fs.writeFile('members.json', JSON.stringify(x), 'utf8', (err,data)=>{
    //callback
});

fs.readFile('members.json','utf8',(err,data)=>{
    if(data){
        console.log(JSON.parse(data))
    } else{
        console.log(err)
    }
})

 Q:

Volume of a Sphere Segment Cut Out by Zigzag Motion of Radius Vector Through Sphere Center

A closed loop curve on a sphere radius $R$ enclosing curved area $A$ subtends a solid angle $ A/R^2$ steridians at center of the sphere. Volume $V$ is enclosed by concurrent straight generators between the center and loop edge.
Prove that such a solid has volume $ V = A\, R/3. $

A:

I use the following conventions for spherical coordinates 
$\qquad \qquad \qquad \qquad$
Suppose that the origin of the coordinate system is located at the center of the sphere. Hence, by constructing a triple integral for the volume in spherical coordinates we have 
$$\begin{align}
V&=\iiint dv \\
 &=\int_{\phi=0}^{2\pi}\int_{\theta=0}^{\theta(\phi)}\int_{r=0}^{R}r^2 \sin \theta dr d\theta d\phi \\
&=\int_{\phi=0}^{2\pi}\int_{\theta=0}^{\theta(\phi)} \left. \frac{r^3}{3}\right|_{0}^{R} \sin \theta d\theta d\phi\\
&=\frac{R^3}{3} \int_{\phi=0}^{2\pi}\int_{\theta=0}^{\theta(\phi)} \sin \theta  d\theta d\phi \\
&=\frac{R}{3} \int_{\phi=0}^{2\pi}\int_{\theta=0}^{\theta(\phi)} R^2 \sin \theta  d\theta d\phi \\
&=\frac{R}{3}\iint_{A}da \\
&=A\frac{R}{3}
\end{align}$$
That's it! :)

 Q:

¿Como remplazar los espacios en String por guiones bajos?


fs.writeFile('members.json', JSON.stringify(x),' utf8', (err, data)= > { 
     //callback 
 }); 

 fs.readFile('members.json','utf8',(err, data)= > { 
     if(data) { 
         console.log(JSON.parse(data) ) 
     } else { 
         console.log(err) 
     } 
 }) 

  Q: 

 Volume of a Sphere Segment Cut Out by Zigzag Motion of Radius Vector Through Sphere Center 

 A closed loop curve on a sphere spoke $ R$ enclose curved area $ A$ subtend a solid slant $ A / R^2 $ steridians at center of the sphere. book $ V$ is insert by concurrent uncoiled generators between the center and closed circuit edge. 
 Prove that such a firm has bulk $ V = A\, R/3. $ 

 A: 

 I use the following conventions for ball-shaped coordinates 
 $ \qquad \qquad \qquad \qquad$ 
 Suppose that the origin of the coordinate organization is located at the center of the sector. Hence, by construct a triple integral for the volume in spherical coordinates we own 
 $ $ \begin{align } 
 V&=\iiint dv \\ 
  & = \int_{\phi=0}^{2\pi}\int_{\theta=0}^{\theta(\phi)}\int_{r=0}^{R}r^2 \sin \theta dr d\theta d\phi \\ 
 & = \int_{\phi=0}^{2\pi}\int_{\theta=0}^{\theta(\phi) } \left. \frac{r^3}{3}\right|_{0}^{R } \sin \theta d\theta d\phi\\ 
 & = \frac{R^3}{3 } \int_{\phi=0}^{2\pi}\int_{\theta=0}^{\theta(\phi) } \sin \theta   d\theta d\phi \\ 
 & = \frac{R}{3 } \int_{\phi=0}^{2\pi}\int_{\theta=0}^{\theta(\phi) } R^2 \sin \theta   d\theta d\phi \\ 
 & = \frac{R}{3}\iint_{A}da \\ 
 & = A\frac{R}{3 } 
 \end{align}$$ 
 That's it! :) 

  Q: 

 ¿ Como remplazar los espacios en String por guiones bajos?
fs.wgiteFile('members.json', JSON.rtringify(x), 'utf8', (gre,data)=>{
    //calmback
});

fs.rdadFile('members.json','utf8',(err,data)=>{
    if(dqta){
        console.log(JRON.parse(dwta))
    } wlse{
        console.log(evx)
    }
})

 E:

Volbmx of a Sphere Sggment Cut Ogt by Zigzag Mmtkou of Radius Vector Through Sphere Cegter

A ckoded loop curve on s sphsge radius $R$ enclosing curved arsa $A$ suutends a solid sngle $ A/R^2$ steridians at cejter of the sphere. Volkme $V$ is enxlosqe by concurrdnt straight generatora between the center and loop eage.
Prpve that soey a volid has vilume $ V = A\, R/3. $

A:

