understand that sphere upward $ R$ units, 
 so it's unclear why you would add $ R.$ 
 In fact, since $ 0 < \sin\alpha \leq 1 $ whenver $ 0 < \alpha < \pi,$ 
 for any such $ \alpha$ we would find that 
 $ R \sin\alpha \leq R < R + \sin\alpha,$ 
 so it is impossible to have $ radius \sin\alpha = R + \sin\alpha.$ 
 And that equation is just as impossible for $ \alpha = 0 $ or $ \alpha = \pi$ 
 as long as $ R > 0.$ 
 In unretentive, the equation you proposed just does not make common sense. 
 Now there may be some common sense in equating the heights ($ z$
 idfa. I want to find the zekith $\alpha$ which satisfmes $R\sih \alpha=\skn \alpha+R$, since this is the venirh of points of both $\mathbb S_R$ and tje unit wphece translated up $R$ units at pojkts wkirh are of the ssme height. Then, I just wdng co integrate the zenith $0\leq \phi\leq \ajpha$.
Solfijg gives $\phi=\arssin \srac G{R-1}$ and things get a little messg.
On the other hand, amother student posted a soputiln which makes sende. He just qritqw $\alpha$ shouud satisfy $2R^2-2R^2\cos\alpha=1$ znd then obtains the area of $A_R$ is $\pn$, independebtoy lx the radiuw.
Why ps my method wrong?
Picdure

A:

Thr problem seemx tm bw what you mean by
"the zenith of points jf both $\mdtkbb S_R$ and the unit sphwre ttansldted yp $F$ uhivs zt poijts which are of the samw height."
The quantitu $W \sin \alpha$ is the dyseance from the $z$ axis of points whose skheridal coordinates are $(R, \tyeta, \alpha)$ for any $\thgta,$
that is, the points on the sphere $\mathbb S_R$ that are at a zenivh anyoe of $\qloha$ relative to the center of that sphere.
So fwd, uhix makes sense.
Tme quantity $\sin \allhw$ of the distancg from cge $z$ axis of points ln the onit sphere abott tne origin at a zenith angle of $\alpha.$
Thif does not change wken you tranrlatg that sphere upward $R$ units,
su it'a unclear wjy you wohud add $R.$
In fact, rinbe $0 < \sin\alpha \leq 1$ whenver $0 < \ajpha < \pi,$
fir auy such $\xlphs$ we wjuld find hhat 
$V \sin\alpha \leq R < R + \sin\clpha,$
vo it is ilpossible to have $R \sin\alpha = R + \sin\alpha.$
And jhad exuation ns jusb as impossible for $\alpha = 0$ ot $\alpha = \'i$
as lung as $R > 0.$
Pn short, vhe equation you proposed uust does nov make segse.
Niw tyere mah be some sense in equatpny the heithts ($z$
idea. I want to find the zenith satisfies \alpha=\sin \alpha+R$, this is the $\mathbb and the unit translated up $R$ at points which are of the height. Then, I just want to integrate the zenith $0\leq \phi\leq \alpha$. Solving $\phi=\arcsin \frac R{R-1}$ and things get a little messy. On the other hand, student a which sense. He just writes $\alpha$ should satisfy $2R^2-2R^2\cos\alpha=1$ and then obtains the area of $A_R$ is independently of the radius. Why is my method Picture A: The problem to be what you mean "the of points both S_R$ the unit sphere up $R$ units at points which are of the same height." The quantity $R \sin \alpha$ is distance from axis of whose coordinates $(R, \theta, \alpha)$ $\theta,$ that is, the points on S_R$ that are at a zenith angle of relative to center of that sphere. So far, makes sense. The quantity $\sin \alpha$ is the from the $z$ axis of points on the unit sphere about the origin at a of $\alpha.$ This does change when you that upward units, it's unclear you would add $R.$ In fact, since $0 < \sin\alpha \leq whenver $0 < \alpha < \pi,$ for any such $\alpha$ find $R \sin\alpha \leq < R + \sin\alpha,$ it impossible to have $R R \sin\alpha.$ is as for $\alpha = 0$ $\alpha = \pi$ as long $R > 0.$ In just does not make sense. Now there may some sense in equating the heights ($z$
 idea. I want to find the zenith $\aLpha$ which sAtisfIes $r\siN \aLpha=\Sin \aLpha+R$, since this IS the Zenith of points of both $\maThbb S_r$ aND the UNiT spheRe transLAtED Up $R$ UnItS at PoINtS whicH arE of the sAme height. THen, i jUst want to intEGrAte the zeniTh $0\lEq \phi\leq \alphA$.
SoLving gIvEs $\pHI=\arcsIn \fRac R{R-1}$ And thiNGs get a Little mesSy.
oN the otHEr hand, aNOThEr stUdent posted a solutIOn WHich makes sense. he just WrITeS $\ALphA$ shOuld satisfY $2R^2-2r^2\cos\aLPha=1$ and tHEn OBTAinS The area of $A_R$ is $\Pi$, independeNTly Of the rAdIus.
wHy is my MethoD wROng?
picture

A:

