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Question Number 5291 by Rasheed Soomro last updated on 05/May/16

$$\mathrm{A}\mathrm{circle}\mathrm{has}\mathrm{been}\mathrm{drawn}\mathrm{with}\mathrm{one}$$$$\mathrm{unit}\mathrm{opened}\mathrm{compass}\mathrm{on}\mathrm{surface}$$$$\mathrm{of}\mathrm{a}\mathrm{sphere}\mathrm{of}\mathrm{one}\mathrm{unit}\mathrm{radius}.$$$$\mathrm{What}\mathrm{will}\mathrm{be}\mathrm{the}\mathrm{area}\mathrm{of}\mathrm{the}\mathrm{circle}?$$

Commented by Yozzii last updated on 05/May/16

$$A=0.75\pi$$

Commented by Yozzii last updated on 05/May/16

$$Placethecompassonthespheresuch$$$$thatitsamplitudeof1unitisunaltered.$$$$Callthelinejoiningthetwopoints$$$$wherethecompassmeetsthesphere$$$$asAB.So\mid AB\mid =1.LetObethecentre$$$$ofthesphere.Itfollowsthat\mid OA\mid =\mid OB\mid =1$$$$sincepointsAandBlieonthesphere.$$$$But,\mid AB\mid =1also.Hence,\mid AB\mid =\mid OA\mid =\mid OB\mid$$$$for△ABO.Andso,△ABOisequilateral.$$$$\Rightarrow \mathrm{\angle }BOA=\mathrm{\angle }OAB=\mathrm{\angle }ABO=\frac{\pi }{3}.$$$$LetJbeapointonthelineBOsuch$$$$thatAJ\mathrm{\perp }BO.Since△ABOisequilateral,$$$$AJisaperpendicularbisectorofBO.$$$$So,wehave\mathrm{\angle }BAJ=\frac{1}{2}\mathrm{\angle }BA0=\frac{\pi }{6}$$$$and\mid AB\mid =1.\Rightarrow \mid BJ\mid =\mid AB\mid cos\mathrm{\angle }BAJ$$$$\mid BJ\mid =1\times cos\frac{\pi }{6}=\frac{\sqrt{3}}{2}.Hence,theradius$$$$ofthecircleinquestionisr=\frac{\sqrt{3}}{2}.$$$$Usingarea,A=\pi r^2\Rightarrow A=\frac{3\pi }{4}squareunits.$$

Commented by Rasheed Soomro last updated on 05/May/16

$$\mathbb{T}{\mathbf{h}}^{\alpha }\mathbf{n}\mathbb{k}\mathbf{S}!\mathrm{I}\mathrm{am}\mathrm{trying}\mathrm{to}\mathrm{understand}!$$$$\mathrm{I}\mathrm{think}\mathrm{that}\mathrm{circle}\mathrm{on}\mathrm{a}\mathrm{sphere}\mathrm{is}$$$$\mathrm{concave}\mathrm{surface}....\mathrm{Anyway}\mathrm{I}\mathrm{like}\mathrm{your}$$$$\mathrm{approach}\mathrm{very}\mathrm{much}.\mathrm{Not}\mathrm{only}\mathrm{in}\mathrm{this}$$$$\mathrm{answer}\mathrm{but}\mathrm{I}\mathbf{always}\mathrm{like}\mathrm{your}\mathrm{approach}!$$

Commented by Yozzii last updated on 05/May/16

$$Ahyes.Thatsasectionofthesurface$$$$ofthesphere.IntegralCalculuscaneasily$$$$derivethesurfaceareagiventhe$$$$centralangle.{You}^{\prime }dthushaveto$$$$inregratethefunctiony=\sqrt{1-x^2}$$$$between0.5⩽x⩽1using$$$$S={\int }_{0.5}^{1}2\pi y\sqrt{1+{\left(\frac{dy}{dx}\right)}^2}dx.$$$$Thatsection,thecross-section$$$$ofthesphere,definedby{x}^2+y^2=1$$$$and0.5⩽x⩽1isrotated2\pi radians$$$$aboutthex-axis.ThegeometrythatIhave$$$$outlinediswhathelpsusfindthe$$$$lowerlimitofx=0.5.$$

Commented by Yozzii last updated on 05/May/16

$$Whatmyinitialanswergavewas$$$$theareaoftheregionofaplanar$$$$cutofthesphereinthecirclegenerated$$$$bythecompass.$$

Commented by Rasheed Soomro last updated on 05/May/16

$$\mathbf{ThankS}\mathbf{again}!$$

Commented by Yozzii last updated on 05/May/16

$$y^2+x^2=1\Rightarrow 2{yy}^{{}^{\prime }}+2x=0\left(implicitdifferentiation\right)$$$$\Rightarrow y^{{}^{\prime }}=-\frac{x}{y}\Rightarrow {\left(y^{{}^{\prime }}\right)}^2+1=\frac{x^2+y^2}{y^2}=\frac{1}{y^2}(y>0for0.5⩽x⩽1)$$$$\therefore \sqrt{1+{\left(y^{\prime }\right)}^2}=\frac{1}{\mid y\mid }=\frac{1}{y}\Rightarrow y\sqrt{1+{\left(y^{{}^{\prime }}\right)}^2}=1$$$$\therefore S=2\pi {\int }_{0.5}^{1}dx=\pi squareunits.$$

Commented by Yozzii last updated on 05/May/16

$${Archimedes}^{\prime }tombstonetheoremis$$$$applicablehere.Sotherequiredsurface$$$$area,afterfindingthat\mid JB\mid =0.5,$$$$AreaofsurfaceofspherebetweenJandB$$$$equalsthesurfaceareaofthesurface,between$$$$twoparallelplanescontainingtheportionofsphere,of$$$$acylindercircumscribingtheentire$$$$sphereinquestion$$$$\therefore S=2\pi rh$$$$wherer=radiusofentiresphere$$$$h=\mid BJ\mid =0.5$$$$\Rightarrow S=2\pi \times 1\times 0.5=\pi squareunitsaswasfound$$$$byintegralcalculus.$$$$ThePappus-Guldintheoremgives$$$$analternativebutperhapslonger$$$$methodsincetheycoordinateofthecetroidoftheregion$$$$boundedbyx=0.5,x=1,y=0and$$$$x^2+y^2=1needstobefound.$$$$\Rightarrow S=2\pi {\overline{ys}}_0$$$$where{s}_0=arclengthofy=r\theta =1\times \frac{\pi }{3}=\frac{\pi }{3}units,$$$$for0.5⩽x⩽1.$$$$$$

Commented by Rasheed Soomro last updated on 06/May/16

$$\mathbb{LOT}\mathcal{OF}\mathrm{Knowledge}!$$$$\mathbb{LOT}\mathcal{OF}\mathbb{T}\mathrm{han}\mathbb{k}\mathcal{S}!$$$$$$

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