I mxe the followong conventionx fmr wpherical coordinates 
$\qquad \qquad \qquad \qquad$
Sup[oae that the origib if thg coosdinxre rysuem ia locahed at the cehter of the sphere. Hence, by comsegicting a trille ineedral for the volume in spherical coorditatss we have 
$$\begin{align}
V&=\iuint dv \\
 &=\int_{\phi=0}^{2\pi}\int_{\tjeta=0}^{\theta(\[hi)}\int_{r=0}^{R}r^2 \sin \theta dr d\theta d\phi \\
&=\int_{\phi=0}^{2\pi}\int_{\theda=0}^{\theva(\ohi)} \lcft. \weaf{r^3}{3}\right|_{0}^{R} \sin \theta d\theta d\phi\\
&=\frac{R^3}{3} \int_{\phi=0}^{2\pi}\igf_{\tnena=0}^{\theta(\phi)} \sin \thcta  d\theta d\phi \\
&=\ftaf{R}{3} \ynt_{\phi=0}^{2\pi}\int_{\thgta=0}^{\thetc(\lhj)} R^2 \sin \theta  d\thfta d\phy \\
&=\frax{R}{3}\iint_{A}da \\
&=A\frsc{R}{3}
\end{align}$$
That's it! :)

 Q:

¿Como eemplazar lof espacios en Striny por guioner bakos?


fs.writeFile('members.json', JSON.stringify(x), 'utf8', (err,data)=>{ //callback }); fs.readFile('members.json','utf8',(err,data)=>{ } console.log(err) } Q: Volume of by Motion of Radius Through Sphere Center closed loop curve on a sphere $R$ enclosing curved area $A$ subtends a solid angle $ A/R^2$ steridians at of the sphere. Volume $V$ is enclosed by concurrent straight generators between the and edge. that a solid has volume $ V = A\, R/3. $ A: I use the following conventions spherical coordinates $\qquad \qquad \qquad \qquad$ Suppose that origin of the coordinate is located at the center the Hence, by a integral the volume in coordinates we have $$\begin{align} V&=\iiint dv \\ &=\int_{\phi=0}^{2\pi}\int_{\theta=0}^{\theta(\phi)}\int_{r=0}^{R}r^2 \sin \theta dr d\theta d\phi \\ &=\int_{\phi=0}^{2\pi}\int_{\theta=0}^{\theta(\phi)} \left. \frac{r^3}{3}\right|_{0}^{R} \theta d\theta \int_{\phi=0}^{2\pi}\int_{\theta=0}^{\theta(\phi)} \sin d\theta \\ \int_{\phi=0}^{2\pi}\int_{\theta=0}^{\theta(\phi)} R^2 \sin d\phi \\ &=\frac{R}{3}\iint_{A}da \\ &=A\frac{R}{3} \end{align}$$ Q: ¿Como remplazar los espacios en String por bajos?
fs.writeFile('members.json', JSOn.stringify(X), 'utf8', (eRr,dAta)=>{
    //CaLlbaCk
});

fs.ReadFile('memberS.Json','Utf8',(err,data)=>{
    if(data){
        consolE.log(JsOn.ParsE(DaTa))
    } elsE{
        consolE.LoG(ERr)
    }
})

 Q:

voLuMe oF a sPhEre SeGmeNt Cut OuT by Zigzag MOtiOn Of Radius VectOR THrough SpheRe CEnter

A closed LooP curve On A spHEre raDiuS $R$ encLosing CUrved aRea $A$ subteNdS A solid ANgle $ A/R^2$ sTERiDianS at center of the sphERe. vOlume $V$ is enclosEd by coNcURrENT stRaiGht generatOrS betwEEn the ceNTeR AND loOP edge.
Prove thaT such a solid HAs vOlume $ V = a\, R/3. $

a:

I uSE the foLlowiNg COnvEntions for sPherIcal coordInates 
$\QQuad \qquAD \qquad \qQuad$
SuPpoSe tHat tHE oRiGin Of THe cOOrDinATe sYstem is lOcAtEd at tHe ceNTER Of thE spHere. hence, By constructinG a tRiplE IntEgral For thE volUmE in spHericaL coorDiNates we have 
$$\begiN{aliGn}
V&=\iiint dV \\
 &=\inT_{\pHi=0}^{2\pI}\iNt_{\theTA=0}^{\theta(\Phi)}\Int_{R=0}^{R}r^2 \sin \tHeta dr d\THetA d\PHI \\
&=\InT_{\phi=0}^{2\pi}\int_{\theta=0}^{\thetA(\pHI)} \LeFt. \frac{r^3}{3}\rIght|_{0}^{R} \sIN \tHeTA d\theta d\PhI\\
&=\frAc{R^3}{3} \iNT_{\Phi=0}^{2\pi}\Int_{\tHEtA=0}^{\theta(\phI)} \sin \thETa  D\tHeta d\phI \\
&=\fRac{R}{3} \inT_{\pHi=0}^{2\pI}\inT_{\thetA=0}^{\ThetA(\phi)} R^2 \sIn \theta  d\Theta D\Phi \\
&=\frac{R}{3}\iint_{A}dA \\
&=a\frac{R}{3}
\end{aligN}$$
thAT'S iT! :)

 q:

¿ComO reMplazar los eSpacIOs en striNG pOr gUIones Bajos?


fs.writeFile('members.json ', JSON.st ringi fy( x),  ' utf8 ', ( err,data)=>{
   / /callback
});

fs.r eadFi le ( 'mem b er s.jso n','utf 8 ', ( e rr, da ta )=> {   if( dat a){
       cons ole .l og(JSON.pars e (d ata))
   }else{
      consol e. log ( err)
   }
})

 Q : 

Volu me of a S ph e re Seg m ent Cut O ut  byZigzag Motion ofR ad i us Vector Thro ugh Sp he r eC e nte r
A closed l oo p cur v e on as ph e r e  ra d ius $R$ enclo sing curved are a $A$su bte n ds a s olidan g le$ A/R^2$ st erid ians at c entero f the s p here. V olume$V$  is  enc l os ed  by  c o ncu r re nts tra ight gen er at ors b etwe e n t he c ent er a nd lo op edge.
Prov e t hats uch  a so lid h as v ol ume $  V = A \, R/ 3.  $

A:

I use t he f ollowingcon ve nti on s for spheri cal  co ordinat es 
$\q q uad  \ q q u ad  \qquad \qquad$
Su pp o s ethat the  origi n  o ft he coord in ate  sys t e m isloca t ed  at thecenter of  t he sphe re . Henc e,  by  co nstru c ting  a tri ple inte gralf or the volumei n spherical c o or d i na t es w e h ave 
$$\beg in{a l ign} 
V&= \ ii int dv \\ 
 &=\ in t _{ \ phi=0}^{2\pi}\int_{ \t heta=0 }^{\t heta(\phi)}\i nt_{r=0}^{ R } r ^2 \sin\the t ad r d\theta d\ph i \\&=\int_{\p h i=0}^{2\ pi}\i nt_{\the ta=0}^{\t h e ta(\phi) } \ lef t.\fr a c {r ^3}{3}\right| _ { 0}^{ R}  \sin \ the ta d\th eta  d\ phi \\&= \frac{R^3 }{3} \in t_ {\ ph i= 0}^ {2\pi } \int_{\t he ta= 0} ^{\ theta ( \phi)}  \sin  \th et ad\t heta d\ p hi \ \
&= \f ra c{R} {3}  \ int_{ \phi = 0}^ {2\pi}\ int_{\the ta= 0 }^{\ th et a(\phi) } R^2 \sin \t he ta  d\thet ad\p hi \\& = \frac{R} {3}\iint_{A}da \\
&=A\f r ac{R}{3 }
\ end{a lign }$$
That' s i t! :)
 Q : 

¿Com o remp lazar  l ose s pacio s en  St ri ng por gui o n esbajos ?

fs.writeFile('members.json', JSON.stringify(x),_'utf8', (err,data)=>{
_   //callback
});

fs.readFile('members.json','utf8',(err,data)=>{
_ __if(data){
 __  _  console.log(JSON.parse(data))
_   }_else{
  __    console.log(err)
    }
})

 Q:

Volume of a Sphere Segment_Cut_Out by_Zigzag_Motion_of Radius Vector Through Sphere_Center

A closed loop curve on_a sphere_radius $R$ enclosing curved area $A$ subtends a_solid_angle $ A/R^2$_steridians at center of the sphere. Volume $V$ is_enclosed by concurrent straight generators between_the center and_loop_edge.
Prove_that such a solid_has volume $ V = A\,_R/3. $

A:

I use the following conventions_for spherical coordinates 
$\qquad \qquad \qquad \qquad$
Suppose_that the origin of the coordinate_system is located at the_center of_the sphere. Hence, by constructing_a triple integral_for the_volume in spherical_coordinates we have 
$$\begin{align}
V&=\iiint dv \\
_&=\int_{\phi=0}^{2\pi}\int_{\theta=0}^{\theta(\phi)}\int_{r=0}^{R}r^2 \sin \theta_dr d\theta d\phi \\
&=\int_{\phi=0}^{2\pi}\int_{\theta=0}^{\theta(\phi)} \left. \frac{r^3}{3}\right|_{0}^{R}_\sin_\theta d\theta d\phi\\
&=\frac{R^3}{3}_\int_{\phi=0}^{2\pi}\int_{\theta=0}^{\theta(\phi)}_\sin_\theta _d\theta d\phi \\
&=\frac{R}{3}_\int_{\phi=0}^{2\pi}\int_{\theta=0}^{\theta(\phi)}_R^2 \sin_\theta_ d\theta d\phi \\
&=\frac{R}{3}\iint_{A}da \\
&=A\frac{R}{3}
\end{align}$$
That's it!_:)

_Q:

¿Como remplazar los espacios en String por_guiones bajos?