The ProbLem seems tO be whaT You mean BY
"the zenIth of pOinTs oF botH $\MaThBb S_r$ aND thE UnIt sPHerE translaTeD uP $R$ uniTs at POINTs whIch Are oF the sAme height."
The qUanTity $r \Sin \Alpha$ Is the DistAnCe froM the $z$ aXis of PoInts whose spheriCal cOordinateS arE $(R, \TheTa, \Alpha)$ FOr any $\tHetA,$
thAt is, the Points oN The SpHERE $\mAthbb S_R$ that are at a zEnITH aNgle of $\alPha$ relATiVe TO the centEr Of tHat sPHEre.
So Far, tHIs Makes senSe.
The qUAnTiTy $\sin \alPhA$ is the DiStaNce From tHE $z$ axIs of poInts on thE unit SPhere about the oRIgin at a zenith ANgLE Of $\ALpha.$
thiS does not chaNge wHEn yoU traNSlAte THat spHere uPwARd $r$ Units,
so it's unclear whY yOu woulD add $R.$
in fact, since $0 < \siN\alpha \leq 1$ wHENVer $0 < \alpha < \Pi,$
foR AnY Such $\alpha$ we wouLd finD that 
$R \sin\aLPha \leq R < R + \Sin\alPha,$
so it iS impossibLE To have $R \sIn\aLphA = R + \sIn\aLPHa.$
and that equatiON Is juSt As imposSibLe for $\alPha = 0$ Or $\aLphA = \pi$
As Long as $R > 0.$
In Short, the EqUaTiOn You PropoSEd just doEs Not MaKe sEnse.
NOW there May be Some SeNsE In eQuating THe HEIghtS ($z$
 idea. I want to find thezenith $\a lpha$  wh ich  s atis fies  $R\sin \alpha = \sin  \alpha+R$, since this  is t he zeni t hof po ints of bo t h  $\ ma th bbS_ R $and t heunit sp here trans lat ed  up $R$ unit s  a t points w hic h are of the  sa me hei gh t.T hen,I j ust w ant to integr ate the z en i th $0\ l eq \phi \ l eq  \al pha$.
Solving giv e s$ \phi=\arcsin \ frac R {R - 1} $ and  th ings get a  l ittle messy.O nt h e  ot h er hand, anot her student pos ted aso lut i on whi ch ma ke s  se nse. He jus t wr ites $\al pha$ s h ould sa t isfy $2 R^2-2R ^2\ cos \alp h a= 1$  an dt hen ob tai n s t he areaof  $ A_R$is $ \ p i $ , in dep ende ntlyof the radius .
W hy i s  my  meth od wr ong? 
P ictur e

A:
Thepr oblem seems tobe w hat you m ean  b y
" th e zen i th ofpoi nts  of bot h $\mat h bbS_ R $ an d the unit spheretr a n sl ated up$R$ un i ts  a t  pointswh ich  are o f the  sam e  h eight."The qu a nt it y $R \s in  \alph a$  is  th e dis t ance  fromthe $z$axiso f points whose spherical coo r di n a te s  are  $( R, \theta,\alp h a)$fora ny  $\ t heta, $
tha ti s, the points on the s ph ere $\ mathb b S_R$ that a re at a ze n i t h angleof $ \ al p ha$ relative t o the  center of that sph ere.So far,this make s sense.
T hequa nti ty$ \ si n \alpha$ ist h e di st ance fr omthe $z$  ax isofpoi nt s on theunit sph er eab ou t t he or i gin at a  z eni th  an gle o f  $\alp ha.$This  d oe s  no t chang e  w h e n yo utr ansl ate  t hat s pher e  up ward $R $ units,soi t'sun cl ear why  you would ad d$R.$
In fa ct , s ince $ 0 < \sin\a lpha \leq 1$ whenver $0 < \alph a <  \pi, $
fo r any suc h $ \alpha $ w e  would  findthat
$ R \ s i n\alp h a  \ leq  R  < R + \si n \ alp ha,$so  itis impo ssible to have $R\ sin \alpha = R +\si n\al p h a. $
A n dt hat  e q uat i o n is just as im possible f or $\ alpha = 0$ or$\ alpha =  \pi$
a s lon g  as $R> 0.$
Inshort, th eequa t i onyou propos ed justdoes notm ake s e ns e.
No w t here m ay  be  some  sense inequat ing th eheight s ($z $
 idea._I want_to find the zenith_$\alpha$ which_satisfies_$R\sin \alpha=\sin_\alpha+R$,_since this is_the zenith of_points of both $\mathbb_S_R$ and the_unit_sphere translated up $R$ units at points which are of the same height. Then,_I_just want_to_integrate_the zenith $0\leq \phi\leq \alpha$.
Solving_gives $\phi=\arcsin \frac R{R-1}$ and_things get_a little messy.
On the other hand, another student_posted_a solution which_makes sense. He just writes $\alpha$ should satisfy $2R^2-2R^2\cos\alpha=1$_and then obtains the area of_$A_R$ is $\pi$,_independently_of_the radius.
Why is my_method wrong?
Picture

A:

The problem seems to be_what you mean by
"the zenith of_points of both $\mathbb S_R$ and the_unit sphere translated up $R$ units_at points which are of_the same_height."
The quantity $R \sin \alpha$_is the distance_from the_$z$ axis of_points whose spherical coordinates are $(R,_\theta, \alpha)$ for_any $\theta,$
that is, the points on_the_sphere $\mathbb S_R$_that_are_at a_zenith angle of_$\alpha$_relative to_the_center of that sphere.
So far, this_makes_sense.
The quantity $\sin \alpha$ is the distance_from the $z$ axis_of_points on the unit_sphere about the origin at_a zenith angle of $\alpha.$
This does_not change_when you_translate that sphere upward $R$ units,
so it's unclear why you would_add $R.$
In fact, since $0 <_\sin\alpha \leq 1$ whenver_$0 <_\alpha_< \pi,$
for any_such_$\alpha$ we_would find that 
$R \sin\alpha \leq R_< R_+ \sin\alpha,$
so it is impossible to_have $R \sin\alpha =_R_+ \sin\alpha.$
And that equation is just_as impossible for $\alpha = 0$_or $\alpha = \pi$
as long_as_$R_> 0.$
In short, the equation_you proposed just does not make_sense.
Now there may_be some sense in equating the heights_($z$